Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef vector<int> vi;

int main(){
  int N,K;cin>>N>>K;
  string S;cin >>S;

  vi zero_endpts;
  vi one_endpts;
  

  for (int i=0;i<N;++i){
    if(i==N-1){
      if(S[i]=='0'){zero_endpts.push_back(i+1);}
      else{one_endpts.push_back(i+1);}
    }
    else{
      if(S[i]=='0'&&S[i+1]=='1'){zero_endpts.push_back(i+1);}
      else if (S[i]=='1'&&S[i+1]=='0'){one_endpts.push_back(i+1);}
    }
  }

  int ans=0;

  int M=zero_endpts.size();
  int L=one_endpts.size();

  if(M<=K){ans=N;}
  else{
    int best_score=0;
    if(S[0]=='0'){
      best_score=one_endpts[K-1];
      for(int i=K;i<M;++i){
        if(zero_endpts[i]!=N-1){
          best_score=max(best_score,one_endpts[i]-zero_endpts[i-K]);
        }
        else{
          best_score=max(best_score, zero_endpts[i]-zero_endpts[i-K]);
        }
      }
    }

    else{
      best_score=one_endpts[K];
      for (int i=K;i<M;++i){
        if(zero_endpts[i]!=N-1){
          best_score=max(best_score,one_endpts[i+1]-zero_endpts[i-K]);
        }
        else{
          best_score=max(best_score, zero_endpts[i]-zero_endpts[i-K]);
        }
      }
      ans=best_score;
    }

  }
  

  cout << ans << endl;
  return 0;
}