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.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        String S=  sc.next();

        List<Integer> list = new ArrayList<>();
        List<Character> list01 = new ArrayList<>();
        char prev = S.charAt(0);
        int count =1;
        for (int i = 1; i < S.length(); i++) {
            char c = S.charAt(i);
            if(prev==c){
                count++;
            }else{
                list.add(count);
                list01.add(prev);
                count=1;
            }
            prev=c;
        }
        list.add(count);
        list01.add(prev);

        if(list01.get(0)=='0'){
            list.add(0,0);
            list01.add(0,'1');
        }
        if(list01.get(list01.size()-1)=='0'){
            list.add(0);
            list01.add('1');
        }
        int ans=0;
        for (int i = 0; i < K; i++) {
            ans = merge(list,list01);
        }
        System.out.println(ans);

    }

    static private int merge(List<Integer> list,List<Character> list01){
        int max=0;
        int maxIdx =-1;
        for (int i = 0; i < list.size() ; i++) {
            if(list01.get(i)=='0') {
                int potentialSize =  list.get(i) ;
                if(i-1>=0){
                    potentialSize +=list.get(i-1);
                }
                if(i+1<list.size()){
                    potentialSize +=list.get(i+1);
                }

                if(potentialSize>max) {
                    max = potentialSize;
                    maxIdx=i;
                }
            }
        }

        //No zero
        if(maxIdx==-1) return list.get(0) ;

        if(maxIdx-1>=0){
            list.set(maxIdx-1,max);
            list.remove(maxIdx);
            list01.remove(maxIdx);
            if(maxIdx<=list.size()-1){
                list.remove(maxIdx);
                list01.remove(maxIdx);
            }
        }else{
            list.set(maxIdx,max);
            if(maxIdx+1<=list.size()-1){
                list.remove(maxIdx+1);
                list01.remove(maxIdx+1);
            }
        }
        return  max;
    }


}