27
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Guidelines

Task

Write a method that takes an array of consecutive (increasing) letters as input and that returns the missing letter in the array (lists in some languages).


Rules

  • This is code golf so the shortest answer in bytes wins!
  • You will always get a valid array
  • There will always be exactly one letter missing
  • The length of the array will always be at least 2.
  • The array will always contain letters in only one case (uppercase or lowercase)
  • You must output in the same case (uppercase or lowercase) that the input is
  • The array will always only go one letter a time (skipping the missing letter)
  • The array length will be between 2 and 25
  • The first or last element of the array will never be missing

Examples

['a','b','c','d','f'] -> 'e'

['O','Q','R','S'] -> 'P'

['x','z'] -> 'y'

['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','w','x','y','z'] -> 'v'

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  • \$\begingroup\$ Can I take a string instead? \$\endgroup\$ – Leaky Nun Jul 17 '17 at 7:41
  • \$\begingroup\$ @LeakyNun Strings are arrays of characters, so yes. \$\endgroup\$ – Amorris Jul 17 '17 at 7:42
  • 1
    \$\begingroup\$ Can the output be an array containing the missing character (e.g: for the input ['a','b','c','d','f','g'], output ['e'], if that makes the code shorter? \$\endgroup\$ – Mr. Xcoder Jul 17 '17 at 8:59
  • 1
    \$\begingroup\$ @Mr.Xcoder A string is just an array of characters, so yes \$\endgroup\$ – Amorris Jul 17 '17 at 9:02
  • 2
    \$\begingroup\$ Rule four is simply a subset of rule eight and can be removed (at least, if you put the word "inclusive" at the end of rule eight). \$\endgroup\$ – NH. Jul 17 '17 at 19:13

49 Answers 49

5
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Pyth, 5 bytes

-rhQe

Try it online!

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11
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C# (.NET Core), 48 47 46 bytes, input as char array

s=>{for(int i=0;s[++i]==++s[0];);return s[0];}

Try it online!

Explanation: the first element in the array is incremented as well as a pointer iterating the following elements. When both the first element and the current element are different, it returns the first element.

C# (.NET Core), 58 56 50 bytes, input as string

s=>{var c=s[0];while(s.IndexOf(++c)>=0);return c;}

Try it online!

Previous 58-byte solution (referenced in the first comment):

s=>{for(int i=1;;i++)if(s[i]-s[0]>i)return(char)(s[i]-1);}

Algorithms using System.Linq

The following algorithms must add using System.Linq; (18 bytes) to the byte count and therefore are longer.

I quite liked this one (52+18 bytes):

s=>{int i=0;return(char)(s.First(c=>c-s[0]>i++)-1);}

And you also have a one-liner (45+18)-byte solution:

s=>(char)(s.Where((c,i)=>c-s[0]>i).First()-1)

And a very clever (37+18)-byte solution, courtesy of Ed'ka:

s=>s.Select(e=>++e).Except(s).First()
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  • 1
    \$\begingroup\$ Won't this fail to compile with not all code paths return a value? But +1 for the comparison check with s[i]-s[0], pretty clever! \$\endgroup\$ – TheLethalCoder Jul 17 '17 at 8:35
  • \$\begingroup\$ @TheLethalCoder It won't fail as the for loop does not have a stop condition, so it will keep iterating until the if condition evaluates to true. \$\endgroup\$ – Charlie Jul 17 '17 at 8:38
  • 1
    \$\begingroup\$ You can save 8 bytes like this: a=>{int i=0;for(;a[++i]-a[0]<=i;);return--a[i];} (when you take the input as char[]). Not thanks to me btw, thanks to @Nevay's comment on my Java 8 answer. \$\endgroup\$ – Kevin Cruijssen Jul 17 '17 at 11:12
  • 1
    \$\begingroup\$ @KevinCruijssen found a way to save two more bytes taking the input as a char array. \$\endgroup\$ – Charlie Jul 17 '17 at 11:51
  • 1
    \$\begingroup\$ Shorter Linq version: s=>s.Select(e=>++e).Except(s).First() \$\endgroup\$ – Ed'ka Jul 20 '17 at 10:57
8
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Alice, 10 bytes

/X.
\ior@/

Try it online!

Explanation

This is just a framework for linear programs that operate entirely in Ordinal (string processing) mode:

/...
\.../

The actual linear code is then:

i.rXo@

Which does:

i   Read all input.
.   Duplicate.
r   Range expansion. If adjacent letters don't have adjacent code points, the
    intermediate code points are filled in between them. E.g. "ae" would turn
    into "abcde". For the inputs in this challenge, this will simply insert
    the missing letter.
X   Symmetric set difference. Drops all the letters that appear in both strings,
    i.e. everything except the one that was inserted by the range expansion.
o   Output the result.
@   Terminate the program.
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7
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Haskell, 33 30 bytes

f a=until(`notElem`a)succ$a!!0

Try it online!

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  • \$\begingroup\$ until saves a byte: f(a:b)=until(`notElem`a:b)succ a \$\endgroup\$ – xnor Jul 17 '17 at 8:31
  • \$\begingroup\$ @xnor 3 bytes, actually. Thanks! \$\endgroup\$ – Anders Kaseorg Jul 17 '17 at 8:45
7
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Ruby, 21 characters

->a{[*a[0]..a[-1]]-a}

Returns a single element array, according to question owner's comment.

Sample run:

irb(main):001:0> ->a{[*a[0]..a[-1]]-a}[['a','b','c','d','f']]
=> ["e"]

Try it online!

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7
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Java 8, 70 57 56 48 46 bytes

a->{for(int i=0;++a[0]==a[++i];);return a[0];}

-14 (70 → 56) and -2 (48 → 46) bytes thanks to @CarlosAlejo.
-8 (56 → 48) bytes thanks to @Nevay.

Explanation:

Try it here.

a->{            // Method with char-array parameter and char return-type
  for(int i=0;  //  Start index-integer at 0 and loop as long as
    ++a[0]      //   the previous character + 1 (by modifying the character at index 0)
    ==a[++i];   //   equals the next character (by raising the index by 1 before checking)
  );            //  End of loop
  return a[0];  //  Return the now modified character at index 0 in the array
}               // End of method
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  • 1
    \$\begingroup\$ You can use an implicit cast instead of an explicit cast to save 8 bytes a->{int i=0;for(;a[++i]-a[0]<=i;);return--a[i];}. \$\endgroup\$ – Nevay Jul 17 '17 at 10:15
6
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C (gcc), 33 35 36 48 60 bytes

All optimizations should be turned off and only on 32-bit GCC.

f(char*v){v=*v+++1-*v?*v-1:f(v);}

Take input as a string.

Try it online!

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  • 2
    \$\begingroup\$ "All optimizations should be turned off and only on 32-bit GCC." is a very roundabout way of saying this doesn't work (only appears to work due to UB) \$\endgroup\$ – sehe Jul 18 '17 at 19:48
  • \$\begingroup\$ I'd say foo(char*a){return*a+1==a[1]?foo(a+1):++*a;} is pretty good; Only 1 char shorter than the more natural foo(char*a){while(*a+1==a[1])a++;return++*a;} \$\endgroup\$ – sehe Jul 18 '17 at 19:54
  • \$\begingroup\$ @sehe constant undefined behavior is considered acceptable on PPCG \$\endgroup\$ – Keyu Gan Jul 19 '17 at 3:07
5
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Python 3, 74 62 58 44 40 bytes

-12 bytes thanks to Erik the Outgolfer. -18 bytes thanks to Leaky Nun. -4 bytes thanks to musicman523.

Takes input as a bytestring.

lambda s:chr(*{*range(s[0],s[-1])}-{*s})

Try it online!

Another cool solution:

lambda s:chr(*{*range(*s[::~-len(s)])}-{*s})
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4
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Mathematica, 46 bytes

Min@Complement[CharacterRange@@#[[{1,-1}]],#]&
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  • \$\begingroup\$ I believe that Min@Complement[CharacterRange@@#[[{1,-1}]],#]& would save a byte. \$\endgroup\$ – LegionMammal978 Jul 17 '17 at 12:27
  • \$\begingroup\$ @LegionMammal978 actually 2! \$\endgroup\$ – J42161217 Jul 17 '17 at 12:45
3
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JavaScript (ES6), 70 bytes

Input as a character array

(a,p)=>a.some(c=>(q=p+1,p=c.charCodeAt(),p>q))&&String.fromCharCode(q)

Less golfed

a=>{
  p = undefined;
  for(i = 0; c = a[i]; i++)
  {
    q = p+1
    p = c.charCodeAt()
    if (p>q)
      return String.fromCharCode(q)
  }
}

Test

F=(a,p)=>a.some(c=>(q=p+1,p=c.charCodeAt(),p>q))&&String.fromCharCode(q)

function update() {
  var a0=A0.value.charCodeAt()
  var a1=A1.value.charCodeAt()
  if (a1>a0) {
    var r = [...Array(a1-a0+1)]
      .map((x,i)=>String.fromCharCode(a0+i))
      .filter(x => x != AX.value)
    I.textContent = r.join('') + " => " + F(r)
  }
  else {
    I.textContent=''
  }
}

update()
input { width: 1em }
Range from <input id=A0 value='O' pattern='[a-zA-Z]' length=1 oninput='update()'>
to <input id=A1 value='T' pattern='[a-zA-Z]' length=1 oninput='update()'>
excluding <input id=AX value='Q' pattern='[a-zA-Z]' length=1 oninput='update()'>
<pre id=I></pre>

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3
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PHP>=7.1, 46 bytes

Take input as string

<?=trim(join(range(($a=$argn)[0],$a[-1])),$a);

PHP Sandbox Online

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3
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Retina, 33 25 bytes

$
¶$_
T`p`_p`.*$
D`.
!`.$

Try it online! Works with any range of ASCII characters. Edit: Saved 8 bytes thanks to @MartinEnder. Explanation: The first stage duplicates the input. The second decreases all of the characters in the copy by 1 code point. The third stage deletes all of the characters in the copy that still appear in the original. This just leaves the original input, the character that precedes the first character of the original input and the missing character. The last stage then just matches the missing character.

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  • \$\begingroup\$ Here is 25, using the same basic idea: tio.run/##K0otycxL/P9fhevQNpV4rpCEgoT4ggQ9LRUulwQ9LsUEPZX///… (I'm decrementing the second line because that saves a byte and then I'm finding the unique char using deduplication.) \$\endgroup\$ – Martin Ender Jul 17 '17 at 14:00
  • \$\begingroup\$ @MartinEnder Deduplication is exactly what I wanted all along, and I already forgot Retina has it, sigh... (I know incrementing the first line takes a byte more than decrementing the second line but it made the match regex shorter.) \$\endgroup\$ – Neil Jul 17 '17 at 14:53
3
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SWI Prolog, 124 bytes

m([H|T]):-n(H,N),c(T,N),!,m(T).
n(I,N):-o(I,C),D is C+1,o(N,D).
c([N|_],N).
c(_,N):-print(N),!,fail.
o(C,O):-char_code(C,O).

Examples:

?- m(['a','b','c','d','f']).
e
false.

?- m(['O','Q','R','S']).
'P'
false.

?- m(['x','z']).
y
false.

?- m(['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','w','x','y','z']).
v
false.

Little explanation:

The m is the "main" procedure,the n produces next expected character in the list. The c does comparison - if expectation matches the next item, continue, else print out expected character and jump out of the window.

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  • 1
    \$\begingroup\$ Shorter than fail: 0=1. \$\endgroup\$ – mat Jul 22 '17 at 22:28
3
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C++14, standard library, generic container type (87 86 bytes)

[](auto a){return++*adjacent_find(begin(a),end(a),[](auto a,auto b){return a+1!=b;});}

Container type from namespace ::std is assumed (e.g. std::string, std::list or std::vector. Otherwise using namespace std; or similar would be assumed.

Thanks to @Ven, with a little bit of preprocessor hacking, you get get it down to 82 bytes (1 newline)

#define x [](auto a,int b=0){return++
x *adjacent_find(begin(a),end(a),x a!=b;});}

See it Live On Coliru

C++14 no standard library (still generic, 64 63 bytes)

[](auto& a){auto p=*begin(a);for(auto c:a)if(c!=p++)return--p;}

Again, need to help name lookup only if container type not from namespace ::std (or associated with it)

Live On Coliru for std::string e.g.

Live On Coliru for char const[] e.g.

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  • \$\begingroup\$ You need to put a space between the strike-out text and the next text. \$\endgroup\$ – CJ Dennis Jul 19 '17 at 7:28
  • \$\begingroup\$ @CJDennis Done. By the way, your current rep (2469) is a beautiful number (being 3*823 and also visually paired as (24)(69) which is (2 2 2 3)(3 23)) \$\endgroup\$ – sehe Jul 19 '17 at 13:53
2
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Charcoal, 18 bytes

Fγ¿¬∨∨‹ι⌊θ›ι⌈θ№θιι

Try it online! Link is to verbose version of code. Takes input as a string. Works with any almost contiguous sequence of ASCII characters.

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2
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C#, 104 bytes

using System.Linq;a=>(char)Enumerable.Range(a.Min(),a.Max()-a.Min()).Except(a.Select(c=>(int)c)).First()

Full/Formatted version:

using System.Linq;

namespace System
{
    class P
    {
        static void Main()
        {
            Func<char[], char> f = a =>
                (char)Enumerable.Range(a.Min(), a.Max() - a.Min())
                                .Except(a.Select(c=>(int)c))
                                .First();

            Console.WriteLine(f(new[] { 'a', 'b', 'c', 'd', 'f' }));

            Console.ReadLine();
        }
    }
}
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  • \$\begingroup\$ A very clever Linq version by Ed'ka: s=>s.Select(e=>++e).Except(s).First() \$\endgroup\$ – Charlie Jul 20 '17 at 11:13
  • \$\begingroup\$ @CarlosAlejo I saw you added it to your answer so I won't update mine but yes it is very clever. A lot shorter than my version of doing it. \$\endgroup\$ – TheLethalCoder Jul 20 '17 at 11:14
2
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MATL, 8 7 bytes

1 byte saved thanks to @Luis

tdqf)Qc

Try it at MATL Online

Explanation

      % Implicitly grab the input as a string
t     % Duplicate the string
d     % Compute the differences between successive characters
q     % Subtract 1 from each element
f     % Get the locations of all non-zero characters (1-based index)
)     % Extract that character from the string
Q     % Add one to get the next character (the missing one)
c     % Convert to character and display
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  • \$\begingroup\$ @LuisMendo Awesome, thanks! \$\endgroup\$ – Suever Jul 17 '17 at 13:31
2
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Excel, 110 + 2 = 112 bytes

=CHAR(CODE(LEFT(A1))-1+MATCH(0,IFERROR(FIND(CHAR(ROW(INDIRECT(CODE(LEFT(A1))&":"&CODE(RIGHT(A1))))),A1),0),0))

Must be entered as an array formula (Ctrl+Shift+Enter) which adds curly brackets { } on each end, adding two bytes. Input is as a string in A1, which is OK per OP.

This is not the shortest answer by far (Excel rarely is) but I like seeing if it can be done.

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2
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Rexx (Regina), 56 bytes

a=arg(1)
say translate(xrange(left(a,1),right(a,1)),'',a)

Try it online!

Finally one that enables REXX to use its strong string-manipulation.

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2
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CJam, 6 bytes (full program) / 7 bytes (code block)

q),^W=

Try it online!

This is a full CJam program that reads the input string from standard input and prints the missing letter to standard output. CJam doesn't actually have "methods", which is what the challenge asks for, but the closest thing would probably be an executable code block, like this:

{),^W=}

Try it online!

This code block, when evaluated, takes the input as a string (i.e. an array of characters) on the stack, and returns the missing character also on the stack.


Explanation: In the full program, q reads the input string and places it on the stack. ) then pops off the last character of the input string, and the range operator , turns it into an array containing all characters with code points below it (including all letters before it in the alphabet). Thus, for example, if the input was cdfgh, then after ), the stack would contain the strings cdfg (i.e. the input with the last letter removed) and ...abcdefg, where ... stands for a bunch of characters with ASCII codes below a (i.e. all characters below the removed last input letter).

The symmetric set difference operator ^ then combines these strings into a single string that contains exactly those characters that appear in one of the strings, but not in both. It preserves the order in which the characters appear in the strings, so for the example input cdfg, the result after ),^ will be ...abe, where ... again stands for a bunch of characters with ASCII codes below a. Finally, W= just extracts the last character of this string, which is exactly the missing character e that we wanted to find (and discards the rest). When the program ends, the CJam interpreter implicitly prints out the contents of the stack.


Bonus: GolfScript, 6 bytes (full program)

),^-1>

Try it online!

It turns out that nearly the same code also works in GolfScript. We save one byte in the full program version due to GolfScript's implicit input, but lose one byte because, unlike CJam's W, GolfScript doesn't have a handy single-letter variable initialized to -1.

Also, CJam has separate integer and character types (and strings are just arrays containing characters), whereas GolfScript only has a single integer type (and has a special string type that behaves somewhat differently from normal arrays). The result of all this is that, if we want the GolfScript interpreter to print out the actual missing letter instead of its ASCII code number, we need to return a single-character string instead of just the character itself. Fortunately, making that change here just requires replacing the indexing operator = with the array/string left truncation operator >.

Of course, thanks to GolfScript's implicit I/O, the code above can also be used as a snippet that reads a string from the stack and returns a single-character string containing the missing letter. Or, rather, any snippet that takes a single string on the stack as an argument, and returns its output as a printable string on the stack, is also a full GolfScript program.

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  • 6
    \$\begingroup\$ Code snippets are not allowed by default; only functions and full programs are. So you probably need that q (program), or {...} (block). +1 for the approach though \$\endgroup\$ – Luis Mendo Jul 17 '17 at 11:18
  • \$\begingroup\$ This is very clever! \$\endgroup\$ – Esolanging Fruit Jul 19 '17 at 9:12
2
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Husk, 6 bytes

→S-(ḣ→

Try it online!

This function takes a string (list of characters) as input, and returns a character as output.

Explanation

→S-(ḣ→
    ḣ→    Get the list of all characters from the null byte to the last character of the input
 S-       Subtract the input from this list
→         Get the last element of the result
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2
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Python 2 - 76 bytes

Loses to existing python 2 solution but it's a slightly different approach so I thought I'd post it anyway:

lambda c:[chr(x)for x in range(ord(c[0]),ord(c[0]+26)if chr(x)not in c][0]
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2
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8th, 99 bytes

Rationale

If the distance between letters is greater than two, then there is a missing letter. Letter distance is obtained by computing the difference between ASCII code of each letter.

Code

: f ' nip s:each repeat over n:- 2 n:= if n:1+ "" swap s:+ . reset 1 then depth n:1- while! reset ;

Ungolfed version

: f \ s -- c 
  ' nip s:each    \ convert each letter into its ASCII code and put them on stack
  repeat
    over
    n:- 2 n:=     \ check if there is a missing letter 
    if            
      n:1+        \ compute the ASCII code of missing letter
      "" swap s:+ \ convert ASCII code into printable character
      .           \ print out the missing letter
      reset 1     \ set condition to exit from while!
    then
    depth n:1-    \ verify if there are letters to check
  while!          
  reset           \ clean stack
;

Usage and examples

ok> "abcdf" f
e
ok> "OQRS" f
P
ok> "xz" f
y
ok> "abcdefghijklmnopqrstuwxyz" f
v
ok> "ab" f

ok> "def" f

ok>
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2
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JavaScript (ES6), 64 bytes

Takes input as a string.

s=>(g=p=>(c=String.fromCharCode(n++))<s[p]?p?c:g(p):g(p+1))(n=0)

How?

  • Initialization: We start with n = 0 and p = 0 and call the recursive function g().

    g = p =>                                   // given p
      (c = String.fromCharCode(n++)) < s[p] ?  // if the next char. c is not equal to s[p]:
        p ?                                    //   if p is not equal to zero:
          c                                    //     step #3
        :                                      //   else:
          g(p)                                 //     step #1
      :                                        // else:
        g(p + 1)                               //   step #2
    
  • Step #1: We increment n until c = String.fromCharCode(n) is equal to the first character of the input string s[0].

  • Step #2: Now that we're synchronized, we increment both n and p at the same time until c = String.fromCharCode(n) is not equal to s[p] anymore.

  • Step #3: We return c: the expected character which was not found.

Test cases

let f =

s=>(g=p=>(c=String.fromCharCode(n++))<s[p]?p?c:g(p):g(p+1))(n=0)

console.log(f('abcdf'))                     // -> 'e'
console.log(f('OQRS'))                      // -> 'P'
console.log(f('xz'))                        // -> 'y'
console.log(f('abcdefghijklmnopqrstuwxyz')) // -> 'v'

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1
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J, 20 bytes

{&a.>:I.1 0 1&E.a.e.
  • a.e. boolean mask for the input letters across ascii charset
  • 1 0 1&E. new boolean mask indicating if the sequence 101 begins at that index, ie, find any place a "skip" sequence begins
  • I. the index of that match, ie, the character before the skipped one
  • >: increment by 1, ie, the index of the skipped char within ascii charset
  • {&a. pick that index from the ascii charset, ie, return the skipped char

Try it online!

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  • \$\begingroup\$ That looks like a snippet to me. \$\endgroup\$ – Adám Jul 17 '17 at 11:31
  • \$\begingroup\$ @Adám It's written in a tacit (point-free) style, which I believe counts as "function-like" as opposed to a snippet. As best as I can tell, it's no more of a snippet than your APL solution is (but I do not know dyalog, so take what I say with a grain of salt). \$\endgroup\$ – zgrep Jul 17 '17 at 14:01
  • \$\begingroup\$ @Adám yes it is, in the sense that it can’t be assigned to a variable but assumes input on its right side. is this not legal? i asked about it somewhere and was told it was fine \$\endgroup\$ – Jonah Jul 17 '17 at 14:39
  • \$\begingroup\$ My understanding for APL/J/K is that the code must be able to reside in a name, whether by assignment or as the body an explicit verb/function (however, the explicit form must then have explicit input too). Snippet is code which assumes values in variables and/or needs pasting into a line, but cannot stand on their own. \$\endgroup\$ – Adám Jul 17 '17 at 15:09
  • \$\begingroup\$ @zgrep No, this code is explicit (non-tacit), but is missing the reference to its argument on the far right. My APL function is a complete tacit function which can assigned or put in a parenthesis. \$\endgroup\$ – Adám Jul 17 '17 at 15:10
1
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ES6, 125 bytes:

(a=>((s,f)=>(r=(i,b)=>a[i]?r(i+1,b||(s[f](i)-s[f](i-1)-1&&String.fromCharCode(s[f](i-1)+1))):b)(1,0))(a.join(""),"charCodeAt"))

http://jsbin.com/vasoqidawe/edit?console

The returned function needs to be called with an array

(["a","c"])

one could save another 9 bytes through removing .join("") and passing a string:

("ac")

ES6, 108 bytes:

(a=>((s,f,o)=>(a.find((_,i)=>(o?++o:o=s[f](i))!==s[f](i)),String.fromCharCode(o)))(a.join(""),'charCodeAt'),0))

http://jsbin.com/tudiribiye/edit?console

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  • 1
    \$\begingroup\$ bind ??? in code golf ? \$\endgroup\$ – edc65 Jul 17 '17 at 9:23
  • \$\begingroup\$ @edc65 whats wrong with it ? ( sorry if this is n00b, but thats my first golf :)) \$\endgroup\$ – Jonas Wilms Jul 17 '17 at 9:27
  • \$\begingroup\$ @edc65 but youre probably right, removing it saved 4 bytes... \$\endgroup\$ – Jonas Wilms Jul 17 '17 at 9:42
  • \$\begingroup\$ a.join("") could be a.join`` \$\endgroup\$ – user2428118 Jul 17 '17 at 10:48
1
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Common Lisp, 88 bytes

(lambda(s)(loop as(x y)on s if(>(#1=char-code y)(1+(#1#x)))return(code-char(1+(#1#x)))))

Try it online!

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1
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Python 2, 69 bytes

lambda a:chr((ord(a[0])+ord(a[-1]))*-~len(a)/2-sum(ord(x)for x in a))

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Some explanations As we know the first and the last elements of the list, we can easily compute the sum of the codes of all the chars in the list + the missed char (using summary formulas of arithmetic progression). The difference between this sum and the sum of the codes of all the chars in the list gives the code of the missed letter.

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1
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05AB1E, 9 7 bytes

ǤÝsKçθ

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  • \$\begingroup\$ I'm going to judge from the 2 that we're using the same algorithm, even though I hardly know 05AB1E :) \$\endgroup\$ – Leaky Nun Jul 17 '17 at 7:51
  • \$\begingroup\$ @LeakyNun Well, I thought of the algorithm too... \$\endgroup\$ – Erik the Outgolfer Jul 17 '17 at 7:53
  • \$\begingroup\$ I just changed my algorithm though. \$\endgroup\$ – Leaky Nun Jul 17 '17 at 7:53
  • \$\begingroup\$ @LeakyNun It'd be longer in 05AB1E anyways. \$\endgroup\$ – Erik the Outgolfer Jul 17 '17 at 7:57
  • \$\begingroup\$ I just thought of another algorithm that contains 2, might be yours... \$\endgroup\$ – Leaky Nun Jul 17 '17 at 7:59
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APL (Dyalog), 17 bytes

(⊃⎕AV/⍨∨\∧~)⎕AV∘∊

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⎕AV∘∊ Boolean: each character in the Atomic Vector (character set) member of the argument?

() apply the following tacit function:

 the first element of

⎕AV the Atomic Vector (the character set)

/⍨ which

∨\ follows the initial (member of the argument)

 but

~ is not (a member of the argument)

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