Given an input of an all-lowercase string [a-z], output the total distance between the letters.


Input: golf

Distance from g to o : 8
Distance from o to l : 3
Distance from l to f : 6

Output: 17


  • Standard loopholes forbidden
  • This is - shortest answer in bytes wins.
  • The alphabet can be traversed from either direction. You must always use the shortest path. (i.e the distance between x and c is 5).


Test cases

Input: aa
Output: 0

Input: stack
Output: 18

Input: zaza
Output: 3

Input: valleys
Output: 35

23 Answers 23


Jelly, 11 8 bytes


Saved 3 bytes thanks to @Martin Ender.

Try it online! or Verify all test cases.


OIæ%13AS  Input: string Z
O         Ordinal. Convert each char in Z to its ASCII value
 I        Increments. Find the difference between each pair of values
  æ%13    Symmetric mod. Maps each to the interval (-13, 13]
      A   Absolute value of each
       S  Sum
          Return implicitly
  • 6
    \$\begingroup\$ I came across æ% while reading through the built-ins the other day, and it was pretty much made for this (type of) problem: OIæ%13AS \$\endgroup\$ Commented Sep 17, 2016 at 15:44
  • \$\begingroup\$ I think this is 9 bytes (æ is two). \$\endgroup\$ Commented Sep 18, 2016 at 9:48
  • 1
    \$\begingroup\$ @elmigranto Jelly has a codepage that encodes each of its characters in one byte: github.com/DennisMitchell/jelly/wiki/Code-page \$\endgroup\$
    – ruds
    Commented Sep 18, 2016 at 15:54

Haskell, 57 56 bytes

sum.q.q.(zipWith(-)=<<tail).map fromEnum

Usage example: sum.q.q.(zipWith(-)=<<tail).map fromEnum $ "valleys" -> 35.

How it works:

q=map$(-)13.abs                -- helper function.
                               -- Non-pointfree: q l = map (\e -> 13 - abs e) l
                               -- foreach element e in list l: subtract the
                               -- absolute value of e from 13

               map fromEnum    -- convert to ascii values
      zipWith(-)=<<tail        -- build differences of neighbor elements
  q.q                          -- apply q twice on every element
sum                            -- sum it up

Edit: @Damien saved one byte. Thanks!

  • \$\begingroup\$ thanks for the rotational distance trick (q.q) \$\endgroup\$ Commented Sep 17, 2016 at 20:18
  • \$\begingroup\$ Wow nice one! You can add map in the definition of q for one byte less \$\endgroup\$
    – Damien
    Commented Sep 21, 2016 at 11:21
  • \$\begingroup\$ @Damien: well spotted. Thanks! \$\endgroup\$
    – nimi
    Commented Sep 21, 2016 at 15:42

MATL, 14, 10 bytes


Try it online!

Thanks @Suever for saving 4 bytes!


d           % Take the difference between consecutive characters
 t_         % Make a copy of this array, and take the negative of each element
   v        % Join these two arrays together into a matrix with height 2
    26\     % Mod 26 of each element
       X<   % Grab the minimum of each column
         s  % Sum these. Implicitly print

Previous version:


Python 3, 69 68 bytes

lambda s:sum([13-abs(13-abs(ord(a)-ord(b)))for a,b in zip(s,s[1:])])


lambda s:
         sum(                                                      )
             [                             for a,b in zip(s,s[1:])]
  • 1
    \$\begingroup\$ You can lose one byte by removing the space before for \$\endgroup\$
    – Daniel
    Commented Sep 17, 2016 at 15:25
  • \$\begingroup\$ @Dopapp Oh yes, thanks! \$\endgroup\$
    – busukxuan
    Commented Sep 17, 2016 at 15:26
  • 2
    \$\begingroup\$ You could take the input as a list of characters and use recursion to save 3 bytes: f=lambda a,b,*s:13-abs(13-abs(ord(a)-ord(b)))+(s and f(b,*s)or 0) \$\endgroup\$ Commented Sep 17, 2016 at 20:46

Java, 126 120 117 bytes

int f(String s){byte[]z=s.getBytes();int r=0,i=0,e;for(;++i<z.length;r+=(e=(26+z[i]-z[i-1])%26)<14?e:26-e);return r;}

Thanks to @KevinCruijssen for pointing out a bug in the original version and suggesting to make the for-loop empty.

The use of (26 + z[i] - z[i - 1]) % 26) is inspired from a comment by @Neil on another answer. (26 + ...)%26 serves the same purpose as Math.abs(...) because of ...? e : 26 - e.


int f(String s) {
    byte[]z = s.getBytes();
    int r = 0, i = 0, e;
    for (; ++i < z.length; r += (e = (26 + z[i] - z[i - 1]) % 26) < 14 ? e : 26 - e);
    return r;
  • \$\begingroup\$ Welcome to the site! What language is this? How many characters/bytes is it? You should [edit] those details into the top of your post, with this markdown: #Language, n bytes` \$\endgroup\$
    – DJMcMayhem
    Commented Sep 17, 2016 at 19:12
  • \$\begingroup\$ Ok. Thanks. I've edited it. Any improvement? :) \$\endgroup\$
    – todeale
    Commented Sep 17, 2016 at 19:16
  • 1
    \$\begingroup\$ You're missing a - before an e in your ungolfed version. \$\endgroup\$
    – Neil
    Commented Sep 19, 2016 at 9:02
  • 2
    \$\begingroup\$ Welcome to PPCG! Hmm, I'm getting a "Type mismatch: cannot convert from int to byte" error at e=z[i]-z[i-1]; So you either need a cast to (byte) or change the e to int. Also, you can remove the for-loop brackets by placing everything inside the for-loop, like this: int f(String s){byte[]z=s.getBytes();int r=0,i=0,e;for(;++i<z.length;r+=(e=z[i]-z[i-1])>0?e<14?e:26-e:-e<14?-e:e+26);return r;} (PS: The reversed the for-loop is unfortunately the same length: int f(String s){byte[]z=s.getBytes();int r=0,i=z.length-1,e;for(;i>0;r+=(e=z[i]-z[--i])>0?e<14?e:26-e:-e<14?-e:e+26);return r;}. \$\endgroup\$ Commented Sep 20, 2016 at 13:55
  • 1
    \$\begingroup\$ Thanks a lot @KevinCruijssen :D. Your suggestion has helped a lot. \$\endgroup\$
    – todeale
    Commented Sep 20, 2016 at 15:00

JavaScript (ES6), 84 82 79 bytes

Saved 3 bytes thanks to Cyoce:



  [d,...s],                    //Destructured input, separates first char from the rest
  p=parseInt,                  //p used as parseInt
  v=(26+p(s[0],36)-p(d,36))%26 //v is the absolute value of the difference using base 36 to get number from char
  s[0]?                        //If there is at least two char in the input
    f(s)                       //sum recursive call
    +                          //added to
    (v>13?26-v:v)              //the current shortest path
  :                            //else
    0                          //ends the recursion, returns 0

Call: f('golf')
Output: 17

Previous solutions :

82 bytes thanks to Neil:


84 bytes:

  • 1
    \$\begingroup\$ Instead of Math.abs(...) you can use (26+...)%26; this works because you're flipping values above 13 anyway. (I think this is how the MATL answer works.) \$\endgroup\$
    – Neil
    Commented Sep 17, 2016 at 10:44
  • 1
    \$\begingroup\$ Save some bytes by prepending the code with p=parseInt; and then using p() instead of parseInt() \$\endgroup\$
    – Cyoce
    Commented Sep 20, 2016 at 15:21

Ruby, 73 bytes

->x{eval x.chars.each_cons(2).map{|a,b|13-(13-(a.ord-b.ord).abs).abs}*?+}

PHP, 93 Bytes


05AB1E, 12 bytes



SÇ                   # convert to list of ascii values
  ¥                  # take delta's
   YF    }           # 2 times do
     Ä5Ø-            # for x in list: abs(x) - 13
          (O         # negate and sum

Try it online!

  • \$\begingroup\$ It's 12 symbols, not bytes. Byte-length would be 16 for UTF-8. \$\endgroup\$ Commented Sep 18, 2016 at 9:49
  • \$\begingroup\$ @elmigranto: Indeed. In UTF-8 that would be the case, but 05AB1E uses CP-1252 where this is 12 bytes. \$\endgroup\$
    – Emigna
    Commented Sep 18, 2016 at 10:06

Perl, 46 bytes

Includes +3 for -p (code contains ')

Give input on STDIN without final newline:

echo -n zaza | stringd.pl


#!/usr/bin/perl -p
s%.%$\+=13-abs 13-abs ord($&)-ord$'.$&%eg}{

Racket 119 bytes

(λ(s)(for/sum((i(sub1(string-length s))))(abs(-(char->integer
(string-ref s i))(char->integer(string-ref s(+ 1 i)))))))


(f "golf")



Detailed version:

(define(f s)
  (for/sum((i(sub1(string-length s))))
    (abs(-(char->integer(string-ref s i))
          (char->integer(string-ref s(+ 1 i)))))))
  • \$\begingroup\$ You could replace (define(f s) with (lambda(s), 2 bytes shorter (anonymous functions are fine). \$\endgroup\$
    – fede s.
    Commented Sep 17, 2016 at 21:15
  • 1
    \$\begingroup\$ Wait, Racket should take (λ(s) too, which if in utf8 is 6 bytes i think \$\endgroup\$
    – fede s.
    Commented Sep 17, 2016 at 21:17

C#, 87 85 bytes

Improved solution - replaced Math.Abs() with the add & modulo trick to save 2 bytes:

s=>{int l=0,d,i=0;for(;i<s.Length-1;)l+=(d=(s[i]-s[++i]+26)%26)>13?26-d:d;return l;};

Initial solution:

s=>{int l=0,d,i=0;for(;i<s.Length-1;)l+=(d=Math.Abs(s[i]-s[++i]))>13?26-d:d;return l;};

Try it online!

Full source, including test cases:

using System;

namespace StringDistance
    class Program
        static void Main(string[] args)
            Func<string,int>f= s=>{int l=0,d,i=0;for(;i<s.Length-1;)l+=(d=Math.Abs(s[i]-s[++i]))>13?26-d:d;return l;};

            Console.WriteLine(f("golf"));   //17
            Console.WriteLine(f("aa"));     //0
            Console.WriteLine(f("stack"));  //18
            Console.WriteLine(f("zaza"));   //3

Actually, 21 bytes

Based partially on cia_rana's Ruby answer.

There was a bug with O (in this case, map ord() over a string) where it would not work with d (dequeue bottom element) and p (pop first element) without first converting the map to a list with #. This bug has been fixed, but as that fix is newer than this challenge, so I've kept # in.

Edit: And the byte count has been wrong since September. Whoops.

Golfing suggestions welcome. Try it online!



         Implicit input string.
          The string should already be enclosed in quotation marks.
O#       Map ord() over the string and convert the map to a list. Call it ords.
;        Duplicate ords.
dX       Dequeue the last element and discard it.
@        Swap the with the duplicate ords.
pX       Pop the last element and discard it. Stack: ords[:-1], ords[1:]
♀-       Subtract each element of the second list from each element of the first list.
          This subtraction is equivalent to getting the first differences of ords.
`...`M   Map the following function over the first differences. Variable i.
  A;       abs(i) and duplicate.
  úl       Push the lowercase alphabet and get its length. A golfy way to push 26.
  -        26-i
  k        Pop all elements from stack and convert to list. Stack: [i, 26-i]
  m        min([i, 26-i])
Σ        Sum the result of the map.
         Implicit return.

Java 7,128 bytes

 int f(String s){char[]c=s.toCharArray();int t=0;for(int i=1,a;i<c.length;a=Math.abs(c[i]-c[i++-1]),t+=26-a<a?26-a:a);return t;}


 int f(String s){
 int t=0;
 for(int i=1,a;
return t;

Pyth, 20 bytes


A program that takes input of a quoted string on STDIN and prints the result.

Try it online

How it works

Lm-13.adbsyy-M.:CMQ2  Program. Input: Q
L                     def y(b) ->
 m      b              Map over b with variable d:
  -13                   13-
     .ad                abs(d)
                CMQ   Map code-point over Q
              .:   2  All length 2 sublists of that
            -M        Map subtraction over that
          yy          y(y(that))
         s            Sum of that
                      Implicitly print

dc + od, 65 bytes

od -tuC|dc -e'?dsN0sT[lNrdsNr-d*vdD[26-]sS<Sd*vlT+sTd0<R]dsRxlTp'


Because in dc you can't access a string's characters, I used od to get the ASCII values. These will be processed in reverse order from the stack (LIFO container) like so:

dsN0sT             # initialize N (neighbor) = top ASCII value, and T (total) = 0
[lNrdsNr-          # loop 'R': calculate difference between current value and N,
                   #updating N (on the first iteration the difference is 0)
   d*vdD[26-]sS<S  # get absolute value (d*v), push 13 (D) and call 'S' to subtract
                   #26 if the difference is greater than 13
   d*vlT+sT        # get absolute value again and add it to T
d0<R]dsR           # repeat loop for the rest of the ASCII values
xlTp               # the main: call 'R' and print T at the end


echo -n "golf" | ./string_distance.sh



C, 82 86 83 76 bytes

t,u;f(char*s){for(t=0;*++s;u=*s-s[-1],t+=(u=u<0?-u:u)>13?26-u:u);return t;}

Assumes input string is at least one character long. This doesn't require #include<stdlib.h>

Edit: Argh, sequence points!

Try it on Ideone

  • \$\begingroup\$ in ideone compiler the string "nwlrbb" and all the rand string i try 6 len return all 0 but it seems not 0 the result.... \$\endgroup\$
    – user58988
    Commented Sep 23, 2016 at 8:56
  • \$\begingroup\$ yes now it seems ok... \$\endgroup\$
    – user58988
    Commented Sep 23, 2016 at 23:02

C, 70 bytes 76 bytes

k,i;f(char *s){for(i=0;*++s;i+=(k=abs(*s-s[-1]))>13?26-k:k);return i;}

Scala, 68 bytes

def f(s:String)=(for(i<-0 to s.length-2)yield (s(i)-s(i+1)).abs).sum

Criticism is welcome.


C#, 217 bytes


IEnumerable<int>g(string k){Func<Char,int>x=(c)=>int.Parse(""+Convert.ToByte(c))-97;for(int i=0;i<k.Length-1;i++){var f=x(k[i]);var s=x(k[i+1]);var d=Math.Abs(f-s);yield return d>13?26-Math.Max(f,s)+Math.Min(f,s):d;}}


IEnumerable<int> g(string k)
  Func<Char, int> x = (c) => int.Parse("" + Convert.ToByte(c)) - 97;
  for (int i = 0; i < k.Length - 1; i++)
    var f = x(k[i]);
    var s = x(k[i + 1]);
    var d = Math.Abs(f - s);
    yield return d > 13 ? 26 - Math.Max(f, s) + Math.Min(f, s) : d;


aa: 0
stack: 18
zaza: 3
valleys: 35

'a' is 97 when converted to bytes, so 97 is subtracted from each one. If the difference is greater than 13 (ie, half of the alphabet), then subtract the differences between each character (byte value) from 26. A last minute addition of "yield return" saved me a few bytes!

  • 1
    \$\begingroup\$ Two useless whitespaces: both before 's'. \$\endgroup\$
    – Yytsi
    Commented Oct 27, 2016 at 17:03

Python 3, 126 bytes

With list incomprehension.

print(sum([min(abs(x-y),x+26-y)for x,y in[map(lambda x:(ord(x)-97),sorted(d[i:i+2]))for i in range(len(d))][:-1]]))
  • \$\begingroup\$ Nice answer. You could replace abs(x-y) by y-x since the call to sorted make x < y. \$\endgroup\$
    – todeale
    Commented Sep 24, 2016 at 7:58

PHP, 79 bytes


Java, 109 bytes

int f(String s){int x=0,t,a=0;for(byte b:s.getBytes()){t=a>0?(a-b+26)%26:0;t=t>13?26-t:t;x+=t;a=b;}return x;

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