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.*;

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

      int n = Integer.parseInt(sc.next());
      int k = Integer.parseInt(sc.next());

      String s = sc.next();
      int lng = s.length();
      char[] c = new char[lng];
      for(int i=0;i<lng;i++){
        c[i] = s.charAt(i);
      }

      int[] kore = new int[n];
      int[] or01 = new int[n];
      int cnt = 0;
      for(int i=0;i<n-1;i++){
        if(c[i]==c[i+1]){
          kore[cnt]++;
        }
        else{
          if(c[i]=='0'){
            or01[cnt] = 0;
          }
          else{
            or01[cnt] = 1;
          }
          cnt++;
        }
      }
      for(int i=0;i<=cnt;i++){
        kore[i]++;
      }

//////main////////////////////////////////////
int range = 2*k+1;
int ans = 0;
int max = 0;
for(int i=-k;i<=lng-range;i++){

if(i<0){
//////////////////////////////////
for(int j=0;j<range+i;j++){
  max += kore[j];
}
//////////////////////////////////
}
else if(0<=i && i<=lng-range){
//////////////////////////////////
  if(or01[i]==1){
    for(int j=i;j<i+range;j++){
      max += kore[j];
    }
  }
//////////////////////////////////
}
else if(lng-range<i){
//////////////////////////////////
for(int j=i;j<lng;j++){
  max += kore[j];
}
//////////////////////////////////
}
  ans = Math.max(ans,max);
  max = 0;
}
if(cnt==0){
  ans=n;
}
System.out.println(ans);
//////////////////////////////////////////////

      /*for(int kores : kore){
        System.out.println(kores);
      }*/

    }
}