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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int K = sc.nextInt();
        int max = 0;
        char[] S = sc.next().toCharArray();
        int oneCount = 0;
        int zeroCount = 0;
        boolean zeroChainFlag = false;
        int totalChain = 0;
        int oneCountList[] = new int[N];
        int indexForMinus = 1;

        // 尺取り法で実装する
        // そもそも反転させるのは近い0同士が1の連結数が最大となると仮定
        for (int i = 0; i < S.length; i++) {
            // 1のとき
            if (S[i] == '1') {
                // 前までが0の場合、リストに0反転の1を含む、1連結数のリストに挿入
                // リストは、反転回数を超えた場合に、減算する用に使用する
                if (zeroChainFlag) {
                    oneCountList[zeroCount] = oneCount;
                    oneCount = 0;
                    zeroChainFlag = false;
                }
                oneCount++;
                totalChain++;
                // 終端が1のときは、最大値更新をする
                if (i == S.length - 1) {
                    max = Math.max(max, totalChain);
                }
                // 0のとき
            } else {
                // 前までが0のとき、反転し続ける
                if (zeroChainFlag) {
                    oneCount++;
                    totalChain++;
                }
                // 前までが0ではないときの処理
                else {
                    zeroChainFlag = true;
                    zeroCount++;
                    // 指定回数K以下の場合
                    if (zeroCount <= K) {
                        oneCount++;
                        totalChain++;
                        // 指定回数Kを超えた場合
                        // ここがミソ。一番古い反転0を含む1の固まりを合計値から引き算する。
                    } else {
                        // 最大値更新
                        max = Math.max(max, totalChain);
                        // 一番古い反転0を含む1の固まりを合計値から引き算する。
                        totalChain = totalChain - oneCountList[indexForMinus];
                        indexForMinus++;
                        // 現在位置の0の反転分を足す
                        totalChain++;
                        oneCount++;
                    }
                }
            }
        }
        System.out.println(max);
        sc.close();
    }
}