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 Java (OpenJDK 11.0.6) (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();
		char[] S = sc.next().toCharArray();
		sc.close();

		List<Integer> list = new ArrayList<Integer>();
		int cnt = 0;
		int prev = S[0];

		// 連続している回数を数える
		for (int i = 0; i < N; i++) {
			if (S[i] == prev) {
				cnt++;
			} else {
				if (prev == '0') {
					prev = '1';
				} else {
					prev = '0';
				}
				// 値の追加
				list.add(cnt);
				cnt = 1;
			}
		}
		// for last
		list.add(cnt);

		// 最初と最後が0なら、listに0を足しておく
		if (S[0] == '0') {
			list.add(0, 0);
		}
		if (S[N - 1] == '0') {
			list.add(0);
		}

		int max = 0;
		int score = 0;
		int idx = S[0] == '0' ? 1 : 0;
		for (int i = 0; i < list.size(); i++) {
			// 最初の2K+1回はただ足し算
			if (i < 2 * K + 1) {
				score += list.get(i);
			} else {
				if (i % 2 != 0) {
					int minus = list.get(idx) + list.get(idx + 1);
					int plus = list.get(i) + list.get(i + 1);
					score = score - minus + plus;
					idx += 2;
				}
			}
			max = Math.max(max, score);
		}

		System.out.println(max);
	}
}