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Crack the bank account's password!

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Crack the bank account's password!

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Introduction

In order to prevent keyloggers from stealing a user's password, a certain bank account system has implemented the following security measure: only certain digits are prompted to be entered each time.

For example, say your target's password is 89097, the system may prompt them to enter the 2nd, 4th and 5th digit:

997

Or it might prompt them to enter the 1st, 3rd and 5th digit:

807

All you know is that your target entered the digits in order, but you don't know which position they belong to in the actual password. All you know is there are two 9s, which must come before 7; and that 8 comes before 0, and 0 before 7. Therefore, there are six possible passwords:

The keylogger in your target's computer has been collecting password inputs for months now, so let's hack in!

Challenge

Given a list of three-digit inputs, output all the possible passwords that are valid for all inputs. In order to reduce computational complexity and to keep the amount of possible results low, the password is guaranteed to be numerical and have a fixed size of 5. The digits in every input are in order: if it's 123, the target typed 1 first, then 2, then 3.

Input/Output examples

|----------------------|--------------------------------------------|

|         Input        |                   Output                   |

|----------------------|--------------------------------------------|

| [320, 723, 730]      | [37230, 72320, 73203, 73230]               |

| [374, 842]           | [37842, 38742, 83742]                      |

| [010, 103, 301]      | [30103]                                    |

| [123, 124, 125, 235] | [12345, 12354, 12435]                      |

| [239, 944]           | [23944]                                    |

| [111, 120]           | [11201, 11120, 11210, 12011, 12110, 12101] |

| [456, 789]           | []                                         |

| [756, 586]           | [07586, 17586, 27586, 37586, 47586, 57586, 57856, 58756, 67586, 70586, 71586, 72586, 73586, 74586, 75086, 75186, 75286, 75386, 75486, 75586, 75686, 75786, 75806, 75816, 75826, 75836, 75846, 75856, 75860, 75861, 75862, 75863, 75864, 75865, 75866, 75867, 75868, 75869, 75876, 75886, 75896, 75986, 76586, 77586, 78586, 79586, 87586, 97586] |

| [123]                | [00123, 01023, 01123, 01203, 01213, 01223, 01230, 01231, 01232, 01233, 01234, 01235, 01236, 01237, 01238, 01239, 01243, 01253, 01263, 01273, 01283, 01293, 01323, 01423, 01523, 01623, 01723, 01823, 01923, 02123, 03123, 04123, 05123, 06123, 07123, 08123, 09123, 10023, 10123, 10203, 10213, 10223, 10230, 10231, 10232, 10233, 10234, 10235, 10236, 10237, 10238, 10239, 10243, 10253, 10263, 10273, 10283, 10293, 10323, 10423, 10523, 10623, 10723, 10823, 10923, 11023, 11123, 11203, 11213, 11223, 11230, 11231, 11232, 11233, 11234, 11235, 11236, 11237, 11238, 11239, 11243, 11253, 11263, 11273, 11283, 11293, 11323, 11423, 11523, 11623, 11723, 11823, 11923, 12003, 12013, 12023, 12030, 12031, 12032, 12033, 12034, 12035, 12036, 12037, 12038, 12039, 12043, 12053, 12063, 12073, 12083, 12093, 12103, 12113, 12123, 12130, 12131, 12132, 12133, 12134, 12135, 12136, 12137, 12138, 12139, 12143, 12153, 12163, 12173, 12183, 12193, 12203, 12213, 12223, 12230, 12231, 12232, 12233, 12234, 12235, 12236, 12237, 12238, 12239, 12243, 12253, 12263, 12273, 12283, 12293, 12300, 12301, 12302, 12303, 12304, 12305, 12306, 12307, 12308, 12309, 12310, 12311, 12312, 12313, 12314, 12315, 12316, 12317, 12318, 12319, 12320, 12321, 12322, 12323, 12324, 12325, 12326, 12327, 12328, 12329, 12330, 12331, 12332, 12333, 12334, 12335, 12336, 12337, 12338, 12339, 12340, 12341, 12342, 12343, 12344, 12345, 12346, 12347, 12348, 12349, 12350, 12351, 12352, 12353, 12354, 12355, 12356, 12357, 12358, 12359, 12360, 12361, 12362, 12363, 12364, 12365, 12366, 12367, 12368, 12369, 12370, 12371, 12372, 12373, 12374, 12375, 12376, 12377, 12378, 12379, 12380, 12381, 12382, 12383, 12384, 12385, 12386, 12387, 12388, 12389, 12390, 12391, 12392, 12393, 12394, 12395, 12396, 12397, 12398, 12399, 12403, 12413, 12423, 12430, 12431, 12432, 12433, 12434, 12435, 12436, 12437, 12438, 12439, 12443, 12453, 12463, 12473, 12483, 12493, 12503, 12513, 12523, 12530, 12531, 12532, 12533, 12534, 12535, 12536, 12537, 12538, 12539, 12543, 12553, 12563, 12573, 12583, 12593, 12603, 12613, 12623, 12630, 12631, 12632, 12633, 12634, 12635, 12636, 12637, 12638, 12639, 12643, 12653, 12663, 12673, 12683, 12693, 12703, 12713, 12723, 12730, 12731, 12732, 12733, 12734, 12735, 12736, 12737, 12738, 12739, 12743, 12753, 12763, 12773, 12783, 12793, 12803, 12813, 12823, 12830, 12831, 12832, 12833, 12834, 12835, 12836, 12837, 12838, 12839, 12843, 12853, 12863, 12873, 12883, 12893, 12903, 12913, 12923, 12930, 12931, 12932, 12933, 12934, 12935, 12936, 12937, 12938, 12939, 12943, 12953, 12963, 12973, 12983, 12993, 13023, 13123, 13203, 13213, 13223, 13230, 13231, 13232, 13233, 13234, 13235, 13236, 13237, 13238, 13239, 13243, 13253, 13263, 13273, 13283, 13293, 13323, 13423, 13523, 13623, 13723, 13823, 13923, 14023, 14123, 14203, 14213, 14223, 14230, 14231, 14232, 14233, 14234, 14235, 14236, 14237, 14238, 14239, 14243, 14253, 14263, 14273, 14283, 14293, 14323, 14423, 14523, 14623, 14723, 14823, 14923, 15023, 15123, 15203, 15213, 15223, 15230, 15231, 15232, 15233, 15234, 15235, 15236, 15237, 15238, 15239, 15243, 15253, 15263, 15273, 15283, 15293, 15323, 15423, 15523, 15623, 15723, 15823, 15923, 16023, 16123, 16203, 16213, 16223, 16230, 16231, 16232, 16233, 16234, 16235, 16236, 16237, 16238, 16239, 16243, 16253, 16263, 16273, 16283, 16293, 16323, 16423, 16523, 16623, 16723, 16823, 16923, 17023, 17123, 17203, 17213, 17223, 17230, 17231, 17232, 17233, 17234, 17235, 17236, 17237, 17238, 17239, 17243, 17253, 17263, 17273, 17283, 17293, 17323, 17423, 17523, 17623, 17723, 17823, 17923, 18023, 18123, 18203, 18213, 18223, 18230, 18231, 18232, 18233, 18234, 18235, 18236, 18237, 18238, 18239, 18243, 18253, 18263, 18273, 18283, 18293, 18323, 18423, 18523, 18623, 18723, 18823, 18923, 19023, 19123, 19203, 19213, 19223, 19230, 19231, 19232, 19233, 19234, 19235, 19236, 19237, 19238, 19239, 19243, 19253, 19263, 19273, 19283, 19293, 19323, 19423, 19523, 19623, 19723, 19823, 19923, 20123, 21023, 21123, 21203, 21213, 21223, 21230, 21231, 21232, 21233, 21234, 21235, 21236, 21237, 21238, 21239, 21243, 21253, 21263, 21273, 21283, 21293, 21323, 21423, 21523, 21623, 21723, 21823, 21923, 22123, 23123, 24123, 25123, 26123, 27123, 28123, 29123, 30123, 31023, 31123, 31203, 31213, 31223, 31230, 31231, 31232, 31233, 31234, 31235, 31236, 31237, 31238, 31239, 31243, 31253, 31263, 31273, 31283, 31293, 31323, 31423, 31523, 31623, 31723, 31823, 31923, 32123, 33123, 34123, 35123, 36123, 37123, 38123, 39123, 40123, 41023, 41123, 41203, 41213, 41223, 41230, 41231, 41232, 41233, 41234, 41235, 41236, 41237, 41238, 41239, 41243, 41253, 41263, 41273, 41283, 41293, 41323, 41423, 41523, 41623, 41723, 41823, 41923, 42123, 43123, 44123, 45123, 46123, 47123, 48123, 49123, 50123, 51023, 51123, 51203, 51213, 51223, 51230, 51231, 51232, 51233, 51234, 51235, 51236, 51237, 51238, 51239, 51243, 51253, 51263, 51273, 51283, 51293, 51323, 51423, 51523, 51623, 51723, 51823, 51923, 52123, 53123, 54123, 55123, 56123, 57123, 58123, 59123, 60123, 61023, 61123, 61203, 61213, 61223, 61230, 61231, 61232, 61233, 61234, 61235, 61236, 61237, 61238, 61239, 61243, 61253, 61263, 61273, 61283, 61293, 61323, 61423, 61523, 61623, 61723, 61823, 61923, 62123, 63123, 64123, 65123, 66123, 67123, 68123, 69123, 70123, 71023, 71123, 71203, 71213, 71223, 71230, 71231, 71232, 71233, 71234, 71235, 71236, 71237, 71238, 71239, 71243, 71253, 71263, 71273, 71283, 71293, 71323, 71423, 71523, 71623, 71723, 71823, 71923, 72123, 73123, 74123, 75123, 76123, 77123, 78123, 79123, 80123, 81023, 81123, 81203, 81213, 81223, 81230, 81231, 81232, 81233, 81234, 81235, 81236, 81237, 81238, 81239, 81243, 81253, 81263, 81273, 81283, 81293, 81323, 81423, 81523, 81623, 81723, 81823, 81923, 82123, 83123, 84123, 85123, 86123, 87123, 88123, 89123, 90123, 91023, 91123, 91203, 91213, 91223, 91230, 91231, 91232, 91233, 91234, 91235, 91236, 91237, 91238, 91239, 91243, 91253, 91263, 91273, 91283, 91293, 91323, 91423, 91523, 91623, 91723, 91823, 91923, 92123, 93123, 94123, 95123, 96123, 97123, 98123, 99123] |

|----------------------|--------------------------------------------|

Rules

Input is guaranteed non-empty.

Leading and trailing zeros matter: 01234 is different from 12340, and 1234 doesn't crack either password. Think of how real passwords work!

Standard I/O rules apply.

No standard loopholes.

This is code-golf, so the shortest answer in bytes wins. Non-codegolfing languages are welcome!

edited 2 days ago

Hat

355

asked Feb 22 at 14:40

cefel

37628

New contributor

5

$begingroup$
Are the digits always in order? Based on the test cases I assume they are, but I couldn't see it mentioned in the rules unless I read past it.
$endgroup$
– Kevin Cruijssen
Feb 22 at 14:44

13

$begingroup$
Welcome to PPCG! This is a nice, well-structured, and neatly formatted first challenge. You've clearly done your homework as far as getting that all down. I'm looking forward to answering it (if someone doesn't answer it in R first!). In the future, we suggest using the sandbox to get feedback before posting to main. Hope you enjoy your time on PPCG!
$endgroup$
– Giuseppe
Feb 22 at 15:04

1

$begingroup$
@Giuseppe thanks! I've been anonymously reading the questions on this site for years, and I've been writing and tweaking and actually solving this specific problem for a couple months: I liked it enough to skip the sandbox. I'll post there first next time!
$endgroup$
– cefel
Feb 22 at 15:07

2

$begingroup$
@Arnauld Well, if your password is 01234 or 12340 you shouldn't be able to log in by typing 1234. Passwords are more a string than a number even if composed by numbers, at least in that sense. So yes, leading and trailing zeros are mandatory.
$endgroup$
– cefel
Feb 22 at 15:15

2

$begingroup$
The final test case appears to be missing 22123... unless I'm misunderstanding something?
$endgroup$
– Jonah
2 days ago

|
show 12 more comments

Introduction

In order to prevent keyloggers from stealing a user's password, a certain bank account system has implemented the following security measure: only certain digits are prompted to be entered each time.

For example, say your target's password is 89097, the system may prompt them to enter the 2nd, 4th and 5th digit:

997

Or it might prompt them to enter the 1st, 3rd and 5th digit:

807

The keylogger in your target's computer has been collecting password inputs for months now, so let's hack in!

Challenge

Input/Output examples

|----------------------|--------------------------------------------|

|         Input        |                   Output                   |

|----------------------|--------------------------------------------|

| [320, 723, 730]      | [37230, 72320, 73203, 73230]               |

| [374, 842]           | [37842, 38742, 83742]                      |

| [010, 103, 301]      | [30103]                                    |

| [123, 124, 125, 235] | [12345, 12354, 12435]                      |

| [239, 944]           | [23944]                                    |

| [111, 120]           | [11201, 11120, 11210, 12011, 12110, 12101] |

| [456, 789]           | []                                         |

| [756, 586]           | [07586, 17586, 27586, 37586, 47586, 57586, 57856, 58756, 67586, 70586, 71586, 72586, 73586, 74586, 75086, 75186, 75286, 75386, 75486, 75586, 75686, 75786, 75806, 75816, 75826, 75836, 75846, 75856, 75860, 75861, 75862, 75863, 75864, 75865, 75866, 75867, 75868, 75869, 75876, 75886, 75896, 75986, 76586, 77586, 78586, 79586, 87586, 97586] |

| [123]                | [00123, 01023, 01123, 01203, 01213, 01223, 01230, 01231, 01232, 01233, 01234, 01235, 01236, 01237, 01238, 01239, 01243, 01253, 01263, 01273, 01283, 01293, 01323, 01423, 01523, 01623, 01723, 01823, 01923, 02123, 03123, 04123, 05123, 06123, 07123, 08123, 09123, 10023, 10123, 10203, 10213, 10223, 10230, 10231, 10232, 10233, 10234, 10235, 10236, 10237, 10238, 10239, 10243, 10253, 10263, 10273, 10283, 10293, 10323, 10423, 10523, 10623, 10723, 10823, 10923, 11023, 11123, 11203, 11213, 11223, 11230, 11231, 11232, 11233, 11234, 11235, 11236, 11237, 11238, 11239, 11243, 11253, 11263, 11273, 11283, 11293, 11323, 11423, 11523, 11623, 11723, 11823, 11923, 12003, 12013, 12023, 12030, 12031, 12032, 12033, 12034, 12035, 12036, 12037, 12038, 12039, 12043, 12053, 12063, 12073, 12083, 12093, 12103, 12113, 12123, 12130, 12131, 12132, 12133, 12134, 12135, 12136, 12137, 12138, 12139, 12143, 12153, 12163, 12173, 12183, 12193, 12203, 12213, 12223, 12230, 12231, 12232, 12233, 12234, 12235, 12236, 12237, 12238, 12239, 12243, 12253, 12263, 12273, 12283, 12293, 12300, 12301, 12302, 12303, 12304, 12305, 12306, 12307, 12308, 12309, 12310, 12311, 12312, 12313, 12314, 12315, 12316, 12317, 12318, 12319, 12320, 12321, 12322, 12323, 12324, 12325, 12326, 12327, 12328, 12329, 12330, 12331, 12332, 12333, 12334, 12335, 12336, 12337, 12338, 12339, 12340, 12341, 12342, 12343, 12344, 12345, 12346, 12347, 12348, 12349, 12350, 12351, 12352, 12353, 12354, 12355, 12356, 12357, 12358, 12359, 12360, 12361, 12362, 12363, 12364, 12365, 12366, 12367, 12368, 12369, 12370, 12371, 12372, 12373, 12374, 12375, 12376, 12377, 12378, 12379, 12380, 12381, 12382, 12383, 12384, 12385, 12386, 12387, 12388, 12389, 12390, 12391, 12392, 12393, 12394, 12395, 12396, 12397, 12398, 12399, 12403, 12413, 12423, 12430, 12431, 12432, 12433, 12434, 12435, 12436, 12437, 12438, 12439, 12443, 12453, 12463, 12473, 12483, 12493, 12503, 12513, 12523, 12530, 12531, 12532, 12533, 12534, 12535, 12536, 12537, 12538, 12539, 12543, 12553, 12563, 12573, 12583, 12593, 12603, 12613, 12623, 12630, 12631, 12632, 12633, 12634, 12635, 12636, 12637, 12638, 12639, 12643, 12653, 12663, 12673, 12683, 12693, 12703, 12713, 12723, 12730, 12731, 12732, 12733, 12734, 12735, 12736, 12737, 12738, 12739, 12743, 12753, 12763, 12773, 12783, 12793, 12803, 12813, 12823, 12830, 12831, 12832, 12833, 12834, 12835, 12836, 12837, 12838, 12839, 12843, 12853, 12863, 12873, 12883, 12893, 12903, 12913, 12923, 12930, 12931, 12932, 12933, 12934, 12935, 12936, 12937, 12938, 12939, 12943, 12953, 12963, 12973, 12983, 12993, 13023, 13123, 13203, 13213, 13223, 13230, 13231, 13232, 13233, 13234, 13235, 13236, 13237, 13238, 13239, 13243, 13253, 13263, 13273, 13283, 13293, 13323, 13423, 13523, 13623, 13723, 13823, 13923, 14023, 14123, 14203, 14213, 14223, 14230, 14231, 14232, 14233, 14234, 14235, 14236, 14237, 14238, 14239, 14243, 14253, 14263, 14273, 14283, 14293, 14323, 14423, 14523, 14623, 14723, 14823, 14923, 15023, 15123, 15203, 15213, 15223, 15230, 15231, 15232, 15233, 15234, 15235, 15236, 15237, 15238, 15239, 15243, 15253, 15263, 15273, 15283, 15293, 15323, 15423, 15523, 15623, 15723, 15823, 15923, 16023, 16123, 16203, 16213, 16223, 16230, 16231, 16232, 16233, 16234, 16235, 16236, 16237, 16238, 16239, 16243, 16253, 16263, 16273, 16283, 16293, 16323, 16423, 16523, 16623, 16723, 16823, 16923, 17023, 17123, 17203, 17213, 17223, 17230, 17231, 17232, 17233, 17234, 17235, 17236, 17237, 17238, 17239, 17243, 17253, 17263, 17273, 17283, 17293, 17323, 17423, 17523, 17623, 17723, 17823, 17923, 18023, 18123, 18203, 18213, 18223, 18230, 18231, 18232, 18233, 18234, 18235, 18236, 18237, 18238, 18239, 18243, 18253, 18263, 18273, 18283, 18293, 18323, 18423, 18523, 18623, 18723, 18823, 18923, 19023, 19123, 19203, 19213, 19223, 19230, 19231, 19232, 19233, 19234, 19235, 19236, 19237, 19238, 19239, 19243, 19253, 19263, 19273, 19283, 19293, 19323, 19423, 19523, 19623, 19723, 19823, 19923, 20123, 21023, 21123, 21203, 21213, 21223, 21230, 21231, 21232, 21233, 21234, 21235, 21236, 21237, 21238, 21239, 21243, 21253, 21263, 21273, 21283, 21293, 21323, 21423, 21523, 21623, 21723, 21823, 21923, 22123, 23123, 24123, 25123, 26123, 27123, 28123, 29123, 30123, 31023, 31123, 31203, 31213, 31223, 31230, 31231, 31232, 31233, 31234, 31235, 31236, 31237, 31238, 31239, 31243, 31253, 31263, 31273, 31283, 31293, 31323, 31423, 31523, 31623, 31723, 31823, 31923, 32123, 33123, 34123, 35123, 36123, 37123, 38123, 39123, 40123, 41023, 41123, 41203, 41213, 41223, 41230, 41231, 41232, 41233, 41234, 41235, 41236, 41237, 41238, 41239, 41243, 41253, 41263, 41273, 41283, 41293, 41323, 41423, 41523, 41623, 41723, 41823, 41923, 42123, 43123, 44123, 45123, 46123, 47123, 48123, 49123, 50123, 51023, 51123, 51203, 51213, 51223, 51230, 51231, 51232, 51233, 51234, 51235, 51236, 51237, 51238, 51239, 51243, 51253, 51263, 51273, 51283, 51293, 51323, 51423, 51523, 51623, 51723, 51823, 51923, 52123, 53123, 54123, 55123, 56123, 57123, 58123, 59123, 60123, 61023, 61123, 61203, 61213, 61223, 61230, 61231, 61232, 61233, 61234, 61235, 61236, 61237, 61238, 61239, 61243, 61253, 61263, 61273, 61283, 61293, 61323, 61423, 61523, 61623, 61723, 61823, 61923, 62123, 63123, 64123, 65123, 66123, 67123, 68123, 69123, 70123, 71023, 71123, 71203, 71213, 71223, 71230, 71231, 71232, 71233, 71234, 71235, 71236, 71237, 71238, 71239, 71243, 71253, 71263, 71273, 71283, 71293, 71323, 71423, 71523, 71623, 71723, 71823, 71923, 72123, 73123, 74123, 75123, 76123, 77123, 78123, 79123, 80123, 81023, 81123, 81203, 81213, 81223, 81230, 81231, 81232, 81233, 81234, 81235, 81236, 81237, 81238, 81239, 81243, 81253, 81263, 81273, 81283, 81293, 81323, 81423, 81523, 81623, 81723, 81823, 81923, 82123, 83123, 84123, 85123, 86123, 87123, 88123, 89123, 90123, 91023, 91123, 91203, 91213, 91223, 91230, 91231, 91232, 91233, 91234, 91235, 91236, 91237, 91238, 91239, 91243, 91253, 91263, 91273, 91283, 91293, 91323, 91423, 91523, 91623, 91723, 91823, 91923, 92123, 93123, 94123, 95123, 96123, 97123, 98123, 99123] |

|----------------------|--------------------------------------------|

Rules

Input is guaranteed non-empty.

Leading and trailing zeros matter: 01234 is different from 12340, and 1234 doesn't crack either password. Think of how real passwords work!

Standard I/O rules apply.

No standard loopholes.

This is code-golf, so the shortest answer in bytes wins. Non-codegolfing languages are welcome!

edited 2 days ago

Hat

355

asked Feb 22 at 14:40

cefel

37628

New contributor

5

$begingroup$
Are the digits always in order? Based on the test cases I assume they are, but I couldn't see it mentioned in the rules unless I read past it.
$endgroup$
– Kevin Cruijssen
Feb 22 at 14:44

13

$begingroup$
Welcome to PPCG! This is a nice, well-structured, and neatly formatted first challenge. You've clearly done your homework as far as getting that all down. I'm looking forward to answering it (if someone doesn't answer it in R first!). In the future, we suggest using the sandbox to get feedback before posting to main. Hope you enjoy your time on PPCG!
$endgroup$
– Giuseppe
Feb 22 at 15:04

1

$begingroup$
@Giuseppe thanks! I've been anonymously reading the questions on this site for years, and I've been writing and tweaking and actually solving this specific problem for a couple months: I liked it enough to skip the sandbox. I'll post there first next time!
$endgroup$
– cefel
Feb 22 at 15:07

2

$begingroup$
@Arnauld Well, if your password is 01234 or 12340 you shouldn't be able to log in by typing 1234. Passwords are more a string than a number even if composed by numbers, at least in that sense. So yes, leading and trailing zeros are mandatory.
$endgroup$
– cefel
Feb 22 at 15:15

2

$begingroup$
The final test case appears to be missing 22123... unless I'm misunderstanding something?
$endgroup$
– Jonah
2 days ago

|
show 12 more comments

Introduction

In order to prevent keyloggers from stealing a user's password, a certain bank account system has implemented the following security measure: only certain digits are prompted to be entered each time.

For example, say your target's password is 89097, the system may prompt them to enter the 2nd, 4th and 5th digit:

997

Or it might prompt them to enter the 1st, 3rd and 5th digit:

807

The keylogger in your target's computer has been collecting password inputs for months now, so let's hack in!

Challenge

Input/Output examples

|----------------------|--------------------------------------------|

|         Input        |                   Output                   |

|----------------------|--------------------------------------------|

| [320, 723, 730]      | [37230, 72320, 73203, 73230]               |

| [374, 842]           | [37842, 38742, 83742]                      |

| [010, 103, 301]      | [30103]                                    |

| [123, 124, 125, 235] | [12345, 12354, 12435]                      |

| [239, 944]           | [23944]                                    |

| [111, 120]           | [11201, 11120, 11210, 12011, 12110, 12101] |

| [456, 789]           | []                                         |

| [756, 586]           | [07586, 17586, 27586, 37586, 47586, 57586, 57856, 58756, 67586, 70586, 71586, 72586, 73586, 74586, 75086, 75186, 75286, 75386, 75486, 75586, 75686, 75786, 75806, 75816, 75826, 75836, 75846, 75856, 75860, 75861, 75862, 75863, 75864, 75865, 75866, 75867, 75868, 75869, 75876, 75886, 75896, 75986, 76586, 77586, 78586, 79586, 87586, 97586] |

| [123]                | [00123, 01023, 01123, 01203, 01213, 01223, 01230, 01231, 01232, 01233, 01234, 01235, 01236, 01237, 01238, 01239, 01243, 01253, 01263, 01273, 01283, 01293, 01323, 01423, 01523, 01623, 01723, 01823, 01923, 02123, 03123, 04123, 05123, 06123, 07123, 08123, 09123, 10023, 10123, 10203, 10213, 10223, 10230, 10231, 10232, 10233, 10234, 10235, 10236, 10237, 10238, 10239, 10243, 10253, 10263, 10273, 10283, 10293, 10323, 10423, 10523, 10623, 10723, 10823, 10923, 11023, 11123, 11203, 11213, 11223, 11230, 11231, 11232, 11233, 11234, 11235, 11236, 11237, 11238, 11239, 11243, 11253, 11263, 11273, 11283, 11293, 11323, 11423, 11523, 11623, 11723, 11823, 11923, 12003, 12013, 12023, 12030, 12031, 12032, 12033, 12034, 12035, 12036, 12037, 12038, 12039, 12043, 12053, 12063, 12073, 12083, 12093, 12103, 12113, 12123, 12130, 12131, 12132, 12133, 12134, 12135, 12136, 12137, 12138, 12139, 12143, 12153, 12163, 12173, 12183, 12193, 12203, 12213, 12223, 12230, 12231, 12232, 12233, 12234, 12235, 12236, 12237, 12238, 12239, 12243, 12253, 12263, 12273, 12283, 12293, 12300, 12301, 12302, 12303, 12304, 12305, 12306, 12307, 12308, 12309, 12310, 12311, 12312, 12313, 12314, 12315, 12316, 12317, 12318, 12319, 12320, 12321, 12322, 12323, 12324, 12325, 12326, 12327, 12328, 12329, 12330, 12331, 12332, 12333, 12334, 12335, 12336, 12337, 12338, 12339, 12340, 12341, 12342, 12343, 12344, 12345, 12346, 12347, 12348, 12349, 12350, 12351, 12352, 12353, 12354, 12355, 12356, 12357, 12358, 12359, 12360, 12361, 12362, 12363, 12364, 12365, 12366, 12367, 12368, 12369, 12370, 12371, 12372, 12373, 12374, 12375, 12376, 12377, 12378, 12379, 12380, 12381, 12382, 12383, 12384, 12385, 12386, 12387, 12388, 12389, 12390, 12391, 12392, 12393, 12394, 12395, 12396, 12397, 12398, 12399, 12403, 12413, 12423, 12430, 12431, 12432, 12433, 12434, 12435, 12436, 12437, 12438, 12439, 12443, 12453, 12463, 12473, 12483, 12493, 12503, 12513, 12523, 12530, 12531, 12532, 12533, 12534, 12535, 12536, 12537, 12538, 12539, 12543, 12553, 12563, 12573, 12583, 12593, 12603, 12613, 12623, 12630, 12631, 12632, 12633, 12634, 12635, 12636, 12637, 12638, 12639, 12643, 12653, 12663, 12673, 12683, 12693, 12703, 12713, 12723, 12730, 12731, 12732, 12733, 12734, 12735, 12736, 12737, 12738, 12739, 12743, 12753, 12763, 12773, 12783, 12793, 12803, 12813, 12823, 12830, 12831, 12832, 12833, 12834, 12835, 12836, 12837, 12838, 12839, 12843, 12853, 12863, 12873, 12883, 12893, 12903, 12913, 12923, 12930, 12931, 12932, 12933, 12934, 12935, 12936, 12937, 12938, 12939, 12943, 12953, 12963, 12973, 12983, 12993, 13023, 13123, 13203, 13213, 13223, 13230, 13231, 13232, 13233, 13234, 13235, 13236, 13237, 13238, 13239, 13243, 13253, 13263, 13273, 13283, 13293, 13323, 13423, 13523, 13623, 13723, 13823, 13923, 14023, 14123, 14203, 14213, 14223, 14230, 14231, 14232, 14233, 14234, 14235, 14236, 14237, 14238, 14239, 14243, 14253, 14263, 14273, 14283, 14293, 14323, 14423, 14523, 14623, 14723, 14823, 14923, 15023, 15123, 15203, 15213, 15223, 15230, 15231, 15232, 15233, 15234, 15235, 15236, 15237, 15238, 15239, 15243, 15253, 15263, 15273, 15283, 15293, 15323, 15423, 15523, 15623, 15723, 15823, 15923, 16023, 16123, 16203, 16213, 16223, 16230, 16231, 16232, 16233, 16234, 16235, 16236, 16237, 16238, 16239, 16243, 16253, 16263, 16273, 16283, 16293, 16323, 16423, 16523, 16623, 16723, 16823, 16923, 17023, 17123, 17203, 17213, 17223, 17230, 17231, 17232, 17233, 17234, 17235, 17236, 17237, 17238, 17239, 17243, 17253, 17263, 17273, 17283, 17293, 17323, 17423, 17523, 17623, 17723, 17823, 17923, 18023, 18123, 18203, 18213, 18223, 18230, 18231, 18232, 18233, 18234, 18235, 18236, 18237, 18238, 18239, 18243, 18253, 18263, 18273, 18283, 18293, 18323, 18423, 18523, 18623, 18723, 18823, 18923, 19023, 19123, 19203, 19213, 19223, 19230, 19231, 19232, 19233, 19234, 19235, 19236, 19237, 19238, 19239, 19243, 19253, 19263, 19273, 19283, 19293, 19323, 19423, 19523, 19623, 19723, 19823, 19923, 20123, 21023, 21123, 21203, 21213, 21223, 21230, 21231, 21232, 21233, 21234, 21235, 21236, 21237, 21238, 21239, 21243, 21253, 21263, 21273, 21283, 21293, 21323, 21423, 21523, 21623, 21723, 21823, 21923, 22123, 23123, 24123, 25123, 26123, 27123, 28123, 29123, 30123, 31023, 31123, 31203, 31213, 31223, 31230, 31231, 31232, 31233, 31234, 31235, 31236, 31237, 31238, 31239, 31243, 31253, 31263, 31273, 31283, 31293, 31323, 31423, 31523, 31623, 31723, 31823, 31923, 32123, 33123, 34123, 35123, 36123, 37123, 38123, 39123, 40123, 41023, 41123, 41203, 41213, 41223, 41230, 41231, 41232, 41233, 41234, 41235, 41236, 41237, 41238, 41239, 41243, 41253, 41263, 41273, 41283, 41293, 41323, 41423, 41523, 41623, 41723, 41823, 41923, 42123, 43123, 44123, 45123, 46123, 47123, 48123, 49123, 50123, 51023, 51123, 51203, 51213, 51223, 51230, 51231, 51232, 51233, 51234, 51235, 51236, 51237, 51238, 51239, 51243, 51253, 51263, 51273, 51283, 51293, 51323, 51423, 51523, 51623, 51723, 51823, 51923, 52123, 53123, 54123, 55123, 56123, 57123, 58123, 59123, 60123, 61023, 61123, 61203, 61213, 61223, 61230, 61231, 61232, 61233, 61234, 61235, 61236, 61237, 61238, 61239, 61243, 61253, 61263, 61273, 61283, 61293, 61323, 61423, 61523, 61623, 61723, 61823, 61923, 62123, 63123, 64123, 65123, 66123, 67123, 68123, 69123, 70123, 71023, 71123, 71203, 71213, 71223, 71230, 71231, 71232, 71233, 71234, 71235, 71236, 71237, 71238, 71239, 71243, 71253, 71263, 71273, 71283, 71293, 71323, 71423, 71523, 71623, 71723, 71823, 71923, 72123, 73123, 74123, 75123, 76123, 77123, 78123, 79123, 80123, 81023, 81123, 81203, 81213, 81223, 81230, 81231, 81232, 81233, 81234, 81235, 81236, 81237, 81238, 81239, 81243, 81253, 81263, 81273, 81283, 81293, 81323, 81423, 81523, 81623, 81723, 81823, 81923, 82123, 83123, 84123, 85123, 86123, 87123, 88123, 89123, 90123, 91023, 91123, 91203, 91213, 91223, 91230, 91231, 91232, 91233, 91234, 91235, 91236, 91237, 91238, 91239, 91243, 91253, 91263, 91273, 91283, 91293, 91323, 91423, 91523, 91623, 91723, 91823, 91923, 92123, 93123, 94123, 95123, 96123, 97123, 98123, 99123] |

|----------------------|--------------------------------------------|

Rules

Input is guaranteed non-empty.

Leading and trailing zeros matter: 01234 is different from 12340, and 1234 doesn't crack either password. Think of how real passwords work!

Standard I/O rules apply.

No standard loopholes.

This is code-golf, so the shortest answer in bytes wins. Non-codegolfing languages are welcome!

edited 2 days ago

Hat

355

asked Feb 22 at 14:40

cefel

37628

New contributor

Introduction

In order to prevent keyloggers from stealing a user's password, a certain bank account system has implemented the following security measure: only certain digits are prompted to be entered each time.

For example, say your target's password is 89097, the system may prompt them to enter the 2nd, 4th and 5th digit:

997

Or it might prompt them to enter the 1st, 3rd and 5th digit:

807

The keylogger in your target's computer has been collecting password inputs for months now, so let's hack in!

Challenge

Input/Output examples

|----------------------|--------------------------------------------|

|         Input        |                   Output                   |

|----------------------|--------------------------------------------|

| [320, 723, 730]      | [37230, 72320, 73203, 73230]               |

| [374, 842]           | [37842, 38742, 83742]                      |

| [010, 103, 301]      | [30103]                                    |

| [123, 124, 125, 235] | [12345, 12354, 12435]                      |

| [239, 944]           | [23944]                                    |

| [111, 120]           | [11201, 11120, 11210, 12011, 12110, 12101] |

| [456, 789]           | []                                         |

| [756, 586]           | [07586, 17586, 27586, 37586, 47586, 57586, 57856, 58756, 67586, 70586, 71586, 72586, 73586, 74586, 75086, 75186, 75286, 75386, 75486, 75586, 75686, 75786, 75806, 75816, 75826, 75836, 75846, 75856, 75860, 75861, 75862, 75863, 75864, 75865, 75866, 75867, 75868, 75869, 75876, 75886, 75896, 75986, 76586, 77586, 78586, 79586, 87586, 97586] |

| [123]                | [00123, 01023, 01123, 01203, 01213, 01223, 01230, 01231, 01232, 01233, 01234, 01235, 01236, 01237, 01238, 01239, 01243, 01253, 01263, 01273, 01283, 01293, 01323, 01423, 01523, 01623, 01723, 01823, 01923, 02123, 03123, 04123, 05123, 06123, 07123, 08123, 09123, 10023, 10123, 10203, 10213, 10223, 10230, 10231, 10232, 10233, 10234, 10235, 10236, 10237, 10238, 10239, 10243, 10253, 10263, 10273, 10283, 10293, 10323, 10423, 10523, 10623, 10723, 10823, 10923, 11023, 11123, 11203, 11213, 11223, 11230, 11231, 11232, 11233, 11234, 11235, 11236, 11237, 11238, 11239, 11243, 11253, 11263, 11273, 11283, 11293, 11323, 11423, 11523, 11623, 11723, 11823, 11923, 12003, 12013, 12023, 12030, 12031, 12032, 12033, 12034, 12035, 12036, 12037, 12038, 12039, 12043, 12053, 12063, 12073, 12083, 12093, 12103, 12113, 12123, 12130, 12131, 12132, 12133, 12134, 12135, 12136, 12137, 12138, 12139, 12143, 12153, 12163, 12173, 12183, 12193, 12203, 12213, 12223, 12230, 12231, 12232, 12233, 12234, 12235, 12236, 12237, 12238, 12239, 12243, 12253, 12263, 12273, 12283, 12293, 12300, 12301, 12302, 12303, 12304, 12305, 12306, 12307, 12308, 12309, 12310, 12311, 12312, 12313, 12314, 12315, 12316, 12317, 12318, 12319, 12320, 12321, 12322, 12323, 12324, 12325, 12326, 12327, 12328, 12329, 12330, 12331, 12332, 12333, 12334, 12335, 12336, 12337, 12338, 12339, 12340, 12341, 12342, 12343, 12344, 12345, 12346, 12347, 12348, 12349, 12350, 12351, 12352, 12353, 12354, 12355, 12356, 12357, 12358, 12359, 12360, 12361, 12362, 12363, 12364, 12365, 12366, 12367, 12368, 12369, 12370, 12371, 12372, 12373, 12374, 12375, 12376, 12377, 12378, 12379, 12380, 12381, 12382, 12383, 12384, 12385, 12386, 12387, 12388, 12389, 12390, 12391, 12392, 12393, 12394, 12395, 12396, 12397, 12398, 12399, 12403, 12413, 12423, 12430, 12431, 12432, 12433, 12434, 12435, 12436, 12437, 12438, 12439, 12443, 12453, 12463, 12473, 12483, 12493, 12503, 12513, 12523, 12530, 12531, 12532, 12533, 12534, 12535, 12536, 12537, 12538, 12539, 12543, 12553, 12563, 12573, 12583, 12593, 12603, 12613, 12623, 12630, 12631, 12632, 12633, 12634, 12635, 12636, 12637, 12638, 12639, 12643, 12653, 12663, 12673, 12683, 12693, 12703, 12713, 12723, 12730, 12731, 12732, 12733, 12734, 12735, 12736, 12737, 12738, 12739, 12743, 12753, 12763, 12773, 12783, 12793, 12803, 12813, 12823, 12830, 12831, 12832, 12833, 12834, 12835, 12836, 12837, 12838, 12839, 12843, 12853, 12863, 12873, 12883, 12893, 12903, 12913, 12923, 12930, 12931, 12932, 12933, 12934, 12935, 12936, 12937, 12938, 12939, 12943, 12953, 12963, 12973, 12983, 12993, 13023, 13123, 13203, 13213, 13223, 13230, 13231, 13232, 13233, 13234, 13235, 13236, 13237, 13238, 13239, 13243, 13253, 13263, 13273, 13283, 13293, 13323, 13423, 13523, 13623, 13723, 13823, 13923, 14023, 14123, 14203, 14213, 14223, 14230, 14231, 14232, 14233, 14234, 14235, 14236, 14237, 14238, 14239, 14243, 14253, 14263, 14273, 14283, 14293, 14323, 14423, 14523, 14623, 14723, 14823, 14923, 15023, 15123, 15203, 15213, 15223, 15230, 15231, 15232, 15233, 15234, 15235, 15236, 15237, 15238, 15239, 15243, 15253, 15263, 15273, 15283, 15293, 15323, 15423, 15523, 15623, 15723, 15823, 15923, 16023, 16123, 16203, 16213, 16223, 16230, 16231, 16232, 16233, 16234, 16235, 16236, 16237, 16238, 16239, 16243, 16253, 16263, 16273, 16283, 16293, 16323, 16423, 16523, 16623, 16723, 16823, 16923, 17023, 17123, 17203, 17213, 17223, 17230, 17231, 17232, 17233, 17234, 17235, 17236, 17237, 17238, 17239, 17243, 17253, 17263, 17273, 17283, 17293, 17323, 17423, 17523, 17623, 17723, 17823, 17923, 18023, 18123, 18203, 18213, 18223, 18230, 18231, 18232, 18233, 18234, 18235, 18236, 18237, 18238, 18239, 18243, 18253, 18263, 18273, 18283, 18293, 18323, 18423, 18523, 18623, 18723, 18823, 18923, 19023, 19123, 19203, 19213, 19223, 19230, 19231, 19232, 19233, 19234, 19235, 19236, 19237, 19238, 19239, 19243, 19253, 19263, 19273, 19283, 19293, 19323, 19423, 19523, 19623, 19723, 19823, 19923, 20123, 21023, 21123, 21203, 21213, 21223, 21230, 21231, 21232, 21233, 21234, 21235, 21236, 21237, 21238, 21239, 21243, 21253, 21263, 21273, 21283, 21293, 21323, 21423, 21523, 21623, 21723, 21823, 21923, 22123, 23123, 24123, 25123, 26123, 27123, 28123, 29123, 30123, 31023, 31123, 31203, 31213, 31223, 31230, 31231, 31232, 31233, 31234, 31235, 31236, 31237, 31238, 31239, 31243, 31253, 31263, 31273, 31283, 31293, 31323, 31423, 31523, 31623, 31723, 31823, 31923, 32123, 33123, 34123, 35123, 36123, 37123, 38123, 39123, 40123, 41023, 41123, 41203, 41213, 41223, 41230, 41231, 41232, 41233, 41234, 41235, 41236, 41237, 41238, 41239, 41243, 41253, 41263, 41273, 41283, 41293, 41323, 41423, 41523, 41623, 41723, 41823, 41923, 42123, 43123, 44123, 45123, 46123, 47123, 48123, 49123, 50123, 51023, 51123, 51203, 51213, 51223, 51230, 51231, 51232, 51233, 51234, 51235, 51236, 51237, 51238, 51239, 51243, 51253, 51263, 51273, 51283, 51293, 51323, 51423, 51523, 51623, 51723, 51823, 51923, 52123, 53123, 54123, 55123, 56123, 57123, 58123, 59123, 60123, 61023, 61123, 61203, 61213, 61223, 61230, 61231, 61232, 61233, 61234, 61235, 61236, 61237, 61238, 61239, 61243, 61253, 61263, 61273, 61283, 61293, 61323, 61423, 61523, 61623, 61723, 61823, 61923, 62123, 63123, 64123, 65123, 66123, 67123, 68123, 69123, 70123, 71023, 71123, 71203, 71213, 71223, 71230, 71231, 71232, 71233, 71234, 71235, 71236, 71237, 71238, 71239, 71243, 71253, 71263, 71273, 71283, 71293, 71323, 71423, 71523, 71623, 71723, 71823, 71923, 72123, 73123, 74123, 75123, 76123, 77123, 78123, 79123, 80123, 81023, 81123, 81203, 81213, 81223, 81230, 81231, 81232, 81233, 81234, 81235, 81236, 81237, 81238, 81239, 81243, 81253, 81263, 81273, 81283, 81293, 81323, 81423, 81523, 81623, 81723, 81823, 81923, 82123, 83123, 84123, 85123, 86123, 87123, 88123, 89123, 90123, 91023, 91123, 91203, 91213, 91223, 91230, 91231, 91232, 91233, 91234, 91235, 91236, 91237, 91238, 91239, 91243, 91253, 91263, 91273, 91283, 91293, 91323, 91423, 91523, 91623, 91723, 91823, 91923, 92123, 93123, 94123, 95123, 96123, 97123, 98123, 99123] |

|----------------------|--------------------------------------------|

Rules

Input is guaranteed non-empty.

Leading and trailing zeros matter: 01234 is different from 12340, and 1234 doesn't crack either password. Think of how real passwords work!

Standard I/O rules apply.

No standard loopholes.

This is code-golf, so the shortest answer in bytes wins. Non-codegolfing languages are welcome!

code-golf

edited 2 days ago

Hat

355

asked Feb 22 at 14:40

cefel

37628

New contributor

edited 2 days ago

Hat

355

asked Feb 22 at 14:40

cefel

37628

New contributor

edited 2 days ago

Hat

355

edited 2 days ago

Hat

355

edited 2 days ago

Hat

355

asked Feb 22 at 14:40

cefel

37628

New contributor

asked Feb 22 at 14:40

cefel

37628

asked Feb 22 at 14:40

cefel

37628

New contributor

cefel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

5

$begingroup$
Are the digits always in order? Based on the test cases I assume they are, but I couldn't see it mentioned in the rules unless I read past it.
$endgroup$
– Kevin Cruijssen
Feb 22 at 14:44

13

$begingroup$
Welcome to PPCG! This is a nice, well-structured, and neatly formatted first challenge. You've clearly done your homework as far as getting that all down. I'm looking forward to answering it (if someone doesn't answer it in R first!). In the future, we suggest using the sandbox to get feedback before posting to main. Hope you enjoy your time on PPCG!
$endgroup$
– Giuseppe
Feb 22 at 15:04

1

$begingroup$
@Giuseppe thanks! I've been anonymously reading the questions on this site for years, and I've been writing and tweaking and actually solving this specific problem for a couple months: I liked it enough to skip the sandbox. I'll post there first next time!
$endgroup$
– cefel
Feb 22 at 15:07

2

$begingroup$
@Arnauld Well, if your password is 01234 or 12340 you shouldn't be able to log in by typing 1234. Passwords are more a string than a number even if composed by numbers, at least in that sense. So yes, leading and trailing zeros are mandatory.
$endgroup$
– cefel
Feb 22 at 15:15

2

$begingroup$
The final test case appears to be missing 22123... unless I'm misunderstanding something?
$endgroup$
– Jonah
2 days ago

|
show 12 more comments

5

$begingroup$
Are the digits always in order? Based on the test cases I assume they are, but I couldn't see it mentioned in the rules unless I read past it.
$endgroup$
– Kevin Cruijssen
Feb 22 at 14:44

13

$begingroup$
Welcome to PPCG! This is a nice, well-structured, and neatly formatted first challenge. You've clearly done your homework as far as getting that all down. I'm looking forward to answering it (if someone doesn't answer it in R first!). In the future, we suggest using the sandbox to get feedback before posting to main. Hope you enjoy your time on PPCG!
$endgroup$
– Giuseppe
Feb 22 at 15:04

1

$begingroup$
@Giuseppe thanks! I've been anonymously reading the questions on this site for years, and I've been writing and tweaking and actually solving this specific problem for a couple months: I liked it enough to skip the sandbox. I'll post there first next time!
$endgroup$
– cefel
Feb 22 at 15:07

2

$begingroup$
@Arnauld Well, if your password is 01234 or 12340 you shouldn't be able to log in by typing 1234. Passwords are more a string than a number even if composed by numbers, at least in that sense. So yes, leading and trailing zeros are mandatory.
$endgroup$
– cefel
Feb 22 at 15:15

2

$begingroup$
The final test case appears to be missing 22123... unless I'm misunderstanding something?
$endgroup$
– Jonah
2 days ago

Are the digits always in order? Based on the test cases I assume they are, but I couldn't see it mentioned in the rules unless I read past it.

– Kevin Cruijssen
Feb 22 at 14:44

Welcome to PPCG! This is a nice, well-structured, and neatly formatted first challenge. You've clearly done your homework as far as getting that all down. I'm looking forward to answering it (if someone doesn't answer it in R first!). In the future, we suggest using the sandbox to get feedback before posting to main. Hope you enjoy your time on PPCG!

– Giuseppe
Feb 22 at 15:04

@Giuseppe thanks! I've been anonymously reading the questions on this site for years, and I've been writing and tweaking and actually solving this specific problem for a couple months: I liked it enough to skip the sandbox. I'll post there first next time!

– cefel
Feb 22 at 15:07

@Arnauld Well, if your password is 01234 or 12340 you shouldn't be able to log in by typing 1234. Passwords are more a string than a number even if composed by numbers, at least in that sense. So yes, leading and trailing zeros are mandatory.

– cefel
Feb 22 at 15:15

The final test case appears to be missing 22123... unless I'm misunderstanding something?

– Jonah
2 days ago

|
show 12 more comments

17 Answers
17

active

oldest

votes

Python, 101 bytes

lambda e:['%05d'%i for i in range(10**5)if all(re.search('.*'.join(x),'%05d'%i)for x in e)]

import re

Try it online! Works in Python 2 as well as Python 3.

(97 bytes in Python 3.8:)

lambda e:[p for i in range(10**5)if all(re.search('.*'.join(x),p:='%05d'%i)for x in e)]

import re

edited 11 hours ago

answered 2 days ago

Lynn

49.5k795228

1

$begingroup$
This is a lovely solution...
$endgroup$
– Jonah
2 days ago

add a comment |

05AB1E, 11 9 bytes

žh5ãʒæIåP

Try it online!

Explanation

žh          # push 0123456789

  5ã        # 5 times cartesian product

    ʒ       # filter, keep only values are true under:

     æ      # powerset of value

      Iå    # check if each of the input values are in this list

        P   # product

edited 2 days ago

answered 2 days ago

Emigna

46.6k432142

add a comment |

JavaScript (ES6), 88 bytes

Prints the results with alert().

a=>{for(k=n=1e5;n--;)a.every(x=>(s=([k]+n).slice(-5)).match([...x].join`.*`))&&alert(s)}

Try it online!

Commented

a => {                    // a[] = input array of 3-character strings

  for(k = n = 1e5; n--;)  // initialize k to 100000; for n = 99999 to 0:

    a.every(x =>          // for each string x = 'XYZ' in a[]:

      ( s =               //   define s as the concatenation of

          ([k] + n)       //   '100000' and n; e.g. '100000' + 1337 -> '1000001337'

          .slice(-5)      //   keep the last 5 digits; e.g. '01337'

      ).match(            //   test whether this string is matching

        [...x].join`.*`   //   the pattern /X.*Y.*Z/

      )                   //

    ) &&                  // end of every(); if all tests were successful:

      alert(s)            //   output s

}                         //

edited 2 days ago

answered Feb 22 at 15:06

Arnauld

76.7k693322

add a comment |

Haskell, 81 80 78 76 bytes

f x=[p|p<-mapM(:['1'..'9'])"00000",all(`elem`(concat.words<$>mapM(:" ")p))x]

The obvious brute force approach in Haskell: built a list of all possible passwords and keep those where all elements from the input list are in respective list of subsequences.

Try it online!

Edit: -1 byte thanks to @xnor, -2 -4 bytes thanks to @H.PWiz

edited 15 hours ago

answered yesterday

nimi

31.9k32285

1

$begingroup$
Looks like you can compute the subseqs yourself a little shorter.
$endgroup$
– xnor
yesterday

1

$begingroup$
concat.words<$>mapM(:" ")p is shorter
$endgroup$
– H.PWiz
yesterday

2

$begingroup$
use p<-mapM(:['1'..'9'])"00000" to save 2 more bytes
$endgroup$
– H.PWiz
16 hours ago

add a comment |

Jelly, 11 bytes

9Żṗ5ŒPiⱮẠɗƇ

Try it online!

answered 2 days ago

Erik the Outgolfer

32k429103

add a comment |

Pyth, 11 bytes

f<QyT^s`MT5

Takes input as a set of strings.
Try it here

Explanation

f<QyT^s`MT5

      s`MT      Take the digits as a string.

     ^    5     Take the Cartesian product with itself 5 times.

f   T           Filter the ones...

 <Qy            ... where the input is a subset of the power set.

answered 2 days ago

Mnemonic

4,7851730

add a comment |

Ruby, 54 bytes

->a{(?0*5..?9*5).select{|x|a.all?{|y|x=~/#{y*'.*'}/}}}

Try it online!

Takes input as an arrray of character arrays.

answered 2 days ago

Kirill L.

4,6151523

$begingroup$
Well done! You beat me by 25 bytes. Should I delete my answer?
$endgroup$
– Eric Duminil
2 days ago

1

$begingroup$
No, as long as you have a valid answer, there's no need to delete it.
$endgroup$
– Kirill L.
2 days ago

add a comment |

Python 3, 98 bytes

f=lambda l,s='':any(l)or print(s)if s[4:]else[f([x[x[:1]==c:]for x in l],s+c)for c in'0123456789']

Try it online!

Recursively tries building every five-digit number string in s, tracking the subsequences in l still remaining to be hit. If all are empty by the end, prints the result.

Python 3.8 (pre-release), 94 bytes

lambda l:[s for n in range(10**5)if all(''in[x:=x[x[:1]==c:]for c in(s:='%05d'%n)]for x in l)]

Try it online!

Behold the power of assignment expressions!. Uses the method from here for checking subsequences.

edited yesterday

answered yesterday

xnor

91.1k18186442

add a comment |

Perl 5 `-a`, 80 bytes

map{s||($t=0)x(5-y///c)|e;for$b(map s//.*/gr,@F){/$b/&&$t++}$t==@F&&say}0..99999

Try it online!

answered 2 days ago

Xcali

5,375520

1

$begingroup$
$t+=/$b/ instead of /$b/&&$t++
$endgroup$
– Nahuel Fouilleul
2 days ago

add a comment |

Retina, 53 bytes

~(`

.$*

m`^

G`

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Try it online! Explanation:

~(`

After executing the script, take the result as a new script, and execute that, too.

.$*

Insert .* everywhere. This results in .*3.*2.*0.* although we only need 3.*2.*0, not that it matters.

m`^

G`

Insert a G` at the start of each line. This turns it into a Retina Grep command.

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Prefix two more Retina commands. The resulting script will therefore look something like this:

K`

Clear the buffer (which contains the original input).

5+

Repeat 5 times...

%`$

... append to each line...

0$"1$"2$"3$"4$"5$"6$"7$"8$"9

... the digit 0, then a copy of the line, then the digit 1, etc. until 9. This means that after n loops you will have all n-digit numbers.

G`.*3.*2.*0.*

G`.*7.*2.*3.*

G`.*7.*3.*0.*

Filter out the possible numbers based on the input.

answered 2 days ago

Neil

81k744178

add a comment |

R, 80 82 bytes

Reduce(intersect,lapply(gsub("",".*",scan(,"")),grep,sprintf("%05d",0:99999),v=T))

Here’s a base R solution using regex. Writing this nested series of functions made me realise how much I’ve learned to appreciate the magrittr package!

Initially hadn’t read rules on input, so now reads from stdin (thanks @KirillL).

Try it online!

edited 2 days ago

answered 2 days ago

Nick Kennedy

21114

$begingroup$
@digEmAll it says it’s 82 bytes doesn’t it? You can’t use integers because of potential for leading zeroes in the input.
$endgroup$
– Nick Kennedy
2 days ago

$begingroup$
Sorry, I read the title and unconsciously I picked the minimum number without noticing that it was strikedthrough... and yes, sorry again you're right about the string input ;)
$endgroup$
– digEmAll
2 days ago

add a comment |

Ruby, 79 77 bytes

->x{(0...1e5).map{|i|'%05d'%i}.select{|i|x.all?{|c|i=~/#{c.gsub('','.*')}/}}}

Try it online!

Input is an array of strings.

Here's a more readable version of the same code:

def f(codes)

  (0...10**5).map{|i| '%05d'%i}.select do |i|

    codes.all? do |code|

      i =~ Regexp.new(code.chars.join('.*'))

    end

  end

end

edited 2 days ago

answered 2 days ago

Eric Duminil

43128

$begingroup$
BTW, your approach can also be made shorter by switching to char array input, as in my version, and -2 more bytes by formatting the upper value as 1e5, like this
$endgroup$
– Kirill L.
2 days ago

$begingroup$
@KirillL. Thanks for the -2 bytes. I won't change the input format because my answer would look too much like yours. Cheers!
$endgroup$
– Eric Duminil
2 days ago

add a comment |

J, 52 bytes

(>,{;^:4~i.10)([#~]*/@e."2((#~3=+/"1)#:i.32)#"_ 1[)]

Try it online!

edited yesterday

answered 2 days ago

Jonah

2,331916

add a comment |

C# (Visual C# Interactive Compiler), 111 bytes

x=>{for(int i=0;i<1e5;){var s=$"{i++:D5}";if(x.All(t=>t.Aggregate(0,(a,c)=>a<0?a:s.IndexOf(c,a))>0))Print(s);}}

Try it online!

// x: input list of strings

x=>{

  // generate all numbers under 100k

  for(int i=0;i<1e5;){

    // convert the current number to

    // a 5 digit string padded with 0's

    var s=$"{i++:D5}";

    // test all inputs against the 5 digit

    // string using an aggregate.

    // each step of the aggregate gets

    // the index of the next occurrence

    // of the current character starting

    // at the previously found character.

    // a negative index indicates error.

    if(x.All(t=>t

             .Aggregate(0,(a,c)=>

               a<0

                 ?a

                 :s.IndexOf(c,a))>0))

      // output to STDOUT

      Print(s);

  }

}

edited yesterday

answered yesterday

dana

1,171166

add a comment |

Japt, 21 bytes

1e5o ù'0 f@e_XèZË+".*

Try it!

1e5o ù'0 f@e_XèZË+".*    # full program



1e5o                     # generate numbers under 100k

     ù'0                 # left pad with 0's

         f@              # filter array

           e_            # check every element of input array

             Xè          # X is the number to be tested.

                         # test it against a regex.

               ZË+".*    # the regex is an element from the input array

                         # with wildcards injected between each character

-2 bytes thanks to @Shaggy!

edited 1 hour ago

answered 17 hours ago

dana

1,171166

$begingroup$
less useless variables: 1e5o ù'0 fA{Ue@AèX®+".* :P
$endgroup$
– ASCII-only
11 hours ago

$begingroup$
also 23: 1e5o ù'0 fA{Ue@AèX¬q".*
$endgroup$
– ASCII-only
10 hours ago

$begingroup$
21 bytes
$endgroup$
– Shaggy
5 hours ago

$begingroup$
Interesting... I didn't realize return X,Y,Z would pick the last term. Thanks for the tips :)
$endgroup$
– dana
2 hours ago

$begingroup$
@dana; Yup, that's a feature of JavaScript: tio.run/##y0osSyxOLsosKNHNy09J/Z9m@1@jQqdS09auuii1pLQoTwHIq/…
$endgroup$
– Shaggy
6 mins ago

add a comment |

PHP 128 bytes

for(;$i++<1e5;$k>$argc||print$s)for($k=0;$n=$argv[++$k];)preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i-1))||$k=$argc;

for(;$i<1e5;$i+=$k<$argc||print$s)for($k=0;$n=$argv[++$k];)if(!preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i)))break;

take input from command line arguments. Run with -nr or try them online.

answered 2 days ago

Titus

13.2k11238

add a comment |

Clean, 113 bytes

import StdEnv,Data.List

$l=[s\s<-iter 5(p=[[c:e]\e<-p,c<-['0'..'9']])[[]]|all(flip any(subsequences s)o(==))l]

Try it online!

answered 16 hours ago

Οurous

7,20111035

add a comment |

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17 Answers
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17 Answers
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votes

Python, 101 bytes

lambda e:['%05d'%i for i in range(10**5)if all(re.search('.*'.join(x),'%05d'%i)for x in e)]

import re

Try it online! Works in Python 2 as well as Python 3.

(97 bytes in Python 3.8:)

lambda e:[p for i in range(10**5)if all(re.search('.*'.join(x),p:='%05d'%i)for x in e)]

import re

edited 11 hours ago

answered 2 days ago

Lynn

49.5k795228

1

$begingroup$
This is a lovely solution...
$endgroup$
– Jonah
2 days ago

add a comment |

Python, 101 bytes

lambda e:['%05d'%i for i in range(10**5)if all(re.search('.*'.join(x),'%05d'%i)for x in e)]

import re

Try it online! Works in Python 2 as well as Python 3.

(97 bytes in Python 3.8:)

lambda e:[p for i in range(10**5)if all(re.search('.*'.join(x),p:='%05d'%i)for x in e)]

import re

edited 11 hours ago

answered 2 days ago

Lynn

49.5k795228

1

$begingroup$
This is a lovely solution...
$endgroup$
– Jonah
2 days ago

add a comment |

Python, 101 bytes

lambda e:['%05d'%i for i in range(10**5)if all(re.search('.*'.join(x),'%05d'%i)for x in e)]

import re

Try it online! Works in Python 2 as well as Python 3.

(97 bytes in Python 3.8:)

lambda e:[p for i in range(10**5)if all(re.search('.*'.join(x),p:='%05d'%i)for x in e)]

import re

edited 11 hours ago

answered 2 days ago

Lynn

49.5k795228

Python, 101 bytes

lambda e:['%05d'%i for i in range(10**5)if all(re.search('.*'.join(x),'%05d'%i)for x in e)]

import re

Try it online! Works in Python 2 as well as Python 3.

(97 bytes in Python 3.8:)

lambda e:[p for i in range(10**5)if all(re.search('.*'.join(x),p:='%05d'%i)for x in e)]

import re

edited 11 hours ago

answered 2 days ago

Lynn

49.5k795228

edited 11 hours ago

answered 2 days ago

Lynn

49.5k795228

answered 2 days ago

Lynn

49.5k795228

answered 2 days ago

Lynn

49.5k795228

1

$begingroup$
This is a lovely solution...
$endgroup$
– Jonah
2 days ago

add a comment |

1

$begingroup$
This is a lovely solution...
$endgroup$
– Jonah
2 days ago

This is a lovely solution...

– Jonah
2 days ago

add a comment |

05AB1E, 11 9 bytes

žh5ãʒæIåP

Try it online!

Explanation

žh          # push 0123456789

  5ã        # 5 times cartesian product

    ʒ       # filter, keep only values are true under:

     æ      # powerset of value

      Iå    # check if each of the input values are in this list

        P   # product

edited 2 days ago

answered 2 days ago

Emigna

46.6k432142

add a comment |

05AB1E, 11 9 bytes

žh5ãʒæIåP

Try it online!

Explanation

žh          # push 0123456789

  5ã        # 5 times cartesian product

    ʒ       # filter, keep only values are true under:

     æ      # powerset of value

      Iå    # check if each of the input values are in this list

        P   # product

edited 2 days ago

answered 2 days ago

Emigna

46.6k432142

add a comment |

05AB1E, 11 9 bytes

žh5ãʒæIåP

Try it online!

Explanation

žh          # push 0123456789

  5ã        # 5 times cartesian product

    ʒ       # filter, keep only values are true under:

     æ      # powerset of value

      Iå    # check if each of the input values are in this list

        P   # product

edited 2 days ago

answered 2 days ago

Emigna

46.6k432142

05AB1E, 11 9 bytes

žh5ãʒæIåP

Try it online!

Explanation

žh          # push 0123456789

  5ã        # 5 times cartesian product

    ʒ       # filter, keep only values are true under:

     æ      # powerset of value

      Iå    # check if each of the input values are in this list

        P   # product

edited 2 days ago

answered 2 days ago

Emigna

46.6k432142

edited 2 days ago

answered 2 days ago

Emigna

46.6k432142

answered 2 days ago

Emigna

46.6k432142

answered 2 days ago

Emigna

46.6k432142

add a comment |

JavaScript (ES6), 88 bytes

Prints the results with alert().

a=>{for(k=n=1e5;n--;)a.every(x=>(s=([k]+n).slice(-5)).match([...x].join`.*`))&&alert(s)}

Try it online!

Commented

a => {                    // a[] = input array of 3-character strings

  for(k = n = 1e5; n--;)  // initialize k to 100000; for n = 99999 to 0:

    a.every(x =>          // for each string x = 'XYZ' in a[]:

      ( s =               //   define s as the concatenation of

          ([k] + n)       //   '100000' and n; e.g. '100000' + 1337 -> '1000001337'

          .slice(-5)      //   keep the last 5 digits; e.g. '01337'

      ).match(            //   test whether this string is matching

        [...x].join`.*`   //   the pattern /X.*Y.*Z/

      )                   //

    ) &&                  // end of every(); if all tests were successful:

      alert(s)            //   output s

}                         //

edited 2 days ago

answered Feb 22 at 15:06

Arnauld

76.7k693322

add a comment |

JavaScript (ES6), 88 bytes

Prints the results with alert().

a=>{for(k=n=1e5;n--;)a.every(x=>(s=([k]+n).slice(-5)).match([...x].join`.*`))&&alert(s)}

Try it online!

Commented

a => {                    // a[] = input array of 3-character strings

  for(k = n = 1e5; n--;)  // initialize k to 100000; for n = 99999 to 0:

    a.every(x =>          // for each string x = 'XYZ' in a[]:

      ( s =               //   define s as the concatenation of

          ([k] + n)       //   '100000' and n; e.g. '100000' + 1337 -> '1000001337'

          .slice(-5)      //   keep the last 5 digits; e.g. '01337'

      ).match(            //   test whether this string is matching

        [...x].join`.*`   //   the pattern /X.*Y.*Z/

      )                   //

    ) &&                  // end of every(); if all tests were successful:

      alert(s)            //   output s

}                         //

edited 2 days ago

answered Feb 22 at 15:06

Arnauld

76.7k693322

add a comment |

JavaScript (ES6), 88 bytes

Prints the results with alert().

a=>{for(k=n=1e5;n--;)a.every(x=>(s=([k]+n).slice(-5)).match([...x].join`.*`))&&alert(s)}

Try it online!

Commented

a => {                    // a[] = input array of 3-character strings

  for(k = n = 1e5; n--;)  // initialize k to 100000; for n = 99999 to 0:

    a.every(x =>          // for each string x = 'XYZ' in a[]:

      ( s =               //   define s as the concatenation of

          ([k] + n)       //   '100000' and n; e.g. '100000' + 1337 -> '1000001337'

          .slice(-5)      //   keep the last 5 digits; e.g. '01337'

      ).match(            //   test whether this string is matching

        [...x].join`.*`   //   the pattern /X.*Y.*Z/

      )                   //

    ) &&                  // end of every(); if all tests were successful:

      alert(s)            //   output s

}                         //

edited 2 days ago

answered Feb 22 at 15:06

Arnauld

76.7k693322

JavaScript (ES6), 88 bytes

Prints the results with alert().

a=>{for(k=n=1e5;n--;)a.every(x=>(s=([k]+n).slice(-5)).match([...x].join`.*`))&&alert(s)}

Try it online!

Commented

a => {                    // a[] = input array of 3-character strings

  for(k = n = 1e5; n--;)  // initialize k to 100000; for n = 99999 to 0:

    a.every(x =>          // for each string x = 'XYZ' in a[]:

      ( s =               //   define s as the concatenation of

          ([k] + n)       //   '100000' and n; e.g. '100000' + 1337 -> '1000001337'

          .slice(-5)      //   keep the last 5 digits; e.g. '01337'

      ).match(            //   test whether this string is matching

        [...x].join`.*`   //   the pattern /X.*Y.*Z/

      )                   //

    ) &&                  // end of every(); if all tests were successful:

      alert(s)            //   output s

}                         //

edited 2 days ago

answered Feb 22 at 15:06

Arnauld

76.7k693322

edited 2 days ago

answered Feb 22 at 15:06

Arnauld

76.7k693322

answered Feb 22 at 15:06

Arnauld

76.7k693322

answered Feb 22 at 15:06

Arnauld

76.7k693322

add a comment |

Haskell, 81 80 78 76 bytes

f x=[p|p<-mapM(:['1'..'9'])"00000",all(`elem`(concat.words<$>mapM(:" ")p))x]

The obvious brute force approach in Haskell: built a list of all possible passwords and keep those where all elements from the input list are in respective list of subsequences.

Try it online!

Edit: -1 byte thanks to @xnor, -2 -4 bytes thanks to @H.PWiz

edited 15 hours ago

answered yesterday

nimi

31.9k32285

1

$begingroup$
Looks like you can compute the subseqs yourself a little shorter.
$endgroup$
– xnor
yesterday

1

$begingroup$
concat.words<$>mapM(:" ")p is shorter
$endgroup$
– H.PWiz
yesterday

2

$begingroup$
use p<-mapM(:['1'..'9'])"00000" to save 2 more bytes
$endgroup$
– H.PWiz
16 hours ago

add a comment |

Haskell, 81 80 78 76 bytes

f x=[p|p<-mapM(:['1'..'9'])"00000",all(`elem`(concat.words<$>mapM(:" ")p))x]

The obvious brute force approach in Haskell: built a list of all possible passwords and keep those where all elements from the input list are in respective list of subsequences.

Try it online!

Edit: -1 byte thanks to @xnor, -2 -4 bytes thanks to @H.PWiz

edited 15 hours ago

answered yesterday

nimi

31.9k32285

1

$begingroup$
Looks like you can compute the subseqs yourself a little shorter.
$endgroup$
– xnor
yesterday

1

$begingroup$
concat.words<$>mapM(:" ")p is shorter
$endgroup$
– H.PWiz
yesterday

2

$begingroup$
use p<-mapM(:['1'..'9'])"00000" to save 2 more bytes
$endgroup$
– H.PWiz
16 hours ago

add a comment |

Haskell, 81 80 78 76 bytes

f x=[p|p<-mapM(:['1'..'9'])"00000",all(`elem`(concat.words<$>mapM(:" ")p))x]

The obvious brute force approach in Haskell: built a list of all possible passwords and keep those where all elements from the input list are in respective list of subsequences.

Try it online!

Edit: -1 byte thanks to @xnor, -2 -4 bytes thanks to @H.PWiz

edited 15 hours ago

answered yesterday

nimi

31.9k32285

Haskell, 81 80 78 76 bytes

f x=[p|p<-mapM(:['1'..'9'])"00000",all(`elem`(concat.words<$>mapM(:" ")p))x]

The obvious brute force approach in Haskell: built a list of all possible passwords and keep those where all elements from the input list are in respective list of subsequences.

Try it online!

Edit: -1 byte thanks to @xnor, -2 -4 bytes thanks to @H.PWiz

edited 15 hours ago

answered yesterday

nimi

31.9k32285

edited 15 hours ago

answered yesterday

nimi

31.9k32285

answered yesterday

nimi

31.9k32285

answered yesterday

nimi

31.9k32285

1

$begingroup$
Looks like you can compute the subseqs yourself a little shorter.
$endgroup$
– xnor
yesterday

1

$begingroup$
concat.words<$>mapM(:" ")p is shorter
$endgroup$
– H.PWiz
yesterday

2

$begingroup$
use p<-mapM(:['1'..'9'])"00000" to save 2 more bytes
$endgroup$
– H.PWiz
16 hours ago

add a comment |

1

$begingroup$
Looks like you can compute the subseqs yourself a little shorter.
$endgroup$
– xnor
yesterday

1

$begingroup$
concat.words<$>mapM(:" ")p is shorter
$endgroup$
– H.PWiz
yesterday

2

$begingroup$
use p<-mapM(:['1'..'9'])"00000" to save 2 more bytes
$endgroup$
– H.PWiz
16 hours ago

Looks like you can compute the subseqs yourself a little shorter.

– xnor
yesterday

concat.words<$>mapM(:" ")p is shorter

– H.PWiz
yesterday

use p<-mapM(:['1'..'9'])"00000" to save 2 more bytes

– H.PWiz
16 hours ago

add a comment |

Jelly, 11 bytes

9Żṗ5ŒPiⱮẠɗƇ

Try it online!

answered 2 days ago

Erik the Outgolfer

32k429103

add a comment |

Jelly, 11 bytes

9Żṗ5ŒPiⱮẠɗƇ

Try it online!

answered 2 days ago

Erik the Outgolfer

32k429103

add a comment |

Jelly, 11 bytes

9Żṗ5ŒPiⱮẠɗƇ

Try it online!

answered 2 days ago

Erik the Outgolfer

32k429103

Jelly, 11 bytes

9Żṗ5ŒPiⱮẠɗƇ

Try it online!

answered 2 days ago

Erik the Outgolfer

32k429103

answered 2 days ago

Erik the Outgolfer

32k429103

answered 2 days ago

Erik the Outgolfer

32k429103

answered 2 days ago

Erik the Outgolfer

32k429103

add a comment |

Pyth, 11 bytes

f<QyT^s`MT5

Takes input as a set of strings.
Try it here

Explanation

f<QyT^s`MT5

      s`MT      Take the digits as a string.

     ^    5     Take the Cartesian product with itself 5 times.

f   T           Filter the ones...

 <Qy            ... where the input is a subset of the power set.

answered 2 days ago

Mnemonic

4,7851730

add a comment |

Pyth, 11 bytes

f<QyT^s`MT5

Takes input as a set of strings.
Try it here

Explanation

f<QyT^s`MT5

      s`MT      Take the digits as a string.

     ^    5     Take the Cartesian product with itself 5 times.

f   T           Filter the ones...

 <Qy            ... where the input is a subset of the power set.

answered 2 days ago

Mnemonic

4,7851730

add a comment |

Pyth, 11 bytes

f<QyT^s`MT5

Takes input as a set of strings.
Try it here

Explanation

f<QyT^s`MT5

      s`MT      Take the digits as a string.

     ^    5     Take the Cartesian product with itself 5 times.

f   T           Filter the ones...

 <Qy            ... where the input is a subset of the power set.

answered 2 days ago

Mnemonic

4,7851730

Pyth, 11 bytes

f<QyT^s`MT5

Takes input as a set of strings.
Try it here

Explanation

f<QyT^s`MT5

      s`MT      Take the digits as a string.

     ^    5     Take the Cartesian product with itself 5 times.

f   T           Filter the ones...

 <Qy            ... where the input is a subset of the power set.

answered 2 days ago

Mnemonic

4,7851730

answered 2 days ago

Mnemonic

4,7851730

answered 2 days ago

Mnemonic

4,7851730

answered 2 days ago

Mnemonic

4,7851730

add a comment |

Ruby, 54 bytes

->a{(?0*5..?9*5).select{|x|a.all?{|y|x=~/#{y*'.*'}/}}}

Try it online!

Takes input as an arrray of character arrays.

answered 2 days ago

Kirill L.

4,6151523

$begingroup$
Well done! You beat me by 25 bytes. Should I delete my answer?
$endgroup$
– Eric Duminil
2 days ago

1

$begingroup$
No, as long as you have a valid answer, there's no need to delete it.
$endgroup$
– Kirill L.
2 days ago

add a comment |

Ruby, 54 bytes

->a{(?0*5..?9*5).select{|x|a.all?{|y|x=~/#{y*'.*'}/}}}

Try it online!

Takes input as an arrray of character arrays.

answered 2 days ago

Kirill L.

4,6151523

$begingroup$
Well done! You beat me by 25 bytes. Should I delete my answer?
$endgroup$
– Eric Duminil
2 days ago

1

$begingroup$
No, as long as you have a valid answer, there's no need to delete it.
$endgroup$
– Kirill L.
2 days ago

add a comment |

Ruby, 54 bytes

->a{(?0*5..?9*5).select{|x|a.all?{|y|x=~/#{y*'.*'}/}}}

Try it online!

Takes input as an arrray of character arrays.

answered 2 days ago

Kirill L.

4,6151523

Ruby, 54 bytes

->a{(?0*5..?9*5).select{|x|a.all?{|y|x=~/#{y*'.*'}/}}}

Try it online!

Takes input as an arrray of character arrays.

answered 2 days ago

Kirill L.

4,6151523

answered 2 days ago

Kirill L.

4,6151523

answered 2 days ago

Kirill L.

4,6151523

answered 2 days ago

Kirill L.

4,6151523

$begingroup$
Well done! You beat me by 25 bytes. Should I delete my answer?
$endgroup$
– Eric Duminil
2 days ago

1

$begingroup$
No, as long as you have a valid answer, there's no need to delete it.
$endgroup$
– Kirill L.
2 days ago

add a comment |

$begingroup$
Well done! You beat me by 25 bytes. Should I delete my answer?
$endgroup$
– Eric Duminil
2 days ago

1

$begingroup$
No, as long as you have a valid answer, there's no need to delete it.
$endgroup$
– Kirill L.
2 days ago

Well done! You beat me by 25 bytes. Should I delete my answer?

– Eric Duminil
2 days ago

No, as long as you have a valid answer, there's no need to delete it.

– Kirill L.
2 days ago

add a comment |

Python 3, 98 bytes

f=lambda l,s='':any(l)or print(s)if s[4:]else[f([x[x[:1]==c:]for x in l],s+c)for c in'0123456789']

Try it online!

Recursively tries building every five-digit number string in s, tracking the subsequences in l still remaining to be hit. If all are empty by the end, prints the result.

Python 3.8 (pre-release), 94 bytes

lambda l:[s for n in range(10**5)if all(''in[x:=x[x[:1]==c:]for c in(s:='%05d'%n)]for x in l)]

Try it online!

Behold the power of assignment expressions!. Uses the method from here for checking subsequences.

edited yesterday

answered yesterday

xnor

91.1k18186442

add a comment |

Python 3, 98 bytes

f=lambda l,s='':any(l)or print(s)if s[4:]else[f([x[x[:1]==c:]for x in l],s+c)for c in'0123456789']

Try it online!

Recursively tries building every five-digit number string in s, tracking the subsequences in l still remaining to be hit. If all are empty by the end, prints the result.

Python 3.8 (pre-release), 94 bytes

lambda l:[s for n in range(10**5)if all(''in[x:=x[x[:1]==c:]for c in(s:='%05d'%n)]for x in l)]

Try it online!

Behold the power of assignment expressions!. Uses the method from here for checking subsequences.

edited yesterday

answered yesterday

xnor

91.1k18186442

add a comment |

Python 3, 98 bytes

f=lambda l,s='':any(l)or print(s)if s[4:]else[f([x[x[:1]==c:]for x in l],s+c)for c in'0123456789']

Try it online!

Recursively tries building every five-digit number string in s, tracking the subsequences in l still remaining to be hit. If all are empty by the end, prints the result.

Python 3.8 (pre-release), 94 bytes

lambda l:[s for n in range(10**5)if all(''in[x:=x[x[:1]==c:]for c in(s:='%05d'%n)]for x in l)]

Try it online!

Behold the power of assignment expressions!. Uses the method from here for checking subsequences.

edited yesterday

answered yesterday

xnor

91.1k18186442

Python 3, 98 bytes

f=lambda l,s='':any(l)or print(s)if s[4:]else[f([x[x[:1]==c:]for x in l],s+c)for c in'0123456789']

Try it online!

Recursively tries building every five-digit number string in s, tracking the subsequences in l still remaining to be hit. If all are empty by the end, prints the result.

Python 3.8 (pre-release), 94 bytes

lambda l:[s for n in range(10**5)if all(''in[x:=x[x[:1]==c:]for c in(s:='%05d'%n)]for x in l)]

Try it online!

Behold the power of assignment expressions!. Uses the method from here for checking subsequences.

edited yesterday

answered yesterday

xnor

91.1k18186442

edited yesterday

answered yesterday

xnor

91.1k18186442

answered yesterday

xnor

91.1k18186442

answered yesterday

xnor

91.1k18186442

add a comment |

Perl 5 `-a`, 80 bytes

map{s||($t=0)x(5-y///c)|e;for$b(map s//.*/gr,@F){/$b/&&$t++}$t==@F&&say}0..99999

Try it online!

answered 2 days ago

Xcali

5,375520

1

$begingroup$
$t+=/$b/ instead of /$b/&&$t++
$endgroup$
– Nahuel Fouilleul
2 days ago

add a comment |

Perl 5 `-a`, 80 bytes

map{s||($t=0)x(5-y///c)|e;for$b(map s//.*/gr,@F){/$b/&&$t++}$t==@F&&say}0..99999

Try it online!

answered 2 days ago

Xcali

5,375520

1

$begingroup$
$t+=/$b/ instead of /$b/&&$t++
$endgroup$
– Nahuel Fouilleul
2 days ago

add a comment |

Perl 5 `-a`, 80 bytes

map{s||($t=0)x(5-y///c)|e;for$b(map s//.*/gr,@F){/$b/&&$t++}$t==@F&&say}0..99999

Try it online!

answered 2 days ago

Xcali

5,375520

Perl 5 `-a`, 80 bytes

map{s||($t=0)x(5-y///c)|e;for$b(map s//.*/gr,@F){/$b/&&$t++}$t==@F&&say}0..99999

Try it online!

answered 2 days ago

Xcali

5,375520

answered 2 days ago

Xcali

5,375520

answered 2 days ago

Xcali

5,375520

answered 2 days ago

Xcali

5,375520

1

$begingroup$
$t+=/$b/ instead of /$b/&&$t++
$endgroup$
– Nahuel Fouilleul
2 days ago

add a comment |

1

$begingroup$
$t+=/$b/ instead of /$b/&&$t++
$endgroup$
– Nahuel Fouilleul
2 days ago

$t+=/$b/ instead of /$b/&&$t++

– Nahuel Fouilleul
2 days ago

add a comment |

Retina, 53 bytes

~(`

.$*

m`^

G`

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Try it online! Explanation:

~(`

After executing the script, take the result as a new script, and execute that, too.

.$*

Insert .* everywhere. This results in .*3.*2.*0.* although we only need 3.*2.*0, not that it matters.

m`^

G`

Insert a G` at the start of each line. This turns it into a Retina Grep command.

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Prefix two more Retina commands. The resulting script will therefore look something like this:

K`

Clear the buffer (which contains the original input).

5+

Repeat 5 times...

%`$

... append to each line...

0$"1$"2$"3$"4$"5$"6$"7$"8$"9

... the digit 0, then a copy of the line, then the digit 1, etc. until 9. This means that after n loops you will have all n-digit numbers.

G`.*3.*2.*0.*

G`.*7.*2.*3.*

G`.*7.*3.*0.*

Filter out the possible numbers based on the input.

answered 2 days ago

Neil

81k744178

add a comment |

Retina, 53 bytes

~(`

.$*

m`^

G`

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Try it online! Explanation:

~(`

After executing the script, take the result as a new script, and execute that, too.

.$*

Insert .* everywhere. This results in .*3.*2.*0.* although we only need 3.*2.*0, not that it matters.

m`^

G`

Insert a G` at the start of each line. This turns it into a Retina Grep command.

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Prefix two more Retina commands. The resulting script will therefore look something like this:

K`

Clear the buffer (which contains the original input).

5+

Repeat 5 times...

%`$

... append to each line...

0$"1$"2$"3$"4$"5$"6$"7$"8$"9

... the digit 0, then a copy of the line, then the digit 1, etc. until 9. This means that after n loops you will have all n-digit numbers.

G`.*3.*2.*0.*

G`.*7.*2.*3.*

G`.*7.*3.*0.*

Filter out the possible numbers based on the input.

answered 2 days ago

Neil

81k744178

add a comment |

Retina, 53 bytes

~(`

.$*

m`^

G`

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Try it online! Explanation:

~(`

After executing the script, take the result as a new script, and execute that, too.

.$*

Insert .* everywhere. This results in .*3.*2.*0.* although we only need 3.*2.*0, not that it matters.

m`^

G`

Insert a G` at the start of each line. This turns it into a Retina Grep command.

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Prefix two more Retina commands. The resulting script will therefore look something like this:

K`

Clear the buffer (which contains the original input).

5+

Repeat 5 times...

%`$

... append to each line...

0$"1$"2$"3$"4$"5$"6$"7$"8$"9

... the digit 0, then a copy of the line, then the digit 1, etc. until 9. This means that after n loops you will have all n-digit numbers.

G`.*3.*2.*0.*

G`.*7.*2.*3.*

G`.*7.*3.*0.*

Filter out the possible numbers based on the input.

answered 2 days ago

Neil

81k744178

Retina, 53 bytes

~(`

.$*

m`^

G`

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Try it online! Explanation:

~(`

After executing the script, take the result as a new script, and execute that, too.

.$*

Insert .* everywhere. This results in .*3.*2.*0.* although we only need 3.*2.*0, not that it matters.

m`^

G`

Insert a G` at the start of each line. This turns it into a Retina Grep command.

^

K`¶5+%`$$¶0"1"2"3"4"5"6"7"8"9¶

"

$$"

Prefix two more Retina commands. The resulting script will therefore look something like this:

K`

Clear the buffer (which contains the original input).

5+

Repeat 5 times...

%`$

... append to each line...

0$"1$"2$"3$"4$"5$"6$"7$"8$"9

... the digit 0, then a copy of the line, then the digit 1, etc. until 9. This means that after n loops you will have all n-digit numbers.

G`.*3.*2.*0.*

G`.*7.*2.*3.*

G`.*7.*3.*0.*

Filter out the possible numbers based on the input.

answered 2 days ago

Neil

81k744178

answered 2 days ago

Neil

81k744178

answered 2 days ago

Neil

81k744178

answered 2 days ago

Neil

81k744178

add a comment |

R, 80 82 bytes

Reduce(intersect,lapply(gsub("",".*",scan(,"")),grep,sprintf("%05d",0:99999),v=T))

Here’s a base R solution using regex. Writing this nested series of functions made me realise how much I’ve learned to appreciate the magrittr package!

Initially hadn’t read rules on input, so now reads from stdin (thanks @KirillL).

Try it online!

edited 2 days ago

answered 2 days ago

Nick Kennedy

21114

$begingroup$
@digEmAll it says it’s 82 bytes doesn’t it? You can’t use integers because of potential for leading zeroes in the input.
$endgroup$
– Nick Kennedy
2 days ago

$begingroup$
Sorry, I read the title and unconsciously I picked the minimum number without noticing that it was strikedthrough... and yes, sorry again you're right about the string input ;)
$endgroup$
– digEmAll
2 days ago

add a comment |

R, 80 82 bytes

Reduce(intersect,lapply(gsub("",".*",scan(,"")),grep,sprintf("%05d",0:99999),v=T))

Here’s a base R solution using regex. Writing this nested series of functions made me realise how much I’ve learned to appreciate the magrittr package!

Initially hadn’t read rules on input, so now reads from stdin (thanks @KirillL).

Try it online!

edited 2 days ago

answered 2 days ago

Nick Kennedy

21114

$begingroup$
@digEmAll it says it’s 82 bytes doesn’t it? You can’t use integers because of potential for leading zeroes in the input.
$endgroup$
– Nick Kennedy
2 days ago

$begingroup$
Sorry, I read the title and unconsciously I picked the minimum number without noticing that it was strikedthrough... and yes, sorry again you're right about the string input ;)
$endgroup$
– digEmAll
2 days ago

add a comment |

R, 80 82 bytes

Reduce(intersect,lapply(gsub("",".*",scan(,"")),grep,sprintf("%05d",0:99999),v=T))

Here’s a base R solution using regex. Writing this nested series of functions made me realise how much I’ve learned to appreciate the magrittr package!

Initially hadn’t read rules on input, so now reads from stdin (thanks @KirillL).

Try it online!

edited 2 days ago

answered 2 days ago

Nick Kennedy

21114

R, 80 82 bytes

Reduce(intersect,lapply(gsub("",".*",scan(,"")),grep,sprintf("%05d",0:99999),v=T))

Here’s a base R solution using regex. Writing this nested series of functions made me realise how much I’ve learned to appreciate the magrittr package!

Initially hadn’t read rules on input, so now reads from stdin (thanks @KirillL).

Try it online!

edited 2 days ago

answered 2 days ago

Nick Kennedy

21114

edited 2 days ago

answered 2 days ago

Nick Kennedy

21114

answered 2 days ago

Nick Kennedy

21114

answered 2 days ago

Nick Kennedy

21114

$begingroup$
@digEmAll it says it’s 82 bytes doesn’t it? You can’t use integers because of potential for leading zeroes in the input.
$endgroup$
– Nick Kennedy
2 days ago

$begingroup$
Sorry, I read the title and unconsciously I picked the minimum number without noticing that it was strikedthrough... and yes, sorry again you're right about the string input ;)
$endgroup$
– digEmAll
2 days ago

add a comment |

$begingroup$
@digEmAll it says it’s 82 bytes doesn’t it? You can’t use integers because of potential for leading zeroes in the input.
$endgroup$
– Nick Kennedy
2 days ago

$begingroup$
Sorry, I read the title and unconsciously I picked the minimum number without noticing that it was strikedthrough... and yes, sorry again you're right about the string input ;)
$endgroup$
– digEmAll
2 days ago

@digEmAll it says it’s 82 bytes doesn’t it? You can’t use integers because of potential for leading zeroes in the input.

– Nick Kennedy
2 days ago

Sorry, I read the title and unconsciously I picked the minimum number without noticing that it was strikedthrough... and yes, sorry again you're right about the string input ;)

– digEmAll
2 days ago

add a comment |

Ruby, 79 77 bytes

->x{(0...1e5).map{|i|'%05d'%i}.select{|i|x.all?{|c|i=~/#{c.gsub('','.*')}/}}}

Try it online!

Input is an array of strings.

Here's a more readable version of the same code:

def f(codes)

  (0...10**5).map{|i| '%05d'%i}.select do |i|

    codes.all? do |code|

      i =~ Regexp.new(code.chars.join('.*'))

    end

  end

end

edited 2 days ago

answered 2 days ago

Eric Duminil

43128

$begingroup$
BTW, your approach can also be made shorter by switching to char array input, as in my version, and -2 more bytes by formatting the upper value as 1e5, like this
$endgroup$
– Kirill L.
2 days ago

$begingroup$
@KirillL. Thanks for the -2 bytes. I won't change the input format because my answer would look too much like yours. Cheers!
$endgroup$
– Eric Duminil
2 days ago

add a comment |

Ruby, 79 77 bytes

->x{(0...1e5).map{|i|'%05d'%i}.select{|i|x.all?{|c|i=~/#{c.gsub('','.*')}/}}}

Try it online!

Input is an array of strings.

Here's a more readable version of the same code:

def f(codes)

  (0...10**5).map{|i| '%05d'%i}.select do |i|

    codes.all? do |code|

      i =~ Regexp.new(code.chars.join('.*'))

    end

  end

end

edited 2 days ago

answered 2 days ago

Eric Duminil

43128

$begingroup$
BTW, your approach can also be made shorter by switching to char array input, as in my version, and -2 more bytes by formatting the upper value as 1e5, like this
$endgroup$
– Kirill L.
2 days ago

$begingroup$
@KirillL. Thanks for the -2 bytes. I won't change the input format because my answer would look too much like yours. Cheers!
$endgroup$
– Eric Duminil
2 days ago

add a comment |

Ruby, 79 77 bytes

->x{(0...1e5).map{|i|'%05d'%i}.select{|i|x.all?{|c|i=~/#{c.gsub('','.*')}/}}}

Try it online!

Input is an array of strings.

Here's a more readable version of the same code:

def f(codes)

  (0...10**5).map{|i| '%05d'%i}.select do |i|

    codes.all? do |code|

      i =~ Regexp.new(code.chars.join('.*'))

    end

  end

end

edited 2 days ago

answered 2 days ago

Eric Duminil

43128

Ruby, 79 77 bytes

->x{(0...1e5).map{|i|'%05d'%i}.select{|i|x.all?{|c|i=~/#{c.gsub('','.*')}/}}}

Try it online!

Input is an array of strings.

Here's a more readable version of the same code:

def f(codes)

  (0...10**5).map{|i| '%05d'%i}.select do |i|

    codes.all? do |code|

      i =~ Regexp.new(code.chars.join('.*'))

    end

  end

end

edited 2 days ago

answered 2 days ago

Eric Duminil

43128

edited 2 days ago

answered 2 days ago

Eric Duminil

43128

answered 2 days ago

Eric Duminil

43128

answered 2 days ago

Eric Duminil

43128

$begingroup$
BTW, your approach can also be made shorter by switching to char array input, as in my version, and -2 more bytes by formatting the upper value as 1e5, like this
$endgroup$
– Kirill L.
2 days ago

$begingroup$
@KirillL. Thanks for the -2 bytes. I won't change the input format because my answer would look too much like yours. Cheers!
$endgroup$
– Eric Duminil
2 days ago

add a comment |

$begingroup$
BTW, your approach can also be made shorter by switching to char array input, as in my version, and -2 more bytes by formatting the upper value as 1e5, like this
$endgroup$
– Kirill L.
2 days ago

$begingroup$
@KirillL. Thanks for the -2 bytes. I won't change the input format because my answer would look too much like yours. Cheers!
$endgroup$
– Eric Duminil
2 days ago

BTW, your approach can also be made shorter by switching to char array input, as in my version, and -2 more bytes by formatting the upper value as 1e5, like this

– Kirill L.
2 days ago

@KirillL. Thanks for the -2 bytes. I won't change the input format because my answer would look too much like yours. Cheers!

– Eric Duminil
2 days ago

add a comment |

J, 52 bytes

(>,{;^:4~i.10)([#~]*/@e."2((#~3=+/"1)#:i.32)#"_ 1[)]

Try it online!

edited yesterday

answered 2 days ago

Jonah

2,331916

add a comment |

J, 52 bytes

(>,{;^:4~i.10)([#~]*/@e."2((#~3=+/"1)#:i.32)#"_ 1[)]

Try it online!

edited yesterday

answered 2 days ago

Jonah

2,331916

add a comment |

J, 52 bytes

(>,{;^:4~i.10)([#~]*/@e."2((#~3=+/"1)#:i.32)#"_ 1[)]

Try it online!

edited yesterday

answered 2 days ago

Jonah

2,331916

J, 52 bytes

(>,{;^:4~i.10)([#~]*/@e."2((#~3=+/"1)#:i.32)#"_ 1[)]

Try it online!

edited yesterday

answered 2 days ago

Jonah

2,331916

edited yesterday

answered 2 days ago

Jonah

2,331916

answered 2 days ago

Jonah

2,331916

answered 2 days ago

Jonah

2,331916

add a comment |

C# (Visual C# Interactive Compiler), 111 bytes

x=>{for(int i=0;i<1e5;){var s=$"{i++:D5}";if(x.All(t=>t.Aggregate(0,(a,c)=>a<0?a:s.IndexOf(c,a))>0))Print(s);}}

Try it online!

// x: input list of strings

x=>{

  // generate all numbers under 100k

  for(int i=0;i<1e5;){

    // convert the current number to

    // a 5 digit string padded with 0's

    var s=$"{i++:D5}";

    // test all inputs against the 5 digit

    // string using an aggregate.

    // each step of the aggregate gets

    // the index of the next occurrence

    // of the current character starting

    // at the previously found character.

    // a negative index indicates error.

    if(x.All(t=>t

             .Aggregate(0,(a,c)=>

               a<0

                 ?a

                 :s.IndexOf(c,a))>0))

      // output to STDOUT

      Print(s);

  }

}

edited yesterday

answered yesterday

dana

1,171166

add a comment |

C# (Visual C# Interactive Compiler), 111 bytes

x=>{for(int i=0;i<1e5;){var s=$"{i++:D5}";if(x.All(t=>t.Aggregate(0,(a,c)=>a<0?a:s.IndexOf(c,a))>0))Print(s);}}

Try it online!

// x: input list of strings

x=>{

  // generate all numbers under 100k

  for(int i=0;i<1e5;){

    // convert the current number to

    // a 5 digit string padded with 0's

    var s=$"{i++:D5}";

    // test all inputs against the 5 digit

    // string using an aggregate.

    // each step of the aggregate gets

    // the index of the next occurrence

    // of the current character starting

    // at the previously found character.

    // a negative index indicates error.

    if(x.All(t=>t

             .Aggregate(0,(a,c)=>

               a<0

                 ?a

                 :s.IndexOf(c,a))>0))

      // output to STDOUT

      Print(s);

  }

}

edited yesterday

answered yesterday

dana

1,171166

add a comment |

C# (Visual C# Interactive Compiler), 111 bytes

x=>{for(int i=0;i<1e5;){var s=$"{i++:D5}";if(x.All(t=>t.Aggregate(0,(a,c)=>a<0?a:s.IndexOf(c,a))>0))Print(s);}}

Try it online!

// x: input list of strings

x=>{

  // generate all numbers under 100k

  for(int i=0;i<1e5;){

    // convert the current number to

    // a 5 digit string padded with 0's

    var s=$"{i++:D5}";

    // test all inputs against the 5 digit

    // string using an aggregate.

    // each step of the aggregate gets

    // the index of the next occurrence

    // of the current character starting

    // at the previously found character.

    // a negative index indicates error.

    if(x.All(t=>t

             .Aggregate(0,(a,c)=>

               a<0

                 ?a

                 :s.IndexOf(c,a))>0))

      // output to STDOUT

      Print(s);

  }

}

edited yesterday

answered yesterday

dana

1,171166

C# (Visual C# Interactive Compiler), 111 bytes

x=>{for(int i=0;i<1e5;){var s=$"{i++:D5}";if(x.All(t=>t.Aggregate(0,(a,c)=>a<0?a:s.IndexOf(c,a))>0))Print(s);}}

Try it online!

// x: input list of strings

x=>{

  // generate all numbers under 100k

  for(int i=0;i<1e5;){

    // convert the current number to

    // a 5 digit string padded with 0's

    var s=$"{i++:D5}";

    // test all inputs against the 5 digit

    // string using an aggregate.

    // each step of the aggregate gets

    // the index of the next occurrence

    // of the current character starting

    // at the previously found character.

    // a negative index indicates error.

    if(x.All(t=>t

             .Aggregate(0,(a,c)=>

               a<0

                 ?a

                 :s.IndexOf(c,a))>0))

      // output to STDOUT

      Print(s);

  }

}

edited yesterday

answered yesterday

dana

1,171166

edited yesterday

answered yesterday

dana

1,171166

answered yesterday

dana

1,171166

answered yesterday

dana

1,171166

add a comment |

Japt, 21 bytes

1e5o ù'0 f@e_XèZË+".*

Try it!

1e5o ù'0 f@e_XèZË+".*    # full program



1e5o                     # generate numbers under 100k

     ù'0                 # left pad with 0's

         f@              # filter array

           e_            # check every element of input array

             Xè          # X is the number to be tested.

                         # test it against a regex.

               ZË+".*    # the regex is an element from the input array

                         # with wildcards injected between each character

-2 bytes thanks to @Shaggy!

edited 1 hour ago

answered 17 hours ago

dana

1,171166

$begingroup$
less useless variables: 1e5o ù'0 fA{Ue@AèX®+".* :P
$endgroup$
– ASCII-only
11 hours ago

$begingroup$
also 23: 1e5o ù'0 fA{Ue@AèX¬q".*
$endgroup$
– ASCII-only
10 hours ago

$begingroup$
21 bytes
$endgroup$
– Shaggy
5 hours ago

$begingroup$
Interesting... I didn't realize return X,Y,Z would pick the last term. Thanks for the tips :)
$endgroup$
– dana
2 hours ago

$begingroup$
@dana; Yup, that's a feature of JavaScript: tio.run/##y0osSyxOLsosKNHNy09J/Z9m@1@jQqdS09auuii1pLQoTwHIq/…
$endgroup$
– Shaggy
6 mins ago

add a comment |

Japt, 21 bytes

1e5o ù'0 f@e_XèZË+".*

Try it!

1e5o ù'0 f@e_XèZË+".*    # full program



1e5o                     # generate numbers under 100k

     ù'0                 # left pad with 0's

         f@              # filter array

           e_            # check every element of input array

             Xè          # X is the number to be tested.

                         # test it against a regex.

               ZË+".*    # the regex is an element from the input array

                         # with wildcards injected between each character

-2 bytes thanks to @Shaggy!

edited 1 hour ago

answered 17 hours ago

dana

1,171166

$begingroup$
less useless variables: 1e5o ù'0 fA{Ue@AèX®+".* :P
$endgroup$
– ASCII-only
11 hours ago

$begingroup$
also 23: 1e5o ù'0 fA{Ue@AèX¬q".*
$endgroup$
– ASCII-only
10 hours ago

$begingroup$
21 bytes
$endgroup$
– Shaggy
5 hours ago

$begingroup$
Interesting... I didn't realize return X,Y,Z would pick the last term. Thanks for the tips :)
$endgroup$
– dana
2 hours ago

$begingroup$
@dana; Yup, that's a feature of JavaScript: tio.run/##y0osSyxOLsosKNHNy09J/Z9m@1@jQqdS09auuii1pLQoTwHIq/…
$endgroup$
– Shaggy
6 mins ago

add a comment |

Japt, 21 bytes

1e5o ù'0 f@e_XèZË+".*

Try it!

1e5o ù'0 f@e_XèZË+".*    # full program



1e5o                     # generate numbers under 100k

     ù'0                 # left pad with 0's

         f@              # filter array

           e_            # check every element of input array

             Xè          # X is the number to be tested.

                         # test it against a regex.

               ZË+".*    # the regex is an element from the input array

                         # with wildcards injected between each character

-2 bytes thanks to @Shaggy!

edited 1 hour ago

answered 17 hours ago

dana

1,171166

Japt, 21 bytes

1e5o ù'0 f@e_XèZË+".*

Try it!

1e5o ù'0 f@e_XèZË+".*    # full program



1e5o                     # generate numbers under 100k

     ù'0                 # left pad with 0's

         f@              # filter array

           e_            # check every element of input array

             Xè          # X is the number to be tested.

                         # test it against a regex.

               ZË+".*    # the regex is an element from the input array

                         # with wildcards injected between each character

-2 bytes thanks to @Shaggy!

edited 1 hour ago

answered 17 hours ago

dana

1,171166

edited 1 hour ago

answered 17 hours ago

dana

1,171166

answered 17 hours ago

dana

1,171166

answered 17 hours ago

dana

1,171166

$begingroup$
less useless variables: 1e5o ù'0 fA{Ue@AèX®+".* :P
$endgroup$
– ASCII-only
11 hours ago

$begingroup$
also 23: 1e5o ù'0 fA{Ue@AèX¬q".*
$endgroup$
– ASCII-only
10 hours ago

$begingroup$
21 bytes
$endgroup$
– Shaggy
5 hours ago

$begingroup$
Interesting... I didn't realize return X,Y,Z would pick the last term. Thanks for the tips :)
$endgroup$
– dana
2 hours ago

$begingroup$
@dana; Yup, that's a feature of JavaScript: tio.run/##y0osSyxOLsosKNHNy09J/Z9m@1@jQqdS09auuii1pLQoTwHIq/…
$endgroup$
– Shaggy
6 mins ago

add a comment |

$begingroup$
less useless variables: 1e5o ù'0 fA{Ue@AèX®+".* :P
$endgroup$
– ASCII-only
11 hours ago

$begingroup$
also 23: 1e5o ù'0 fA{Ue@AèX¬q".*
$endgroup$
– ASCII-only
10 hours ago

$begingroup$
21 bytes
$endgroup$
– Shaggy
5 hours ago

$begingroup$
Interesting... I didn't realize return X,Y,Z would pick the last term. Thanks for the tips :)
$endgroup$
– dana
2 hours ago

$begingroup$
@dana; Yup, that's a feature of JavaScript: tio.run/##y0osSyxOLsosKNHNy09J/Z9m@1@jQqdS09auuii1pLQoTwHIq/…
$endgroup$
– Shaggy
6 mins ago

less useless variables: 1e5o ù'0 fA{Ue@AèX®+".* :P

– ASCII-only
11 hours ago

also 23: 1e5o ù'0 fA{Ue@AèX¬q".*

– ASCII-only
10 hours ago

21 bytes

– Shaggy
5 hours ago

Interesting... I didn't realize return X,Y,Z would pick the last term. Thanks for the tips :)

– dana
2 hours ago

@dana; Yup, that's a feature of JavaScript: tio.run/##y0osSyxOLsosKNHNy09J/Z9m@1@jQqdS09auuii1pLQoTwHIq/…

– Shaggy
6 mins ago

add a comment |

PHP 128 bytes

for(;$i++<1e5;$k>$argc||print$s)for($k=0;$n=$argv[++$k];)preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i-1))||$k=$argc;

for(;$i<1e5;$i+=$k<$argc||print$s)for($k=0;$n=$argv[++$k];)if(!preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i)))break;

take input from command line arguments. Run with -nr or try them online.

answered 2 days ago

Titus

13.2k11238

add a comment |

PHP 128 bytes

for(;$i++<1e5;$k>$argc||print$s)for($k=0;$n=$argv[++$k];)preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i-1))||$k=$argc;

for(;$i<1e5;$i+=$k<$argc||print$s)for($k=0;$n=$argv[++$k];)if(!preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i)))break;

take input from command line arguments. Run with -nr or try them online.

answered 2 days ago

Titus

13.2k11238

add a comment |

PHP 128 bytes

for(;$i++<1e5;$k>$argc||print$s)for($k=0;$n=$argv[++$k];)preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i-1))||$k=$argc;

for(;$i<1e5;$i+=$k<$argc||print$s)for($k=0;$n=$argv[++$k];)if(!preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i)))break;

take input from command line arguments. Run with -nr or try them online.

answered 2 days ago

Titus

13.2k11238

PHP 128 bytes

for(;$i++<1e5;$k>$argc||print$s)for($k=0;$n=$argv[++$k];)preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i-1))||$k=$argc;

for(;$i<1e5;$i+=$k<$argc||print$s)for($k=0;$n=$argv[++$k];)if(!preg_match("/$n[0].*$n[1].*$n[2]/",$s=sprintf("%05d

",$i)))break;

take input from command line arguments. Run with -nr or try them online.

answered 2 days ago

Titus

13.2k11238

answered 2 days ago

Titus

13.2k11238

answered 2 days ago

Titus

13.2k11238

answered 2 days ago

Titus

13.2k11238

add a comment |

Clean, 113 bytes

import StdEnv,Data.List

$l=[s\s<-iter 5(p=[[c:e]\e<-p,c<-['0'..'9']])[[]]|all(flip any(subsequences s)o(==))l]

Try it online!

answered 16 hours ago

Οurous

7,20111035

add a comment |

Clean, 113 bytes

import StdEnv,Data.List

$l=[s\s<-iter 5(p=[[c:e]\e<-p,c<-['0'..'9']])[[]]|all(flip any(subsequences s)o(==))l]

Try it online!

answered 16 hours ago

Οurous

7,20111035

add a comment |

Clean, 113 bytes

import StdEnv,Data.List

$l=[s\s<-iter 5(p=[[c:e]\e<-p,c<-['0'..'9']])[[]]|all(flip any(subsequences s)o(==))l]

Try it online!

answered 16 hours ago

Οurous

7,20111035

Clean, 113 bytes

import StdEnv,Data.List

$l=[s\s<-iter 5(p=[[c:e]\e<-p,c<-['0'..'9']])[[]]|all(flip any(subsequences s)o(==))l]

Try it online!

answered 16 hours ago

Οurous

7,20111035

answered 16 hours ago

Οurous

7,20111035

answered 16 hours ago

Οurous

7,20111035

answered 16 hours ago

Οurous

7,20111035

add a comment |

cefel is a new contributor. Be nice, and check out our Code of Conduct.

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…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

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cefel is a new contributor. Be nice, and check out our Code of Conduct.

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

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This page is only for reference, If you need detailed information, please check here

Crack the bank account's password!Tips for golfing in PythonIs string X a subsequence of string Y?Random...

Introduction

Challenge

Input/Output examples

Rules

Introduction

Challenge

Input/Output examples

Rules

Introduction

Challenge

Input/Output examples

Rules

Introduction

Challenge

Input/Output examples

Rules

17 Answers 17

Python, 101 bytes

05AB1E, 11 9 bytes

JavaScript (ES6), 88 bytes

Commented

Haskell, 81 80 78 76 bytes

Jelly, 11 bytes

Pyth, 11 bytes

Explanation

Ruby, 54 bytes

Python 3, 98 bytes

Python 3.8 (pre-release), 94 bytes

Perl 5 -a, 80 bytes

Retina, 53 bytes

R, 80 82 bytes

Ruby, 79 77 bytes

J, 52 bytes

C# (Visual C# Interactive Compiler), 111 bytes

Japt, 21 bytes

PHP 128 bytes

Clean, 113 bytes

Your Answer

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17 Answers 17

17 Answers 17

Python, 101 bytes

Python, 101 bytes

Python, 101 bytes

Python, 101 bytes

05AB1E, 11 9 bytes

05AB1E, 11 9 bytes

05AB1E, 11 9 bytes

05AB1E, 11 9 bytes

JavaScript (ES6), 88 bytes

Commented

JavaScript (ES6), 88 bytes

Commented

JavaScript (ES6), 88 bytes

Commented

JavaScript (ES6), 88 bytes

Commented

Haskell, 81 80 78 76 bytes

Haskell, 81 80 78 76 bytes

Haskell, 81 80 78 76 bytes

Haskell, 81 80 78 76 bytes

Jelly, 11 bytes

Jelly, 11 bytes

Jelly, 11 bytes

Jelly, 11 bytes

Pyth, 11 bytes

Explanation

Pyth, 11 bytes

Explanation

Pyth, 11 bytes

Explanation

Pyth, 11 bytes

Explanation

Ruby, 54 bytes

Ruby, 54 bytes

Ruby, 54 bytes

Ruby, 54 bytes

17 Answers
17

Perl 5 `-a`, 80 bytes

17 Answers
17

17 Answers
17

Perl 5 `-a`, 80 bytes

Perl 5 `-a`, 80 bytes

Perl 5 `-a`, 80 bytes

Perl 5 `-a`, 80 bytes