Find the Missing Letter

Question

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'

Answer 1

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()

Answer 2

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.

Answer 3

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!

Answer 4

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

Answer 5

Haskell, 33 30 bytes

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

Try it online!

Answer 6

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!

Answer 7

Pyth, 5 bytes

-rhQe

Try it online!

Answer 8

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})

Answer 9

Mathematica, 46 bytes

Min@Complement[CharacterRange@@#[[{1,-1}]],#]&

Answer 10

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.

Answer 11

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.

Answer 12

Husk, 3 bytes

-¹…

Try it online!

I'm slightly nervous about posting this, as it seems wierd to beat all other answers (including one in Husk) by such a significant margin... but it seems to work for the test cases, at least...

How?

-       # output elements that are different between:
 ¹      # 1. the input list, and
  …     # 2. the input list, with all the gaps filled 
        #    with the missing character ranges

Answer 13

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>

Answer 14

PHP>=7.1, 46 bytes

Take input as string

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

PHP Sandbox Online

Answer 15

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.

Answer 16

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

Answer 17

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.

Answer 18

Jelly, 6 bytes

O‘Ọḟ⁸Ḣ

Try it online!

How it works

O‘Ọḟ⁸Ḣ - Main link. Takes a string S on the left
O      - Convert to char codes
 ‘     - Increment each
  Ọ    - Convert from char codes

         This yields the characters that directly follow each character in S
         Call this C

    ⁸  - Yield S
   ḟ   - Remove the characters from C that are in S
     Ḣ - Return the first one

Answer 19

Arturo, 21 bytes

$->s[--@s\0..last<=s]

Try it!

$->s[            ; a function taking an argument s
    <=s          ; duplicate s
    last         ; last letter in s
    ..           ; range
    s\0          ; first letter in s
    @            ; reify range
    --           ; set difference (between the range and duplicated s)
]                ; end function

Answer 20

Uiua ^SBCS, 9 bytes

⊢▽¬⊃∊∘⊸+1

Try it!

⊃∊∘ can be replaced with ⤚∊ for -1 byte once ⤚ comes out of experimental.

Explanation

⊢▽¬⊃∊∘⊸+1­⁡​‎⁠‎⁡⁠⁢⁤‏⁠‎⁡⁠⁣⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁣‏‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁢‏‏​⁡⁠⁡‌⁢⁢​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌­
       +1  # ‎⁡increment input
      ⊸    # ‎⁢keeping original input underneath
   ⊃∊∘     # ‎⁣membership mask, keeping incremented input underneath
  ¬        # ‎⁤invert mask
 ▽         # ‎⁢⁡keep
⊢          # ‎⁢⁢first

Answer 21

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.

Answer 22

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();
        }
    }
}

Answer 23

APL (Dyalog), 17 bytes

(⊃⎕AV/⍨∨\∧~)⎕AV∘∊

Try it online!

⎕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)

Answer 24

Python 3, 42 bytes

f=lambda a,*r:r[0]-a>1and chr(a+1)or f(*r)

Try it online!

Uses unpacked bytestring as input: f(*b'OQRS')

Answer 25

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.

Answer 26

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.

Answer 27

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'

Answer 28

k, 17 13 bytes

-4 bytes thanks to coltim.

{*(`c$1+x)^x}

Try it online!

Explanation:

{           } /def func(x): # x should be a list of chars
      1+x     /  above  = [ord(c)+1 for c in x]
  (`c$   )    /  above  = [chr(i) for i in above]
          ^x  /  result = [e for e in above if e not in x]
 *            /  return result[0]

Answer 29

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

Answer 30

J, 18 bytes

{.@(>:-.])&.(3&u:)

Same approach as many others.

Try it online!

{.@(>:-.])&.(3&u:)
            (3&u:)  NB. convert to char codes
    >:              NB. increment
        ]           NB. input
      -.            NB. set difference
  @                 NB. then
{.                  NB. take the head
          &.        NB. lastly, perform the inverse of 3&u:, which converts back to char