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One aspect of password strength testing is runs of adjacent letters on the keyboard. In this challenge, a program must be created that returns true if a string contains any runs of adjacent letters.

What counts as a run of adjacent letters?

For this simplified version of a password strength tester, a run of adjacent characters is 3 or more letters which are next to each other in a single direction (left, right, above or below) on a QWERTY keyboard. For the purpose of this challenge the layout of the keyboard looks like this:

1234567890
QWERTYUIOP
ASDFGHJKL
ZXCVBNM

In the diagram above Q is below 1 but not below 2, so a string that contains 1qa or aq1 anywhere inside it would make the program return true, but 2qa would not.

Input

The password string to check. It will only contain the characters [0-9a-z] or [0-9A-Z] (your choice).

Output

The program must return a truthy value if the password contains one or more runs of adjacent keys, or falsey if it contains none.

Examples

The following inputs should output true:

  • asd
  • ytrewq
  • ju7
  • abc6yhdef

And these inputs should output false:

  • abc
  • aaa
  • qewretry
  • zse
  • qwdfbn
  • pas

Rules

  • Answers may be complete programs or functions.
  • Standard loopholes are disallowed.
  • This is , lowest score (in bytes) wins!
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3
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Pyth - 66 62 60 bytes

Pretty straightforward approach. Checks if any of substrings len 3 are in any of the rotations of the keyboard. Will be using base encoding for keyboard.

.E}Rjb+J+Kc+jkS9"0
qwertyuiop
asdfghjkl
zxcvbnm"b.tKN_MJ.:z3

Test Suite.

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  • \$\begingroup\$ @user81655 fixed, and saved two bytes. \$\endgroup\$ – Maltysen Nov 27 '15 at 4:58
2
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Japt, 78 bytes

Japt is a shortened version of JavaScript. Interpreter

V=1oA ¬+`0\nqØÆyuiop\n?dfghjkl \nzxcvbnm`;1+¡Y©(((VbX -VbUgY-1)-5 a -5 %A a)bB

Outputs 0 for falsey cases; otherwise, a positive integer. The ? should be replaced with the unprintable Unicode char U+0086, or if you don't want to go to all that trouble, just as.

How it works

V=1oA q +"0\nqwertyuiop\nasdfghjkl \nzxcvbnm";1+Um@Y&&(((VbX -VbUgY-1)-5 a -5 %A a)bB
           // Implicit: U = input string
V=1oA q    // Set variable V to the digits 1-9, plus
+"...";    // this string.
Um@        // Take U and map each character X and its index Y with this function:
Y&&        //  If Y is 0, return Y; otherwise,
VbX -      //  take the index of X in V, subtract
VbUgY-1    //  the index of (char at position Y - 1 in U) in V,
-5 a -5    //  subtract 5, take the absolute value, subtract 5 again,
%A a       //  take modulo by 10, then take the absolute value.
           //  This returns 1 for any pair of characters that is adjacent
           //  within V, horizontally or vertically.
bB +1      // Take the index of 11 in the result and add one.
           // Implicit: output last expression
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2
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C#, 227

int h(string k){var q="1234567890,QWERTYUIOP,ASDFGHJKL,ZXCVBNM,1QAZ,2WSX,3EDC,4RFV,5TGB,6YHN,7UJM,8IK,9OL,";int i=0,j=0;for(;i<k.Length-2;i++)if((q+String.Concat(Enumerable.Reverse(q))).Contains(k.Substring(i,3)))j=1;return j;}

0 is falsey, 1 is truthy. Concatenated all keys horizontal and vertical, and reversed, and checks if any of 3 chars of input is contained within.

C# is really verbose, gotta dive into other languages :(

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0
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PHP, 173+1 bytes

while(~$argn[$i+2])($p=preg_match)($r=_.join(".{10}",str_split(($t=substr($argn,$i++,3))))."|$t"._,$d=_1234567890_qwertyuiop_asdfghjkl__zxcvbnm)||$p($r,strrev($d))?die(1):0;

Run as pipe with -nR with lowercase input or try it online.

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0
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Clojure, 156 bytes

#(some(set(for[R[["1234567890""QWERTYUIOP""ASDFGHJKL.""ZXCVBNM..."]]R[R(apply map list R)]r R p(partition 3 1 r)p((juxt seq reverse)p)]p))(partition 3 1 %))

This was a quite interesting task to implement.

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