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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int main(){
    int n,k;
    string s;
    cin >> n >> k >> s;
    int ans = 0;
    int j;
    
    for(int i = 0;i < n-ans;i ++){
        int cnt = 0;
        if(s.at(i) == '0'){
            continue;
        }
        if(i > 0 &&s.at(i) == '1' && s.at(i-1) == '1'){
            continue;
        }
        for(j = i;j < n-1;j ++){
            if(s.at(j) != s.at(j+1)){
                cnt ++;
            }
            if(cnt == 2*k+1){
                break;
            }
        }
        ans = max(ans,j-i+1);

        if(j == n-1){
            break;
        }

    }

    vector<char> t(n);
    for(int i = 0;i < n;i ++){
        t.at(i) = s.at(n-1-i);
    }

    for(int i = 0;i < n-ans;i ++){
        int cnt = 0;
        if(t.at(i) == '0'){
            continue;
        }
        if(i > 0 &&t.at(i) == '1' && t.at(i-1) == '1'){
            continue;
        }

        for(j = i;j < n-1;j ++){
            if(t.at(j) != t.at(j+1)){
                cnt ++;
            }
            if(cnt == 2*k+1){
                break;
            }

        }
        ans = max(ans,j-i+1);

        if(j == n-1){
            break;
        }

    }

    for(int i = 0;i < n;i ++){
        if(s.at(i) == '1'){
            break;
        }
        if(i == n-1){
            if(k == 1){
                ans = 1;
            }else{
                ans = n;
            }
        }
    }

    cout << ans << endl;
}