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: Accepted)

#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main() {
    int n, k;
    cin >> n >> k;
    
    string s;
    cin >> s;
    
    int cntUp = 0;
    int cntDown = 0;
    vector<int> up;
    vector<int> down;
    
    if(s[0] == '0') {
        down.push_back(0);
    }
    
    for(int i = 0; i < s.size(); i++) {
        if(s[i] == '0') {
            cntUp++;
        }
        else {
            cntDown++;
        }
        
        if(i == s.size() - 1 || s[i] != s[i + 1]) {
            if(s[i] == '0') {
                up.push_back(cntUp);
                cntUp = 0;
            }
            else {
                down.push_back(cntDown);
                cntDown = 0;
            }
        }
    }
    
    if(s[s.size() - 1] == '0') {
        down.push_back(0);
    }
    
    if(k >= up.size()) {
        cout << n << endl;
    }
    else {
        int cnt = 0;
        for(int i = 0; i < k; i++) {
            cnt += up[i];
            cnt += down[i];
        }
        cnt += down[k];
        
        int ans = cnt;
        int index = 0;
        while(k + index != up.size()) {
            cnt -= up[index];
            cnt -= down[index];
            
            cnt += up[k + index];
            if(k + index + 1 < down.size()) {
                cnt += down[k + index + 1];
            }
            
            if(ans < cnt) {
                ans = cnt;
            }
            
            index++;
        }
        
        cout << ans << endl;
    }
    
    return 0;
}