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;
    int ans = 0, j=0, tmp=1, flag = 0, sum=0, ans2 =0, len;
    cin >> N >> K;
    int han1 = 2*K+1;
    int han2 = 2*K+2;
    vector<int> v(N);
    string s;
    cin >> s;
    
    for(int i=1; i <= s.length(); i++){
	
	if(s[i-1] == s[i])
	    tmp++;
	else
	    flag++;
	if(flag == 1){
	    flag--;
	    v[j] = tmp;
	    j++;
	    tmp=1;
	}
    }
    for(int i=0; v[i] != 0; i++){
	// cout << v[i] << endl;
	len = i;
    }

    if(len > han1){
	for(int i=0; i<han1; i++){
	    sum += v[i];
	}
	ans = sum;
    
	for(int i=0; i<N-han1; i++){	
	    sum -= v[i];
	    sum += v[i+han1];
	    ans = max(ans, sum);
	}
    }
    else{
	ans = 0;
	for(int i=0; i < len; i++){
	    // cout << v[i] << endl;
	    ans += v[i];
	}
		
    }
    /////
    // if(len > han2){
    // 	sum=0;
    // 	for(int i=0; i<han2; i++){
    // 	    sum += v[i];
    // 	}
    // 	ans2 = sum;

    // 	for(int i=0; i<N-han2; i++){	
    // 	    sum -= v[i];
    // 	    sum += v[i+han2];
    // 	    ans2 = max(ans, sum);
    // 	}
    // }
    // else{
    // 	for(int i=0; i<len; i++){
    // 	    ans2 += v[i];
    // 	}
	
    // }

    
    cout << ans  << endl;
        
    
 }