18
\$\begingroup\$

Based on this question from Code Review

Given a non-empty string of printable ASCII characters, output the second non-repeating character. For example, for input DEFD, output F.

Input

Output

  • The second character that doesn't repeat, when reading left-to-right, again in a suitable format.
  • The output character is case-insensitive.
  • If no such character exists (e.g., all characters repeat), output an empty string.

Rules

  • The algorithm should ignore case. That is, D and d count as the same character.
  • Either a full program or a function are acceptable.
  • The input string will be guaranteed non-empty (i.e., at least one character in length).
  • The input string is ASCII. Any valid character could repeat, not just alphanumeric (this includes spaces).
  • Standard loopholes are forbidden.
  • This is so all usual golfing rules apply, and the shortest code (in bytes) wins.

Examples

Input is on first line, output is on second line.

DEFD
F

FEED
D

This is an example input sentence.
x

...,,,..,,!@
@

ABCDefgHijklMNOPqrsTuVWxyz
B

AAAAAABBBBB


Thisxthis


This this.
.
\$\endgroup\$
  • 8
    \$\begingroup\$ If it wasn't case-insensitive, I'd consider doing it in Forth. String operations suck in that language, though. \$\endgroup\$ – mbomb007 Jul 1 '16 at 17:05
  • \$\begingroup\$ What if my language doesn't support lowercase letters? \$\endgroup\$ – Adám Jul 3 '16 at 10:32
  • \$\begingroup\$ @Adám Does it utilize a different code page? How would it normally input an ASCII string if it doesn't support lowercase letters? \$\endgroup\$ – AdmBorkBork Jul 4 '16 at 13:37
  • 1
    \$\begingroup\$ The system I had in mind had 7-bit code page; a modified standard code page where uppercase letters occupy the lowercase positions, and the uppercase positions were used for glyphs. This was done on old APL systems so that one could use Shift to access APL glyphs, while unshifted letters were classic coding-style capitals. \$\endgroup\$ – Adám Jul 4 '16 at 13:46

36 Answers 36

10
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MATL, 11 bytes

tk&=s1=)FT)

This exits with an error (allowed by default) if there is no second non-repeated character.

Try it online!

Explanation

t      % Implicitly take input string. Duplicate
k      % Convert to lowercase
&=     % 2D array of equality comparisons
s      % Sum of each column
1=     % True for entries that equal 1
)      % Apply logical index to the input string to keep non-repeated characters
TF)    % Apply logical index to take 2nd element if it exists. Implicitly display 
\$\endgroup\$
  • \$\begingroup\$ The ninja edit strikes again. :P \$\endgroup\$ – Dennis Jul 1 '16 at 16:27
  • \$\begingroup\$ @Dennis Hahaha. Well, I guess you'll remove a couple of bytes soon \$\endgroup\$ – Luis Mendo Jul 1 '16 at 17:22
10
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Retina, 25 bytes

i!2=`(.)(?<!\1.+)(?!.*\1)

Try it online! (The first line enables running the code on a test suite of several inputs.)

Explanation

This is just a single regex match, the regex being:

(.)(?<!\1.+)(?!.*\1)

That is, match a character and ensure it doesn't appear anywhere else in the input. The rest is configuration:

  • i activates case insensitivity.
  • ! tells Retina to print the matches as opposed to counting them.
  • 2= tells Retina to print only the second match as opposed to all of them.
\$\endgroup\$
  • 1
    \$\begingroup\$ Ah, thanks for teaching me about the 2=. \$\endgroup\$ – Leaky Nun Jul 1 '16 at 15:58
6
\$\begingroup\$

05AB1E, 15 12 bytes

l©v®y¢iy}}1@

Explained

l©            # store lower case string in register
  v     }     # for each char in lower case string
   ®y¢iy      # if it occurs once in string, push it to stack
         }    # end if
          1@  # push the 2nd element from stack and implicitly display

Try it online

Saved 3 bytes thanks to @Adnan

\$\endgroup\$
  • \$\begingroup\$ Or for 12 bytes l©v®y¢iy}}1@ :). \$\endgroup\$ – Adnan Jul 2 '16 at 0:37
  • \$\begingroup\$ @Adnan: Nice! Didn't think of using @. \$\endgroup\$ – Emigna Jul 2 '16 at 10:59
5
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Python 2, 59 58 bytes

Returns a list of a single character, or an empty list if no output. (Stupid case-insensitivity...)

s=input().lower();print[c for c in s if s.count(c)<2][1:2]

Try it online

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  • 1
    \$\begingroup\$ @mbomb007 meta.codegolf.stackexchange.com/a/7634 \$\endgroup\$ – Dennis Jul 1 '16 at 16:40
  • \$\begingroup\$ It's not valid input. The user should never have to escape their input. \$\endgroup\$ – mbomb007 Jul 1 '16 at 16:41
  • 4
    \$\begingroup\$ Sure it is. We provide lists on STDIN in the language's list format. Why would strings be any different? \$\endgroup\$ – Dennis Jul 1 '16 at 16:43
5
\$\begingroup\$

Jelly, 11 bytes

Œlµḟœ-Q$Ḋḣ1

Try it online! or verify all test cases.

How it works

Œlµḟœ-Q$Ḋḣ1  Main link. Argument: s (string)

Œl           Convert s to lowercase.
  µ          Begin a new, monadic chain. Argument: s (lowercase string)
       $     Combine the two links to the left into a monadic chain.
      Q        Unique; yield the first occurrence of each character.
    œ-         Perform multiset subtraction, removing the last occurrence of each
               character.
   ḟ         Filterfalse; keep characters that do not appear in the difference.
        Ḋ    Dequeue; remove the first character.
         ḣ1  Head 1; remove everything but the first character.
\$\endgroup\$
4
\$\begingroup\$

Batch, 171 bytes

@echo off
set a=.
set s=%~1
:l
if "%s%"=="" exit/b
set c=%s:~0,1%
call set t=%%s:%c%=%%
if "%s:~1%"=="%t%" set a=%a%%c%
set s=%t%
if "%a:~2%"=="" goto l
echo %c%

Alternative formulation, also 171 bytes:

@echo off
set a=.
set s=%~1
:l
if "%s%"=="" exit/b
set c=%s:~0,1%
set t=%s:~1%
call set s=%%s:%c%=%%
if "%s%"=="%t%" set a=%a%%c%
if "%a:~2%"=="" goto l
echo %c%
\$\endgroup\$
  • \$\begingroup\$ Can't make it to run on W2008R2. Line "call set ..." expands to "call set t=%s:D=%" and aborts with "The syntax of the command is incorrect" message. \$\endgroup\$ – meden Jul 2 '16 at 23:06
  • \$\begingroup\$ @meden Sorry, some typos crept into my post. The dead giveaway was that the post was shorter than I said it was! They're fixed now. \$\endgroup\$ – Neil Jul 3 '16 at 0:09
3
\$\begingroup\$

Pyth, 16 15 bytes

1 byte thanks to @mbomb007

=rz1.xhtfq1/zTzk
=rz1:fq1/zTz1 2

Test suite.

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  • 2
    \$\begingroup\$ I don't even know Pyth, but if you say so. :D \$\endgroup\$ – mbomb007 Jul 1 '16 at 19:26
  • \$\begingroup\$ @mbomb007 You know, the [1:2] trick. \$\endgroup\$ – Leaky Nun Jul 2 '16 at 1:20
  • \$\begingroup\$ You can save a byte with t<…2 instead of :…1 2. You can save another byte by moving the =rz1 to its first use, if you also change 1 to Z (for lowercase instead of uppercase output): t<fq1/zT=rzZ2. \$\endgroup\$ – Anders Kaseorg Jul 4 '16 at 1:23
3
\$\begingroup\$

Actually, 19 bytes

;╗`ù╜ùc1=`░ε;(qq1@E

Try it online!

Explanation:

;╗`ù╜ùc1=`░ε;(qq1@E
;╗                   push a copy of input to reg0
  `ù╜ùc1=`░          [v for v in s if
   ù╜ùc1=              s.lower().count(v.lower()) == 1]
           ε;(qq     append two empty strings to the list
                1@E  element at index 1 (second element)
\$\endgroup\$
3
\$\begingroup\$

C#, 129 128 bytes

char c(string i){var s=i.Where((n,m)=>i.ToLower().Where(o=>o==Char.ToLower(n)).Count()<2).ToArray();return s.Length>1?s[1]:' ';}

works fine. I wish i didnt need to lowercase everything

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  • \$\begingroup\$ Throws an IndexOutOfRangeException when I pass "Thisxthis" as an argument. Other than that, I think that ==1 can be changed to <2. \$\endgroup\$ – Yytsi Jul 3 '16 at 11:10
2
\$\begingroup\$

C# lambda with Linq, 63 bytes

s=>(s=s.ToUpper()).Where(c=>s.Count(C=>c==C)<2).Skip(1).First()
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  • \$\begingroup\$ You should be able to replace .Skip(1).First() with .ElementAt(1) \$\endgroup\$ – aloisdg Jul 3 '16 at 13:46
  • \$\begingroup\$ Even better you can convert to list and use the index .ToList()[1] \$\endgroup\$ – aloisdg Jul 3 '16 at 13:51
  • \$\begingroup\$ This throws an exception for inputs like "", "AABB", and "AABBC", where there is not a matching character in the 2nd position. I think you need FirstOrDefault. \$\endgroup\$ – Grax Dec 28 '16 at 22:29
2
\$\begingroup\$

C#, 141 bytes

void p(){var x=Console.ReadLine().ToLower();var c=0;foreach(char i in x){if(x.Split(i).Length-1<2){if(++c==2){Console.WriteLine(i);break;}}}}

Without break(smallest), 135 bytes

void p(){var x=Console.ReadLine().ToLower();var c=0;foreach(char i in x){if(x.Split(i).Length-1<2){if(++c==2){Console.WriteLine(i);}}}}

With for(;;), 150 bytes

void p(){for(;;){var x=Console.ReadLine().ToLower();var c=0;foreach(char i in x){if(x.Split(i).Length-1<2){if(++c==2){Console.WriteLine(i);break;}}}}}

Ungolfed with comments

void p()
{
    var x=Console.ReadLine().ToLower();//Get lowercase version of input from STDIN
    var c=0; //Create "count" integer
    foreach(char i in x){//For each char in input from STDIN
        if(x.Split(i).Length-1<2)//If current char occurs once in input from STDIN
        {
            if(++c==2){ //Add 1 to count and if count is 2
                Console.WriteLine(i); //Print result to STDOUT
                break; //Exit foreach
            } //End of IF
         } //End of IF
     } //End of FOREACH
} //End of VOID

12 bytes saved by TuukkaX(change count to c).

3 bytes saved by TuukkaX(change string to var).

4 bytes saved by TuukkaX in "With for(;;)"(changed while(true) to for(;;)).

2 bytes saved by TuukkaX(changed c++;if(c==2) to if(++c==2)).

14 bytes saved by Bryce Wagner(changed x.ToCharArray() to x).

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  • \$\begingroup\$ @TuukkaX oh, really. thank you! \$\endgroup\$ – r3pear Jul 1 '16 at 20:13
  • \$\begingroup\$ Welcome to PPCG! That's a nice first post! Since the rules mention that answers to this problem should either be functions or full programs, your codes require little tweaks. You can also save bytes by using var instead of string and having something like c instead of count. \$\endgroup\$ – Yytsi Jul 1 '16 at 20:15
  • \$\begingroup\$ @TuukkaX Thank you again! Shortly I'll modify the code and change string to var. \$\endgroup\$ – r3pear Jul 1 '16 at 20:17
  • \$\begingroup\$ @TuukkaX Should I add something like void program(){}??? \$\endgroup\$ – r3pear Jul 1 '16 at 20:19
  • \$\begingroup\$ Yes, but give a one byte function name to save bytes! :) \$\endgroup\$ – Yytsi Jul 1 '16 at 20:20
2
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x86 machine code, 43 bytes

In hex:

FC31C031C95641AC84C0740E3C6172F63C7A77F28066FFDFEBEC5EAC49740B89F751F2AE5974F44A77F1C3

Function takes a pointer to the input string in (E)SI and an integer in (E)DX and returns the (E)DX-th non-repeating character or zero if there's no such character. As a side-effect it converts the string to upper case.

Disassembly:

fc             cld
31 c0          xor    eax,eax
31 c9          xor    ecx,ecx
56             push   esi
_loop0:                         ;Search for the NULL char,
41             inc    ecx       ;counting the length in the process
ac             lodsb
84 c0          test   al,al
74 0e          je     _break0   ;NULL found, break
3c 61          cmp    al,0x61   ;If char is
72 f6          jb     _loop0    ;between 'a' and 'z'
3c 7a          cmp    al,0x7a   ;convert this char
77 f2          ja     _loop0    ;to uppercase in-place
80 66 ff df    and    byte ptr [esi-0x1],0xdf
eb ec          jmp    _loop0
_break0:
5e             pop    esi       ;Reset pointer to the string
_loop:                          ;ECX=string length with NULL
ac             lodsb            ;Load next char to AL
49             dec    ecx
74 0b          je     _ret      ;End of string found, break (AL==0)
89 f7          mov    edi,esi   ;EDI points to the next char
51             push   ecx
f2 ae          repnz scasb      ;Search for AL in the rest of the string
59             pop    ecx
74 f4          je     _loop     ;ZF==1 <=> another instance found, continue
4a             dec    edx
77 f1          ja     _loop     ;If not yet the EDX-th non-rep char, continue
_ret:
c3             ret
\$\endgroup\$
2
\$\begingroup\$

APL, 32 bytes

{⊃1↓⍵/⍨1=+/∘.=⍨(⎕UCS ⍵)+32×⍵∊⎕A}

Try it || All test cases

Explanation:

                (⎕UCS ⍵)+32×⍵∊⎕A  Add 32 to uppercase letters
            ∘.=⍨                    Make an equality matrix
          +/                        Check how many matches
    ⍵/⍨1=                           Keep elements with 1 match
  1↓                                Drop the first one
⊃                                   Return the second one

I was about to post it with 16 bytes, but the I realized it had to be case-insensitive...

\$\endgroup\$
  • 1
    \$\begingroup\$ (⎕UCS ⍵)+32×⍵∊⎕A819⌶⍵ \$\endgroup\$ – Adám Aug 31 '18 at 7:45
  • \$\begingroup\$ I've never seen that operator before. What version does it work in? \$\endgroup\$ – Woofmao Aug 31 '18 at 17:14
  • \$\begingroup\$ It is called the i-beam. It is an operator in all versions of Dyalog APL. It was originally a function in old versions of IBM's APL for special calls to the IBM system. Get it? I-B-M — i-beam? \$\endgroup\$ – Adám Aug 31 '18 at 17:32
  • \$\begingroup\$ Documentation for in general and for service 819 ("819" ≈ "BIg"). Try it online! \$\endgroup\$ – Adám Aug 31 '18 at 17:40
  • \$\begingroup\$ Well, I've learned something new. tryapl.org doesn't seem to recognize it, so do you mind if I just use your TIO link? \$\endgroup\$ – Woofmao Aug 31 '18 at 18:17
1
\$\begingroup\$

Retina, 43 36 bytes

iM!`(.)(?<!\1.+)(?!.*\1)
!`(?<=^.¶).

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Mathematica, 49 bytes

Cases[Tally@ToUpperCase@#,{_,1}][[2,1]]~Check~""&

Anonymous function. Takes a list of characters as input. Ignore any errors that are generated.

\$\endgroup\$
1
\$\begingroup\$

JavaScript (Firefox 48 or earlier), 60 bytes

f=s=>(m=s.match(/(.).*\1/i))?f(s.replace(m[1],"","gi")):s[1]

Returns undefined if there are only zero or one non-repeating characters. Works by case-insensitively deleting all occurrences of characters that appear more than once in the string. Relies on a non-standard Firefox extension that was removed in Firefox 49. 119 91 byte ES6 version:

f=s=>(m=s.match(/(.).*?(\1)(.*\1)?/i))?f((m[3]?s:s.replace(m[2],"")).replace(m[1],"")):s[1]

Recursively searches for all characters that appear at least twice in the string. If the character appears exactly twice then both occurrences are deleted otherwise only the first occurrence is deleted (the other occurrences will be deleted later). This allows the occurrences to have a difference case.

\$\endgroup\$
  • \$\begingroup\$ I believe you can actually adapt your Firefox 48 answer to be ES6-compliant by replacing m[1] with new RegExp(`${m[1]}`,"gi") \$\endgroup\$ – Value Ink Jul 7 '16 at 1:17
  • \$\begingroup\$ @KevinLau-notKenny That wouldn't work for special characters, and it cost me 33 bytes to special-case them, taking me up to 93, unfortunately. \$\endgroup\$ – Neil Jul 7 '16 at 7:54
  • \$\begingroup\$ noooooo not the special characters! I've had to edit my Ruby answer to accommodate for them as well, now. \$\endgroup\$ – Value Ink Jul 7 '16 at 8:09
1
\$\begingroup\$

J, 25 bytes

(1{2{.]-.]#~1-~:)@tolower

Usage

   f =: (1{2{.]-.]#~1-~:)@tolower
   f 'DEFD'
f
   f 'FEED'
d
   f 'This is an example input sentence.'
x
   f '...,,,..,,!@'
@
   f 'ABCDefgHijklMNOPqrsTuVWxyz'
b
   f 'AAAAAABBBBB'

   f 'Thisxthis'

   f 'This this.'
.

Explanation

(1{2{.]-.]#~1-~:)@tolower  Input: s
                  tolower  Converts the string s to lowercase
              ~:           Mark the indices where the first time a char appears
            1-             Complement it
         ]                 Identity function to get s
          #~               Copy only the chars appearing more than once
      ]                    Identity function to get s
       -.                  Remove all the chars from s appearing more than once
   2{.                     Take the first 2 chars from the result (pad with empty string)
 1{                        Take the second char at index 1 and return it
\$\endgroup\$
1
\$\begingroup\$

Bash, 58 bytes

tr A-Z a-z>t
tr -dc "`fold -1<t|sort|uniq -u`"<t|cut -c2

Caution: This creates a temporary file named t. If it already exists, it will be overwritten.

\$\endgroup\$
1
\$\begingroup\$

C, 174 bytes

int c(char*s){int y=128,z=256,c[384],t;memset(c,0,z*6);for(;t=toupper(*s);s++){c[t]++?c[t]-2?0:c[z+(c[y+c[z+t]]=c[y+t])]=c[z+t]:c[z]=c[y+(c[z+t]=c[z])]=t;}return c[y+c[y]];}

This is not the most short, but quite efficient implementation. In essence it uses double-linked list to maintain ordered set of candidate characters and scans input string just once. Returns character code or zero if none found.

A little bit ungolfed version:

int c(char*s)
{
    int y=128,z=256,c[384],t;
    //It's basically c[3][128], but with linear array the code is shorter

    memset(c,0,z*6);

    for(;t=toupper(*s);s++)
    {
        c[t]++ ?        // c[0][x] - number of char x's occurrence
            c[t] - 2 ?  // > 0
                0       // > 1 - nothing to do  
                : c[z + (c[y + c[z + t]] = c[y + t])] = c[z + t]  // == 1 - remove char from the list
            : c[z] = c[y + (c[z + t] = c[z])] = t; // == 0 - add char to the end of the list
    }
    return c[y + c[y]];
}
\$\endgroup\$
1
\$\begingroup\$

C#, 143 bytes

char c(string s){var l=s.Select(o=>Char.ToLower(o)).GroupBy(x=>x).Where(n=>n.Count()<2).Select(m=>m.Key).ToList();return l.Count()>1?l[1]:' ';}
\$\endgroup\$
1
\$\begingroup\$

TSQL, 128 bytes

Golfed:

DECLARE @ varchar(99)=',,zzzbb@kkkkkkJgg'

,@i INT=99WHILE @i>1SELECT
@i-=1,@=IIF(LEN(@)>LEN(x)+1,x,@)FROM(SELECT
REPLACE(@,SUBSTRING(@,@i,1),'')x)x PRINT SUBSTRING(@,2,1)

Ungolfed:

DECLARE @ varchar(99)=',,zzzbb@kkkkkkJgg'

,@i INT=99

WHILE @i>1
  SELECT
    @i-=1,@=IIF(LEN(@)>LEN(x)+1,x,@)
  FROM
    (SELECT 
       REPLACE(@,SUBSTRING(@,@i,1),'')x
    )x

PRINT SUBSTRING(@,2,1)

Fiddle

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1
\$\begingroup\$

Ruby, 53 bytes

Input is STDIN, output is STDOUT. In Ruby, out-of-index positions in an array or string return nil, which is not printed.

String#count is a strange function in Ruby because instead of counting the number of occurrences for the string that was passed in, it counts the number of occurrences for each letter in that string. It's usually annoying but we can use it to our advantage this time. String#swapcase swaps upper and lower case letters.

$><<gets.chars.reject{|c|$_.count(c+c.swapcase)>1}[1]

Old version that wasn't safe against special characters like . - 46 bytes

$><<gets.chars.reject{|c|$_=~/#{c}.*#{c}/i}[1]
\$\endgroup\$
1
\$\begingroup\$

Java 8, 172 157 bytes

(String s)->{s=s.toLowerCase();for(char i=0,c;s.length()>0;s=s.replace(c+"","")){c=s.charAt(0);if(!s.matches(".*"+c+".*"+c+".*")&&++i>1)return c;}return' ';}

-15 bytes.. Dang I was bad at golfing back then. ;)

Explanation:

Try it here.

(String s)->{                          // Method with String parameter and character return-type
  s=s.toLowerCase();                   // Make the input-String lowercase
  for(char i=0,c;s.length()>0;         // Loop over the characters of `s`
      s=s.replace(c+"","")){           // And after every iteration, remove all occurrences of the previous iteration
    c=s.charAt(0);                     // Get the current first character
    if(!s.matches(".*"+c+".*"+c+".*")  // If it doesn't occur more than once
     &&++i>1)                          // And this was the second one we've found
      return c;                        // Return this second characters
  }                                    // End of loop
  return' ';                           // Else: return an empty character/nothing
}                                      // End of method
\$\endgroup\$
1
\$\begingroup\$

R, 79 bytes

function(z){y=tolower(el(strsplit(z,"")));x=table(y);y[y%in%names(x[x==1])][2]}

Try it online!

I definitely feel like something can be golfed out here. But I really enjoyed this challenge.

This answer splits the string into a vector of characters, changes them all to lower case, and tables them (counts them). Characters that occur once are selected and compared to characters within the aforementioned vector, then the second value that is true is returned as output. An empty string, or a string with no repeating characters outputs NA.

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1
\$\begingroup\$

Perl 6, 38 32 bytes

-6 bytes thanks to nwellnhof by changing .comb to a case-insensitive regex

{~grep({2>m:g:i/$^a/},.comb)[1]}

Try it online!

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  • 1
    \$\begingroup\$ m:g:i/$^a/ for 32 bytes. \$\endgroup\$ – nwellnhof Sep 1 '18 at 13:17
1
\$\begingroup\$

K (oK) / K4, 11 bytes

Solution:

*1_&1=#:'=_

Try it online!

Explanation:

*1_&1=#:'=_ / the solution
          _ / convert input to lowercase
         =  / group alike characters
      #:'   / count (#:) each group
    1=      / 1 equal to length of the group?
   &        / where true
 1_         / drop the first
*           / take the first
\$\endgroup\$
0
\$\begingroup\$

Jelly, 15 bytes

Œlµċ@Ị¥Ðf¹ḊḢȯ“”

Try it online!

Verify all testcases. (Slightly modified to cater for all testcases)

\$\endgroup\$
0
\$\begingroup\$

Perl, 75 bytes

 my$s=<>;chomp$s;my$c;for my$i(split//,$s){my$m=@{[$s=~/$i/gi]};$m<2and++$c>=2and say$i and last}
\$\endgroup\$
0
\$\begingroup\$

Javascript (using external Library) (107 bytes)

Crushed this using a library I wrote. Not sure if I have to count the declaration of variable "s", which is the string in question.

(s)=>_.From(s).ToLookup(y=>y.toLowerCase(),z=>z).Where(g=>g.Value.Count()==1).Select(x=>x.Key).ElementAt(1)

This will handle an empty string input, an input with only one non-repeating character, and an input with 2+ non-repeating characters

Image 1

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  • \$\begingroup\$ Do you have a link to the library in question? Also, this being code golf, you gotta take out whitespace where you can \$\endgroup\$ – Value Ink Jul 7 '16 at 0:24
  • \$\begingroup\$ Hey, yeah it's github.com/mvegh1/Enumerable . No docs yet. Sorry, I will clean up this answer to reduce as much whitespace \$\endgroup\$ – applejacks01 Jul 7 '16 at 0:25
  • \$\begingroup\$ You should probably mention and link it in the answer body. Also, regarding the bytecount, consensus is to put it into an anonymous lambda (so s=> ...) \$\endgroup\$ – Value Ink Jul 7 '16 at 0:41
  • \$\begingroup\$ Ok no problem. I didn't want to offend anybody by linking to my code, but I did mention I used my library. I'll update my answer with the lambda, thanks for letting me know \$\endgroup\$ – applejacks01 Jul 7 '16 at 0:45
0
\$\begingroup\$

Clojure, 109 bytes

#(let[s(clojure.string/lower-case %)](or(second(remove(set(map(fn[[k v]](if(> v 1)k))(frequencies s)))s))""))

Ough, I hope there is a more succinct way.

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