[1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1]

[1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1]

 1, 1, 1
          2, 2
                1, 1, 1, 1
                            2, 2, 2
                                     1, 1, 1

[3, 3, 3, 4, 4, 4, 4, 5, 5, 4, 4, 3, 3, 4, 4]

 3, 3, 3
          4, 4, 4, 4
                      5, 5
                            4, 4
                                  3, 3
                                        4, 4

[3, 3], [4, 4], [4, 4] and [5, 5].

Input: [1, 1, 2, 2, 3, 3, 4]
Output: [4]

Input: [1]
Output: [1]

Input: [1, 1, 10, 10, 10, 100, 100]
Output: [1, 1], [100, 100]

mM_MmhbrQ8

mr]d9.mhbrQ8

MinimalBy[Length]@*Split

import Data.Lists
argmins length.group

void f(string o){var r=Regex.Matches(o,@"([0-9])1{0,}").Cast<Match>().OrderBy(x=>x.Groups[0].Value.Length);foreach(var s in r){foreach(var z in r)if(s.Length>z.Length)return;Console.WriteLine(s.Value);}}

public static void f(string inp)
    {

        var r = Regex.Matches(inp, @"([0-9])1{0,}").Cast<Match>().OrderBy(x => x.Groups[0].Value.Length);

        foreach (Match s in r)
        {
            foreach (Match z in r)
                if (s.Length > z.Length)
                    return;

        Console.WriteLine(s.Value);
        }


    }

x=input()
r=[q[2]for q in filter(lambda l:(len(l[2])>0)&((l[0]<1)or(x[l[0]-1]!=x[l[0]]))&((l[1]>len(x)-1)or(x[l[1]]!=x[l[1]-1]))&(len(filter(lambda k:k==l[2][0],l[2]))==len(l[2])),[(a,b,x[a:b])for a in range(0,len(x))for b in range(0,len(x)+1)])]
print filter(lambda k:len(k)==min([len(s)for s in r]),r)

Y'tX<tb=bw)wTX"

1 100
1 100

Y'    %// Run length encoding, gives 2 vectors of group-lengths and values
t     %// Duplicate group lengths
X<    %// Minimum group length
tb    %// Duplicate and get vector of group lengths to the top
=     %// Find which group lengths are equal to the minimum
bw)   %// And get the values of those groups
wTX"  %// Repeats the matrix of minimum-length-group values by the minimum group length

I0;œṗ¹L=¥ÐfL€Ṃ$$

I0;œṗ¹L=¥ÐfL€Ṃ$$     Main link. List: z = [a,b,c,...]

I                    Compute [b-a, c-b, d-c, ...]
 0;                  Concatenate 0 in front: [0, b-a, c-b, d-c, ...]
   œṗ                Split z where the corresponding item in the above array is not zero.
      L=¥Ðf          Filter sublists whose length equal:
           L€Ṃ$      the minimum length throughout the list.

     ¹         $     (grammar stuffs)

a=>a.map(n=>n==c[0]?c.push(n):b.push(c=[n]),c=b=[])&&b.sort((a,b)=>a[l]-b[l],l='length').filter(e=>e[l]==b[0][l])

M!`b(d+)(,1b)*
(,()|.)+
$#2:$&
O#`.+
s`^(.*b(.+:).*)¶(?!2).+
$1
.+:
<empty-line>

M!`b(d+)(,1b)*

1,1
10,10,10
100,100

(,()|.)+
$#2:$&

1:1,1
2:10,10,10
1:100,100

O#`.+

1:1,1
1:100,100
2:10,10,10

s`^(.*b(.+:).*)¶(?!2).+
$1

1:1,1
1:100,100

.+:
<empty-line>

1,1
100,100

{z/⍨(⊢=⌊/)≢¨z←(1,2≠/⍵)⊂⍵}

[:(#~[:(=<./)#@>)]<;.1~1,2~:/]

f =: [:(#~[:(=<./)#@>)]<;.1~1,2~:/]
   f 1 1 2 2 3 3 4
┌─┐
│4│
└─┘
   f 3 3 3 4 4 4 4 5 5 4 4 3 3 4 4
┌───┬───┬───┬───┐
│5 5│4 4│3 3│4 4│
└───┴───┴───┴───┘

[:(#~[:(=<./)#@>)]<;.1~1,2~:/]  Input: s
                              ]  Identity function, get s
                         2       The constant 2
                                Operate on each overlapping sublist of size 2
                          ~:/      Check if each pair is unequal, 1 if true else 0
                       1,        Prepend a 1 to that list
                 ]               Identity function, get s
                  <;.1~          Using the list above, chop s at each true index
[:(             )                Operate on the sublists
             #@>                 Get the length of each sublist
     [:(    )                    Operate on the length of each sublist
         <./                     Get the minimum length
        =                        Mark each index as 1 if equal to the min length else 0
   #~                            Copy only the sublists with min length and return

#(let[G(group-by count(partition-by + %))](G(apply min(keys G))))

ḅlᵒlᵍh

ḅ         The list of runs of consecutive equal elements of
          the input
 lᵒ       sorted by length
   lᵍ     and grouped by length
          has the output variable
     h    as its first element.