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 int long long
using namespace std;
main()
{
	//freopen("input.txt","rt",stdin);
	int n,k;
	cin>>n>>k;
	string str;
	cin>>str;
	vector<int>v1;
	vector<int>v2;
	int p =0;
	int p1 =0;
	for(int i=0;i<n;i++)
	{
		if(str[i] == '1')
		{
			if(p1 > 0){
				v2.push_back(p1);
				p1 =0;
			}
			p++;
		}
		else
		{
			if(p > 0){
				v1.push_back(p);
				p =0;
			}
			p1++;
		}
	}
	if(p!= 0)
	{
		v1.push_back(p);
	}
	if(p1 != 0)
	{
		v2.push_back(p1);
	}
	vector<int>ans1;
	vector<int>ans2;
	int s = 0;
	int l = v2.size();
	for(int i=0;i<min(k,l);i++)
	{
		s += v2[i];
	}
	ans1.push_back(s);
	int d =0;
	for(int i=k;i<v2.size();i++)
	{
		s -= v2[d];
		s += v2[i];
		d++;
		ans1.push_back(s);
	}
	int s1 = 0;
	int l1 = v1.size();
	for(int i=0;i<min(k+1,l1);i++)
	{
		s1 += v1[i];
	}
	ans2.push_back(s1);
	int d1 =0;
	for(int i=k+1;i<v1.size();i++)
	{
		s1 -= v1[d1];
		s1 += v1[i];
		d1++;
		ans2.push_back(s1);
	}
	int mx =0;
	for(int i=0;i<ans1.size();i++)
	{
		int a = ans1[i] + ans2[i];
		mx = max(mx,a);
	}
	cout<<mx<<endl;

}