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.ArrayList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = Integer.parseInt(sc.next());
		int k = Integer.parseInt(sc.next());
		
		char []c = sc.next().toCharArray();
		ArrayList <Integer> List = new ArrayList <Integer>();
		
		int cnt = 0;
		char now = '1';
		
		for(int i = 0 ; i < n ;i++) {
			if(now == c[i]) {
				cnt++;
			}
			else {
				List.add(cnt);
				now = (char)((now - '0' + 1) % 2 + '0');
				cnt = 1;
			}
		}
		
		List.add(cnt);
		
		if(List.size() % 2 == 0) {
			List.add(0);
		}
		int m = List.size();
	
		int [] sum = new int [m+1];
		
		for(int i = 0 ; i < m ; i++) {
			sum[i+1] = sum[i] + List.get(i);
		}
		
		
		int max = 0;
		int K = 2 * k + 1;
		
		for(int i = 0 ; i <= m ; i+=2) {
			int j = Math.min(i + K , m);
			//System.out.println(i+" "+j+" "+(sum[j] - sum[i]));
			max = Math.max(max, sum[j] - sum[i]);
		}
		
		System.out.println(max);
	}		
}