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) {
        // read inputs
        Scanner in = new Scanner(System.in);
        final int N = in.nextInt();
        final int K = in.nextInt();
        final String s = in.next();

        List<int[]> zeros = new ArrayList<>();
        int len = s.length();
        int idx = 0;
        while (idx < len) {
            int j = idx;
            while (idx < len && s.charAt(idx) == '0') idx++;
            while (idx < len && s.charAt(idx) == '1') idx++;
            zeros.add(new int[] {j, idx});
        }

        int max = 0;
        if (zeros.size() <= K) max = len;
        for (int i = K; i < zeros.size(); i++) {
            int l = zeros.get(i-K)[0];
            int r = zeros.get(i)[1]-1;
            max = Math.max(max, r-l);
        }
        System.out.println(max);
    }
}