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;
    cin >> n >> k;
    string str;
    cin >> str;

    vector<pair<int, int>> zerSegs;

    int st;

    for(int i = 0; i < n; ++i){
        if(str[i] == '1' && i != 0 && str[i - 1] == '0'){
            zerSegs.push_back({st, i - 1});
        }else if(str[i] == '0' && (str[i - 1] == '1' || i == 0)){
            st = i;
        }
        if(str[i] == '0' && i == n - 1){
            zerSegs.push_back({st, i});
        }
    }

    int numSegs = zerSegs.size();
    if(k >= numSegs){
        cout << n << "\n";
        return 0;
    }

    int mxlen = 0;
    for(int i = 0; i + k <= numSegs; ++i){
        if(i == 0){
            mxlen = max(mxlen, zerSegs[i + k].first);
        }else if(i + k == numSegs){
            mxlen = max(mxlen, n - zerSegs[i - 1].second - 1);
        }else{
            mxlen = max(mxlen, zerSegs[i + k].first + 1 - zerSegs[i - 1].second);
        }
    }

    cout << mxlen << "\n";
    return 0;
}