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.Scanner;

public class Main{

	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int length = sc.nextInt();
		int times = sc.nextInt();
		String s = sc.next();
		int counter=0;
		String st[] = s.split("");
		int flag =0;
		if(st[0].equals("1") )
			flag=1;
		
		for(int i=0;i<length-1;i++){
			if(!(st[i].equals(st[i+1]))){
				counter++;
			}
		}
		
		int counter2=0;
		int point[] = new int[counter+1];
		int j=0;
		
		for(int i=0;i<length-1;i++){
			if(!(st[i].equals(st[i+1]))){
				point[j] = i-counter2+1;
				counter2 =i+1;
				j++;
			}
		}
		if(j-1 != counter){
			point[j]=length-1-counter2+1;
		}

		int[] cumulativesum = new int[counter+2];
		cumulativesum[0]=0;
		for(int i=1;i<=counter+1;i++)
			cumulativesum[i]=cumulativesum[i-1]+point[i-1];
		
		int max1=-1;
		int max2=-1;
		int k=0;

		if(counter+1>2*k+1){
			k=2*times+1;
		}else
			k=counter+1;

		if(flag ==1) {
			for(int i=0;i+k<counter+2;i=i+2){
				if( (cumulativesum[i+k]-cumulativesum[i]) >max2) {
					max2=cumulativesum[i+k]-cumulativesum[i];
				}
			}
		}else {
			for(int i=1;i+k<counter+2;i=i+2){
				if( (cumulativesum[i+k]-cumulativesum[i]) >max2) {
					max2=cumulativesum[i+k]-cumulativesum[i];
				}
			}
		}

		System.out.println(Math.max(max1,max2));
	}
}