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>
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
 //Compiler version g++ 6.3.0

 int main()
 {
 	int n,m;
 	cin >> n>>m;
 	string x;
 	cin >> x;
 	vector<int> a;
 	bool s=false;
 	if(x[0]=='1'){
 		s=true;
 	}
 	int r=0;
 	int l=0;
 	a.push_back(0);
 	for (int i = 1; i < n+1; i++){
 		if(x[i]!=x[i-1]){
 			r++;
 			l+=r;
 			a.push_back(l);
 			r=0;
 		}else{
 			r++;
 		}
 	}
 	if(r!=0){
 		r++;
 	}
 	int sz=a.size();
 	if(sz/2<=m&&s){;
 		cout << n << endl;
 		return 0;
 	}
 	if(!s&&(sz+1)/2<=m){
 		cout << n << endl;
 		return 0;
 	}
 	int ans=0;
 	if(s){
 		
 		int sz=a.size();
 		for (int i = 0; i < (sz-1)/2-m; i++){
 			ans=max(ans,a.at((i+m)*2-1)-a.at(i*2));
 		}
 		
 		if(sz%2==0){
 			ans=max(ans,a.back()-a.at(sz-m*2-2));
 		}else{
 			ans=max(ans,a.back()-a.at(sz-m*2-1));
 		
 		}
 	}else{
 		ans=a.at(m*2);
 		for (int i = 1; i < (a.size()+1)/2-m; i++){
 			ans=max(ans,a.at((i+m)*2)-a.at(i*2-1));
 		}
 		if(a.size()%2==0){
 			ans=max(ans,a.back()-a.at(a.size()-m*2-1));
 		}else{
 			ans=max(ans,a.back()-a.at(a.size()-m*2-2));
 		}
 	}
 	cout << ans << endl;
 }