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 n,k,a[100010],ans=0;
int b[100010],c[100010],cnt=0;

int main()
{
	cin>>n>>k;
	b[0]=0;
	for(int i=1;i<=n;i++)
	{
		char ch;
		cin>>ch;
		if(ch=='1') a[i]=1;
		else a[i]=0;
	}
	int ch=2;
	for(int i=1;i<=n;i++)
	{
		if(a[i]!=ch) ch=a[i],c[++cnt]=ch,b[cnt]++;
		else b[cnt]++;
	}
	if(c[1]==1)
	{
		for(int j=1;j<=cnt;j+=2)
		{
			int s=0;
			for(int i=j;i<=min(j+k*2,cnt);i++) s+=b[i];
			if(c[min(j+k*2,cnt)]==0) s-=b[min(j+k*2,cnt)];
			ans=max(ans,s);
		}
	}
	else
	{
		for(int j=2;j<=cnt;j+=2)
		{
			int s=0;
			for(int i=j-2;i<=min(cnt,j+k*2-2);i++) s+=b[i];
			if(c[min(cnt,j+k*2-2)]==0) s-=b[min(cnt,j+k*2-2)];
			ans=max(ans,s);
		}
	}
	cout<<ans<<endl;
	return 0;
}