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: Accepted)


import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        String[] s = sc.next().split("");
        int[] len = new int[n + 3];
        int index = 0;
        boolean is0 = false;
        for (int i = 0; i < s.length; i++) {
            if (s[i].equals("0")) {
                if (!is0)index++;
                is0 = true;
            } else {
                if (is0)index++;
                is0 = false;
            }
            len[index]++;
        }
        int res = 0;
        for(int j=0;j<2*k+1;j++){
            if(j>index)break;
            res += len[j];
        }
        int final_res = res;
        for (int i = 2; i <= index; i += 2) {
            int tmp = 0;
            if(i+2*k-1>index)break;
            tmp = res - len[i-2] - len[i-1] + len[i+2*k] + len[i+2*k-1];
            res = tmp;
            final_res = Math.max(tmp, final_res);
        }
        System.out.println(final_res);
    }
}