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.ArrayList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		int n=scn.nextInt();
		int k=scn.nextInt();
		String str=scn.next();
		ArrayList<Integer> al=new ArrayList<>();
		int cnt=1;
		for(int i=1;i<str.length();i++) {
			if(str.charAt(i)==str.charAt(i-1)) {
				cnt++;
			}else {
				al.add(cnt);
				cnt=1;
			}
		}
		al.add(cnt);
		long []sum=new long[al.size()];
		for(int j=0;j<al.size();j++) {
			if(j>0) {
				sum[j]+=sum[j-1]+al.get(j);
			}else {
				sum[j]=al.get(j);
			}
		}
//		System.out.println(al.size());
//		for(int i=0;i<al.size();i++) {
//			System.out.print(al.get(i)+" ");
//		}
//		System.out.println();
		n=al.size()/2;
		if(str.charAt(0)=='0'&&al.size()%2!=0)
			n++;
		if(k>=n) {
			System.out.println(sum[al.size()-1]);
		}
		else if(str.charAt(0)=='1') {
			
			long max=sum[1+(k-1)*2+1<al.size()?1+(k-1)*2+1:1+(k-1)*2];
			for(int j=3;j<al.size();j+=2) {
				if(j+(k-1)*2+1<al.size()&&sum[j+(k-1)*2+1]-sum[j-2]>max) {
					max=sum[j+(k-1)*2+1]-sum[j-2];
				}
			}
			System.out.println(max);
		}else {
			long max=sum[(k-1)*2+1<al.size()?(k-1)*2+1:(k-1)*2];
			for(int j=2;j<al.size();j+=2) {
				if(j+(k-1)*2+1<al.size()&&sum[j+(k-1)*2+1]-sum[j-2]>max) {
					max=sum[j+(k-1)*2+1]-sum[j-2];
				}
			}
			System.out.println(max);
		}
	}

}