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: Runtime Error)


import java.util.Arrays;
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();
		char[] list = sc.next().toCharArray();
		int number = serch(list, K);
		System.out.println(number);
	}

	static int serch(char[] list, int K) {
		int count = 0;
		int number = 0;
		boolean flag = false;
		char head = list[0];
		boolean induction = false;
		boolean indflag = true;
		int indnum = 0;
		for(int i = 0; i < list.length; i++) {
			if(list[i] == '0') {
				if(indflag) {
					indflag = false;
					induction = true;
				}
				if(!flag) {
					count++;
					flag = true;
				}
				if(count > K) {
					break;
				}
				number++;
			} else {
				if(induction) {
					indnum = serch(Arrays.copyOfRange(list, i, list.length), K);
					induction = false;
				}
				flag = false;
				number++;
			}
		}

		return Integer.max(number, indnum);
	}

}