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;
import java.util.stream.Stream;
import java.util.ArrayList;
public class Main{
    public static void main(String[] args) {
 
 
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt(); 
        int K = sc.nextInt(); 
        String str = sc.next(); 
        String [] strArray = str.split("");
        int[] ia = new int[N];
        int sum = 0;
        int max = 0;
        int kaisu = 0;
        int range = 0;

        for (int i = 0; i < N; i++) {
            ia[i] = Integer.parseInt(strArray[i]); 
        }

        ArrayList<Integer> array = new ArrayList<Integer>();

        int count = 1;

        for(int i = 0; i < N ; i++){
            if( i < (N-1)){
                if(ia[i] == ia[i + 1]){
                    count++; 
                }
                else{
                    array.add(count);
                    count = 1;
                }
            }
            else{
                array.add(count);
            }
        }

//         System.out.println(array);

         if(ia[0] == 1){

            for(int i = 0; i < array.size() ; i = i+2){
                int cnt = i;
                sum = 0;
                for(int j = cnt; j < (range = (cnt + 1 + K * 2) < array.size() ? (cnt + 1 + K * 2) : array.size()); j++){
                    sum = sum + array.get(j) ;
//                    System.out.println(j);
                }
//                System.out.println(i + " " +sum);
               max = Math.max(max, sum);
            }
        }

        if(ia[0] == 0){
            if(array.size() > 1 && K > 1)
                max = array.get(0) + array.get(1);

            if(array.size() == 1 && K == 1)
                max = array.get(0) ;

            if(array.size() == 1 && K == 0)
                max = 0 ;

            for(int i = 1; i < array.size() ; i = i+2){
                int cnt = i;
                sum = 0;
                for(int j = cnt; j < (range = (cnt + 1 + K * 2) < array.size() ? (cnt + 1 + K * 2) : array.size()); j++){
                    sum = sum + array.get(j) ;
//                    System.out.println(j);
                }
//                System.out.println(i + " " +sum);
                max = Math.max(max, sum);
            }
        }

        System.out.println(max);
    }
}