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: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int k = scanner.nextInt();
			scanner.nextLine();
			char[] s = scanner.nextLine().toCharArray();
			List<Result> list = new ArrayList<>();
			boolean isOne;
			if (s[0] == '0') {
				list.add(new Result(k, 0, 0));
				isOne = true;
			} else {
				isOne = false;
			}
			for (int i = 0; i < s.length; i++) {
				if (isOne && ('0' == s[i])) {
					for (Result result : list) {
						if (result.count <= 0) {
							if (result.k > 0) {
								result.k--;
							} else {
								result.count = i - result.start;
							}
						}
					}
					isOne = false;
				} else if (!isOne && ('1' == s[i])) {
					list.add(new Result(k, i, 0));
					isOne = true;
				}
			}
			for (Result result : list) {
				if (result.count <= 0) {
					result.count = n - result.start;
				}
			}
			list.sort((x, y) -> y.count - x.count);
			System.out.println(list.get(0).count);
		}
	}

	private static class Result {
		int k;
		int start;
		int count;

		Result(int k, int start, int count) {
			super();
			this.k = k;
			this.start = start;
			this.count = count;
		}

		@Override
		public String toString() {
			return "k=" + k + ",start=" + start + ",count=" + count;
		}
	}
}