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.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        String s = in.next();
        char first = s.charAt(0);
        List<Integer> l = new ArrayList<>();
        char prevChar = first;
        int currentStreak = 1;
        for(int i=1; i<n; i++) {
            char c = s.charAt(i);
            if(prevChar == c) {
                currentStreak++;
            }
            else {
                l.add(currentStreak);
                currentStreak = 1;
                prevChar = c;
            }
        }
        l.add(currentStreak);

        int result = 0;
        for(int i=0; i<l.size(); i++) {
            boolean subListStartsWith0 = (first == '0') ^ (i%2==1);
            int subListLength = k*2+1;
            if(subListStartsWith0) {
                subListLength = k*2;
            }
            List<Integer> subList = l.subList(i, Math.min(l.size(), i+subListLength));
            int score = 0;
            for(int j=0; j<subList.size(); j++) {
                score += subList.get(j);
            }
            result = Math.max(result, score);
        }
        System.out.println(result);
    }
}