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

public class Main {


	public static void main(String args[]) {

		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		sc.nextLine();
		String s = sc.nextLine();

		List<Integer> stand = new ArrayList<>();
		List<Integer> hand = new ArrayList<>();

		char flag = '0';
		int count = 0;
		for(int i=0;i<N;i++) {

			if(s.charAt(i) == flag) {
				count++;
			}else if(flag == '0'){
				stand.add(count);
				count = 1;
				flag = '1';
			}else {
				hand.add(count);
				count = 1;
				flag = '0';
			}
		}

		if(flag == '0') {
			stand.add(count);
			flag = '1';
			count = 0;
		}
		if(flag == '1') hand.add(count);

		for(int i=0;i<K;i++) {
			int index = maxhandstandindex(stand, hand);

			if(index != 0) {
				hand.set(index-1, hand.get(index-1) + stand.get(index) + hand.get(index));
				hand.remove(index);
				stand.remove(index);
			}else {
				hand.set(index, stand.get(index) + hand.get(index));
				stand.set(index, 0);
			}
		}

		System.out.println(Collections.max(hand));
	}

	public static int maxhandstandindex(List<Integer> stand,List<Integer> hand) {
		List<Integer> sum = new ArrayList<>();

		for(int i=0;i<Math.min(stand.size(), hand.size());i++) {
			sum.add(stand.get(i) + hand.get(i));
		}

		return sum.indexOf(Collections.max(sum));
	}

}