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 = sc.nextInt();
    int K = sc.nextInt();
    String S = sc.next();
    
    int zeroSeqCount = 0;
    int tail = 0;
    // set initial tail
    while (tail < N) {
      char c = S.charAt(tail);
      if (c == '0') {
        zeroSeqCount++;
        if (zeroSeqCount == K+1) {
          break;
        }
        tail++;
        while (tail < N && S.charAt(tail) == '0') {
          tail++;
        }
      } else {
        tail++;
      }
    }
    if (tail == N) {
      System.out.println(N);
      return;
    }

    int head = 0;
    int max = tail - head;
    while (tail < N) {

      //
      // move tail
      //

      // skip seq 0
      while (tail < N && S.charAt(tail) == '0') {
        tail++;
      }
      // move to head of next 0 seq
      while (tail < N && S.charAt(tail) != '0') {
        tail++;
      }
      
      //
      // move head
      //
      while (head < tail && S.charAt(head) != '0') {
        head++;
      }
      while (head < tail && S.charAt(head) == '0') {
        head++;
      }
      
      max = Math.max(max, tail-head);
    }
    
    System.out.println(max);
  }  
}