R, 49 47 44 bytes

function(m,`?`=which)m==names(?table(m)<2)?T

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Takes input as a matrix, returns 1-indexed coordinates

Python 2, 57 bytes

lambda m:[map(len,map(set,a)).index(2)for a in zip(*m),m]

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A port of this to Python 3 can be 62 bytes:

lambda m:[[len(set(v))for v in a].index(2)for a in(zip(*m),m)]

The list comprehension, [len(set(v))for v in a], is shorter than the double map by two bytes now as it would need to be cast to a list like list(map(len,map(set,a)))

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Perl 6, 41 38 37 bytes

3 bytes saved thanks to @nwellnhof.

1 byte saved thanks to Jo King.

{map {[+] ^∞Z*!<<.&[Z~~]},$_,.&[Z]}

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Explanation

It takes the input as a list of lists of characters and returns list of length 2 containing zero-based X and Y coordinates of the needle.

It works by applying the block {[+] ^∞ Z* !<<.&[Z~~]} on the input and on its transpose. .&[Z~~] goes through all columns of the argument and returns True if all the elements are the same, False otherwise. We then negate all the values (so we have a list with one bool per column, where the bool answers the question "Is the needle in that column?"), multiply them element-wise with a sequence 0,1,2,... (True = 1 and False = 0) and sum the list, so the result of the whole block is the 0-based number of the column where the needle was found.

Nwellnhof's better approach, Perl 6, 34 bytes

{map *.first(:k,*.Set>1),.&[Z],$_}

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Explanation

Generally the same approach, just more effective. It still uses a block on the array and its transpose, but now the block converts all rows intoSets and checks for the number of elements. The first function then gives index (due to the :k) of the first row that contained more than 1 element. Because of that, the order of $_ and .&[Z] needed to be swapped.

Brachylog, 20 bytes

c≡ᵍ∋Ȯ&;I∋₎;J∋₎gȮ∧I;J

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Outputs [I,J], where I is the row index and J the column index, both 0-indexed.

Stupidely long, but getting indexes in Brachylog is usually very verbose.

Explanation

c                       Concatenate the Input into a single string

 ≡ᵍ                     Group identical characters together

   ∋Ȯ                   Ȯ is a list of One element, which is the needle character

     &;I∋₎              Take the Ith row of the Input

          ;J∋₎          Take the Jth character of the Ith row

              gȮ        That character, when wrapped in a list, is Ȯ

                ∧I;J    The output is the list [I,J]

05AB1E, 9 6 bytes

Saved 3 bytes switching input format.

Input is taken as a string and a row-length.

Output is a zero-based list of the form [y, x]

D.mks‰

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or as a Test Suite

Explanation

D           # duplicate the input string

 .m         # get the least frequent character

   k        # get its index in the string

    s       # swap the row length to the top of the stack

     ‰      # divmod the index of the least frequent char with the row length

PHP, 99 85 bytes

Using string without newlines and the width (or height) ('########N###########', 5) as input.

• -5 bytes by removing chr() call, props to @Titus


• -9 bytes by taking input as two function args, also props to @Titus

function($a,$l){return[($p=strpos($a,array_flip(count_chars($a,1))[1]))%$l,$p/$l|0];}

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Ungolfed:

function need_hay( $a, $l ) {



    // identify the "needle" by counting the chars and 

    // looking for the char with exactly 1 occurrence

    // note: this is 1 byte shorter than using array_search()

    $n = array_flip( count_chars( $a, 1 ) )[1];



    // find the location in the input string

    $p = strpos( $a, $n );



    // row is location divided by row length, rounded down

    $r = floor( $p / $l );



    // column is remainder of location divided by row length

    $c = $p % $l;



    return array( $c, $r );



}

Output:

#####

###N#

#####

#####

[3,1]



^^^

^^^

^N^

^^^

^^^

^^^

[1,2]



jjjjjj

j@jjjj

jjjjjj

[1,1]

Python 3 + NumPy, 75 66 bytes

-9 bytes thanks to @ASCII-only

lambda x:where(x.view('i')-median(x.view('i')))

from numpy import*

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This assumes that the input is a NumPy array. The output is zero-indexed, and first vertical, then horizontal.

It converts the input from char to int then calculates the median of the array, which will be the haystack character. We subtract that from the array, which makes the needle the only non-zero element. Finally, return the index of that element with numpy.where().

Jelly, 5 bytes

Outputs [height, width] (1-indexed).

ŒĠLÐṂ

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ŒĠLÐṂ – Monadic link / Full program. Takes a list of strings M as input.

ŒĠ    – Group the multidimensional indices by their values (treating M as a matrix).

  LÐṂ – And retrieve the shortest group of indices (those of the unique character).

Jelly, 5 bytes

ŒĠḊÐḟ

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Jelly, 4 bytes

_{Maybe this could've just been a comment for Mr. Xcoder it is pretty similar...}

ŒĠEƇ

A monadic link accepting the matrix of characters which yields a list of one item, the 1-indexed (row, column) co-ordinate from top-left.

(...As a full program given an argument formatted such that parsing results in a list of lists of characters -- that is a list of strings in Python format -- the single coordinate is printed.)

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How?

ŒĠEƇ - Link: matrix, M

ŒĠ   - multi-dimensional indices grouped by Value

     -  ...due to the 2*2 minimum size and one needle this will be a list of two lists one

     -     of which will have length one (the needle coordinates as a pair) and the other

     -     containing all other coordinates as pairs

   Ƈ - filter keeping those for which this is truthy:

  E  -   all equal?

     -   ... 1 for the list of length 1, 0 for the list of at least 3 non-equal coordinates

Java 8, 132 111 bytes

m->{int c=m[0][0],i=0,j;for(c=m[1][0]!=c?m[1][1]:c;;i++)for(j=m[i].length;j-->0;)if(m[i][j]!=c)return i+","+j;}

-8 bytes (and -13 more implicitly) thanks to @dana.

Input as character-matrix.

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Explanation:

m->{                    // Method with char-matrix parameter and String return-type

  int c=m[0][0],        //  Character to check, starting at the one at position 0,0

      i=0,j;            //  Index integers

  for(c=m[1][0]!=c?     //  If the second character does not equal the first:

         m[1][1]        //   Use the character at position 1,1 instead

        :c;             //  Else: keep the character the same

      ;i++)             //  Loop `i` from 0 indefinitely upwards:

    for(j=m[i].length;j-->0;)

                        //   Inner loop `j` in the range (amount_of_columns, 0]:

      if(m[i][j]!=c)    //    If the `i,j`'th character doesn't equal our character to check:

        return i+","+j;}//     Return `i,j` as result

JavaScript (ES6), 55 bytes

Takes input as $(s)(w)$, where $s$ is a string and $w$ is the width of the matrix. Returns $[x,y]$.

s=>w=>[(i=s.indexOf(/(.)1+(.)/.exec(s+s)[2]))%w,i/w|0]

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JavaScript (ES6),  65  64 bytes

Saved 1 byte thanks to @Neil

Takes input as a matrix of characters. Returns $[x,y]$.

m=>m.some((r,y)=>r.some((c,x)=>!m[p=[x,y],~y&1].includes(c)))&&p

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How?

We look for the first character $c$ located at $(x,y)$ which does not appear anywhere in another row $r[Y]$. We can perform this test on any row, as long as $Yne y$. Because the input matrix is guaranteed to be at least $2times 2$, we can simply use $Y=0$ if $y$ is odd or $Y=1$ if $y$ is even.

MATL, 12 8 bytes

tX:XM-&f

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Using the mode function as the majority-detector. Returns 1-based indices.

 t           % duplicate the input

  X:         % turn the copy into a linear array

    XM       % find the arithmetic mode of that (the 'haystack' character)

      -      % Subtract that from the original input

       &f    % find the position of the non-zero value in that result

-4 characters thanks to @LuisMendo

Wolfram Language 37 58 bytes

My earlier entry did not correctly handle the case where the "odd character out" was at the upper left corner of the matrix. This does.

#~Position~Keys[TakeSmallest[Counts@Flatten@#,1]][[1]]&

Counts@Flatten@# lists how many of each character are in the array, #.

TakeSmallest[...,1] returns the least frequent count, in the form of an association rule such as <| "Z"->1|>

Keys...[[1]] returns the "key" to the only item in the association, that of the least used character. ("Z" in the present case)

#~Position~... returns then position of the key in the original matrix, #.

Perl 5 -p00, 52 45 bytes

/^(.)(1*

)*(1*)|^/;$_=$&=~y/

//.$".length$3

45 bytes

52 bytes

How

• 
  -p00 : like -n but also print, paragraph mode


• 
  /^(.)(1*
)*(1*)|^/ : matches either
  
  
  
  
  • from start $1: first character, $2: repetition (not used), $3: characters before the "needle" in the line, $& whole match
  
  
  • or null string (position 0) no capture.
  
  
  
  


• 
  $_= : to assign the default input/argument variable


• so $&=~y/
// the number of newlines of $&
  


• 
  .$". : concatenate with $" (space character by default) and concatenate


• 
  length$3 : the length of $3
  

C# (Visual C# Interactive Compiler), 109 108 107 bytes

^{First() => Last() for -1 byte}

^{currying for -1 byte thanks to Embodiment of Ignorance}

a=>w=>{var d=a.Where(b=>b!=a[0]).Select(b=>a.IndexOf(b));return d.Count()>1?(0,0):(d.Last()%w,d.Last()/w);}

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J, 22 bytes

$#:(i.~.{~1 i.~#/.~)@,

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NB. returns answer in (row, column) format.

Python 2, 53 47 bytes

lambda s,w:divmod(s.find(min(s,key=s.count)),w)

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Call as f("########N###########", 5) (allowed in a comment). Outputs (y, x).

Erik saved 6 bytes, suggesting rearranging the output + using divmod. Thanks!

R 42 bytes

function(m)which(ave(m,m,FUN=length)==1,T)

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Input: a haystack matrix m

Output: (row,col) vector - index starting at 1

Python 3, 93 bytes

def f(s):x=s.find("n")+1;return[(i%x,i//x)for i,c in enumerate(s)if s.count(c)<2and" "<c][0]

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Input is taken as a multiline string. Output is 0-indexed

Octave, 40 bytes

@(x){[r,c]=find(x-mode(+x(:))) [c,r]}{2}

Port of @sundar's MATL answer. Output is a two-element vector with 1-based column and row indices.

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Retina 0.8.2, 41 bytes

s`(?=(.)+1)(.*?¶)*(.*)(?!1|¶).+

$.3,$#2

Try it online! 0-indexed. Explanation:

s`

Allow . to match newlines. This costs 3 bytes (3rd byte is the ? before the ¶) but saves 6 bytes.

(?=(.)+1)

Look ahead for two identical characters. 1 then becomes the hay.

(.*?¶)*

Count the number of newlines before the needle.

(.*)

Capture the hay to the left of the needle.

(?!1|¶)

Ensure that the needle isn't hay or a newline.

.+

Match the rest of the hay so that the result replaces it.

$.3,$#2

Output the width of the left hay and the number of newlines.

C# (Visual C# Interactive Compiler), 82 bytes

x=>w=>{int y=x.IndexOf(x.GroupBy(c=>c).Last(g=>g.Count()<2).Key);return(y%w,y/w);}

Thanks to dana for shaving off 6 bytes!

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Old solution, 106 bytes

n=>m=>{var z=n.Distinct();int d=n.IndexOf(n.Count(c=>c==z.First())>1?z.Last():z.First());return(d%m,d/m);}

Both take input as a string and an integer specifying the amount of columns.

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C (clang), 74 bytes

h(char*s,z,x){for(s+=z--;*s==*--s|*s==s[-1];)z--;printf("%d,%d",z%x,z/x);}

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DEGOLF

int h(char*s,int z,int x){// z = string size, x = row size



 for(s+=z--;

 // move pointer just over the end of the string 

 // and move z counter to the end of string



*s-*--s?   ==>  *s==*--s|  @ceilingcat suggestion 

 // if the previous element is different we will check if the next element is also different

 // if not the result is 1 and the iteration continue

 // in the first iteration it will be different because the pointer is just  over the end



 *s-s[-1]? ==> changed to *s==s[-1]   @ceilingcat suggestion 

 // the second check returns 0 if the char changed again so it was the needle

 // if not it's because in the first iteration the first check finded a difference just because the pointer was just over the end



 /*0:1*/   :1;)z--;





 printf("%d,%d",z%x,z/x);

}

Java 8, 104 Bytes

(x,w)->{int i=0,p=x.length;for(;i<p;i++)if(x[i]!=x[(i+1)%p]&&x[i]!=x[(i+2)%p])break;return i/w+","+i%w;}

Input is array of char, and integer indicating row width.

Output is zero-based, vertical then horizontal (i.e., row number then column number)

Explanation:

(x,w)->{

    int i=0, p=x.length;

    for (;i<p;i++)          //iterate through characters in x

      if (x[i]!=x[(i+1)%p] && x[i]!=x[(i+2)%p])    //compare x[i] with the two subsequent characters in array, wrapping around if necessary

        break;

    return i/w+","+i%w;}  //return row number then column number, zero-based

Python 3, 93 89 85 58 bytes

Complete rewrite taking input as concatenated string, width:

lambda g,w:divmod(g.index({g.count(c):c for c in g}[1]),w)

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Original answer:

def k(g):t=''.join(g);return divmod(t.index({t.count(c):c for c in t}[1]),len(g[0]))

EDIT: Saved 4 bytes by swapping linebreak/indent for semicolons. Saved another 4 bytes by using divmod(thanks @JonathanFrech).

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I know this could be a lot shorter, but I just wanted to try an approach around this dict comprehension.

MATL, 11 bytes

tX:YmyYk-&f

Output is row, then column; 1-based.

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Explanation

t    % Implicit input. Duplicate

X:   % Linearize into a column

Ym   % Compute mean (characters are converted to ASCII codes)

y    % Duplicate from below: pushes input again

Yk   % Closest value: gives the input value that is closest to the mean

-    % Subtract, element-wise. Gives non-zero for the value farthest from the mean

&f   % Two-output find: gives row and column indices of nonzeros. Implicit display

Pyth, 15 14 12 bytes

.Dxz-zh.-z{z

Takes input as the length of the row and the input without lines and outputs as [row, column].
Try it here

Explanation

.Dxz-zh.-z{z

       .-z{z    Subtract one of each character from the input.

      h         Take the first.

    -z          Remove all instances from the input.

  xz            Find the remaining character in the input.

.D          Q   Take the result divmod the (implicit) length of the row.

Old approach

mxJmt{kdeSJ.TB

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Explanation

mxJmt{kdeSJ.TB

           .TBQ   Take the (implicit) input and its transpose...

m      d          ... and for each...

   mt{k           ... deduplicate each row...

 xJ     eSJ       ... and find the index of the largest.     

Charcoal, 40 bytes

≔§⎇⌕θ§θ¹ηθ⁰ζＳθＷ⁼№θζＬθ«⊞υωＳθ»Ｉ⌕Ｅθ⁼ιζ⁰,ＩＬυ

Try it online! Link is to verbose version of code. I must be doing something wrong because this is almost as long as the Retina answer. Explanation:

≔§⎇⌕θ§θ¹ηθ⁰ζ

Check whether the second character in the first string is also the first character, and take the first character of the first string if so otherwise the first character of the second string if not. This is then the hay.

ＳθＷ⁼№θζＬθ«⊞υωＳθ»

Keep reading strings until a string whose hay is less than its length is found.

Ｉ⌕Ｅθ⁼ιζ⁰,ＩＬυ

Output the position of the mismatching element and then the number of strings previously read.

MATLAB, 68 22 bytes

[r,c]=find(v~=v(1));if size(r,1)>1 disp([1,1]);else disp([r,c]);end;

If I could exclude any one case, such as [1,1] in this solution, I could have saved several bytes.

Updated solution:

@(v)find(v-mode(v(:)))

Thanks to @sundar for helping me with the special case problem and saving 42 bytes! Also, thanks to @Luis_Mendo for the suggestions and saving me another 2 bytes!

Röda, 81 bytes

f a{i=indexOf;l=i("

",a)+1;chars a|sort|count|[[_2,_1]]|min|i _[1],a|[_%l,_1//l]}

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Takes input as a string containing newline-terminated lines. Returns a stream containing 0-indexed horizontal and vertical indexes.