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(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;

    vector<int> B, W;
    int cnt = 1;
    for(int i = 1; i < n; i++){
        if(s[i - 1] == s[i]){
            cnt++;
            if(i == n - 1){
                if(s[i - 1] == '0'){
                    B.push_back(cnt);
                }else{
                    W.push_back(cnt);
                }
            }
        }else{
            if(s[i - 1] == '0'){
                B.push_back(cnt);
            }else{
                W.push_back(cnt);
            }
            cnt = 1;
        }
    }
    if(cnt == 1){
        if(s[n - 1] == '0'){
            B.push_back(cnt);
        }else{
            W.push_back(cnt);
        }
    }

    int ans = 0;
    vector<int> Ba, Wa;
    Ba.push_back(0);
    Wa.push_back(0);
    for(int i = 0; i < B.size(); i++){
        Ba.push_back(Ba[Ba.size() - 1] + B[i]);
    }
    for(int i = 0; i < W.size(); i++){
        Wa.push_back(Wa[Wa.size() - 1] + W[i]);
        ans = max(ans, W[i]);
    }

    
    if(Ba.size() == Wa.size()){
        for(int i = k; i < Ba.size(); i++){
            ans = max(ans, Ba[i] - Ba[i - k] + Wa[i] - Wa[i - k]);
        }
    }else if(Ba.size() > Wa.size()){
        for(int i = k + 1; i < Ba.size(); i++){
            ans = max(ans, Ba[i] - Ba[i - k - 1] + Wa[i - 1] - Wa[i - k]);
        }
    }else{
        for(int i = k; i < Ba.size(); i++){
            ans = max(ans, Ba[i] - Ba[i - k] + Wa[i + 1] - Wa[i - k]);
        }
    }

    cout << ans << endl;

    return 0;
}