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(), K = sc.nextInt();
        sc.nextLine();
        String s = sc.nextLine();
        int ans = 0;
        List<Integer> lbs = new ArrayList<>(), rbs = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) == '0') {
                if (i == 0 || s.charAt(i - 1) == '1') {
                    lbs.add(i);
                }
                if (i == s.length() - 1 || s.charAt(i + 1) == '1') {
                    rbs.add(i);
                }
            }
        }
        if (lbs.size() <= K) {
            ans = n;
        } else {
            ans = Math.max(lbs.get(K), n - rbs.get(rbs.size() - K - 1) - 1);
            for (int i = K + 1; i < lbs.size() - 1; ++i) {
                ans = Math.max(ans, lbs.get(i) - rbs.get(i - K - 1) - 1);
            }
        }
        System.out.println(ans);
    }
}