What's the second non-repeating character?

Question

_{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

• A single string, in any suitable format.

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 code-golf 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.
.

Answer 1

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 

Answer 2

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.

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

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%

Answer 7

Pyth, 16 15 bytes

1 byte thanks to @mbomb007

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

Test suite.

Answer 8

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)

Answer 9

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

Answer 10

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).

Answer 11

C# lambda with Linq, 63 bytes

s=>(s=s.ToUpper()).Where(c=>s.Count(C=>c==C)<2).Skip(1).First()

Answer 12

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

Answer 13

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...

Answer 14

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.

Answer 15

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!

Answer 16

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

Answer 17

Retina, 43 36 bytes

iM!`(.)(?<!\1.+)(?!.*\1)
!`(?<=^.¶).

Try it online!

Answer 18

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.

Answer 19

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.

Answer 20

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

Answer 21

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.

Answer 22

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]];
}

Answer 23

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]:' ';}

Answer 24

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

Answer 25

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]

Answer 26

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

Answer 27

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.

Answer 28

JavaScript, 57 bytes

v=>(v+'\0'+v).replace(/(.)(?=.*\1.*\1)|\0.*/gi,"")[1]||""

JavaScript doesn't have Regex lookbehind, so I appended a null character and another copy of the string so that I can use lookahead. I then remove the null character and all characters after it as part of the regex replace.

Answer 29

PowerShell, 55 Bytes

old but good challenge, deserves a PS answer.

([char[]]"$args".ToLower()|group|? Count -eq 1)[1].Name

Throws errors when there are no non-repeating chars, exits with no error or return if there is only one repeating char.

the .ToLower() cost a bit due to the case insensitivity, and using |% ToL*r actually costs bytes since you need to wrap it all in brackets before converting to a char[]

Answer 30

PHP, 76 bytes

for($a=$argn;~$c=$a[$i++];)stripos($a,$c)<strripos($a,$c)?:$r.=$c;echo$r[1];

Try it online!

Answer 31

Java (OpenJDK 8), 133 131 bytes

s->{String t=s.toUpperCase();int[]i=t.chars().filter(c->t.indexOf(c)==t.lastIndexOf(c)).toArray();return(char)(i.length>1?i[1]:0);}

Try it online!

Answer 32

Pyth, 17 bytes

Jrz1.x.(fq/JT1J1k

Test suite

Python 3 translation:

J=input().upper()
try:
    print([T for T in J if J.count(T)==1][1])
except:
    print()

Answer 33

05AB1E, 9 bytes

l©ʒ®s¢}¦¬

Try it online or verify all test cases.

Explanation:

l©           # Take the lowercase input, and store it in the register
  ʒ   }      # Filter the characters by:
   ®s¢       #  Keep those with an occurrence-count in this lowercase input of exactly 1
       ¦     # Remove the first character
        ¬    # And then take the head (which we implicitly output)

Answer 34

Jelly, 15 bytes

Œlµċ@Ị¥Ðf¹ḊḢȯ“”

Try it online!

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

Answer 35

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}

Answer 36

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