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;

int main(){
  int n,k;
  vector<pair<int,int>> s;
  cin>>n>>k;

  int rev=0;
  s.push_back(make_pair(0,0));
  char a,b;
  cin>>a;
  if(a=='0')rev++;
  int l=1;
  for(int i=1; i<n; i++){
    cin>>b;
    if(a=='0' && b=='0')l++;
    if(a=='0' && b=='1'){
        s.push_back(make_pair(l,rev));
        l++;
    }
    if(a=='1' && b=='0'){
      s.push_back(make_pair(l,rev));
      l++;
      rev++;
    }
    if(a=='1' && b=='1')l++;
    a=b;
  }
  s.push_back(make_pair(l,rev));

  int m=s.size(), w=(k+1)*2-1, ans=0;

  if((k+1)*2>m)ans=n;

  if(s[1].second==0){
    for(int i=0; i+w<m; i+=2)ans=max(ans, s[i+w].first-s[i].first);
    if(0<=m-w)ans=max(ans, s[m-1].first-s[m-w].first);
  }else{
    if(m>2)ans=max(ans, s[2].first);
    for(int i=1; i+w<m; i+=2)ans=max(ans, s[i+w].first-s[i].first);
    if(0<=m-w)ans=max(ans, s[m-1].first-s[m-w].first);  
  }
  cout<<ans<<endl;
}