Find the substring with the most 1's in a sequence

Dyalog APL, 11

(-∘1+⍳⌈/)+/

Try it here. Usage:

   f ← (-∘1+⍳⌈/)+/
   4 f 0 1 1 0 1 1 1 0 0 0 0 1 1
1

Explanation

This is a dyadic (meaning binary) function that takes the substring length from the left, and the sequence from the right. Its structure is the following:

   ┌───┴────┐
 ┌─┴──┐     /
 ∘  ┌─┼─┐ ┌─┘
┌┴┐ + ⍳ / +  
- 1   ┌─┘    
      ⌈      

Explanation by explosion:

(-∘1+⍳⌈/)+/
(       )+/  ⍝ Take sums of substrings of given length, and feed to function in parentheses
    + ⌈/     ⍝ The array of sums itself, and its maximum
     ⍳       ⍝ First index of right argument in left
 -∘1         ⍝ Subtract 1 (APL arrays are 1-indexed)

As an example, let's take 4 and 0 1 1 0 1 1 1 0 as inputs. First we apply the function +/ to them and get 2 3 3 3 3. Then, + and ⌈/ applied to this array give itself and 3, and 2 3 3 3 3 ⍳ 3 evaluates to 2, since 3 first occurs as the second element. We subtract 1 and get 1 as the final result.

Ruby, 42

f=->s,n{(0..s.size).max_by{|i|s[i,n].sum}}

Takes input by calling it, e.g.

f['01001010101101111011101001010100010101101010101010101101101010010110110110',5]

This compares substrings using their total ASCII value and returns the index of the maximum. I'm not sure if max_by is required by the Ruby spec to be stable but it appears to be in the C implementation.

Python 2, 56

lambda s,l:max(range(len(s)),key=lambda i:sum(s[i:i+l]))

Accepts an array of integers, then the length.