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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		String S = sc.next();
		int ans1 = 0, ans = 0;
		List<Integer> a = new ArrayList<>();
		int con = 1;
		if(N == 1)
			a.add(1);
		else if(S.charAt(0) != S.charAt(1))
			a.add(1);
		for(int i = 1; i < N; i++) {
			if(S.charAt(i) == S.charAt(i-1))
				con += 1;
			else {
				a.add(con);
				con = 1;
			}
			if(i == N-1)
				a.add(con);
		}
		if(S.charAt(0) == '1') {
			for(int i = 0; i < a.size(); i += 2) {
				ans1 = 0;
				for(int j = i; j <= i + K*2 && j < a.size(); j++) {
					ans1 += a.get(j);
				}
				ans = Math.max(ans, ans1);
			}
		}
		else {
			for(int i = 0; i < K*2 && i < a.size(); i++)
				ans += a.get(i);
			for(int i = 1; i < a.size(); i += 2) {
				ans1 = 0;
				for(int j = i; j <= i + K*2 && j < a.size(); j++) {
					ans1 += a.get(j);
				}
				ans = Math.max(ans, ans1);
			}
		}
		System.out.println(ans);
	}

}