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.Arrays;
import java.util.Scanner;

class Main {
    static int K;

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        K = sc.nextInt();
        sc.nextLine();
        String S = sc.nextLine();
        ArrayList<Integer> sum = new ArrayList<Integer>();
        int f = Integer.parseInt(S.substring(0, 1));
        int cnt = f > 0 ? 1 : -1;
        sum.add(0);

        for (int i = 1; i < S.length(); i++) {
            if (cnt > 0) {
                if (S.charAt(i) == '1') {
                    cnt++;
                } else {
                    sum.add(cnt);
                    cnt = -1;
                }
            } else {
                if (S.charAt(i) == '0') {
                    cnt--;
                } else {
                    sum.add(cnt);
                    cnt = 1;
                }
            }
        }
        sum.add(cnt);
        sum.add(0);

        if(sum.get(1)>0)sum.remove(0);


        // System.out.println(Arrays.toString(sum.toArray()));
        int[] dp = new int[sum.size()+1];
        int val = 0;
        for (int i = 0; i < sum.size(); i++) {
            val += Math.abs(sum.get(i));
            dp[i+1]=val;
        }
        int max=0;
        for (int i = 0; i < sum.size(); i+=2) {
            // int val = 0;
            // int magic = 0;
            // for (int j = i; j <= i + K * 2 && j<sum.size(); j++) {
            //     // System.out.println(i + K * 2);
            //     val += Math.abs(sum.get(j));
            // }
            max=Math.max(max,dp[Math.min(i + K * 2+1,sum.size()-1)]-dp[i]);
        }

        System.out.println(max);
    }

    // static int hand(ArrayList<Integer> sum, int tg, int start, int cnt, int val) {
    //     if (cnt >= k)
    //         return val;

    //     if (tg < -1 || tg < sum.size())
    //         return val;
    //     if (tg == -1)
    //         return val + Math.abs(sum.get(0));
    //     else if (tg == sum.size())
    //         return val + Math.abs(sum.get(sum.size() - 1));

    //     return Math.max(hand(sum, tg - 2, start, cnt + 1, val), hand(sum, tg + 2, start, cnt + 1, val));
    // }
}