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 Scala (2.11.7) (judge verdict: Accepted)

import scala.collection.mutable.ArrayBuffer

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)
    val N, K = sc.nextInt
    val S = sc.next

    val starts = ArrayBuffer[Int](0)
    val ends = new ArrayBuffer[Int]()
    var prev = ' '
    for (i <- 0 until S.length) {
      if (prev == '0' && S(i) == '1') {
        starts += i
      }
      if (prev == '1' && S(i) == '0') {
        ends += i
      }
      prev = S(i)
    }
    ends += S.length

    var result = 0
    for (i <- ends.indices) {
      val count =
        if (S(0) == '1') {
          if (i - K >= 0)
            ends(i) - starts(i - K)
          else
            0
        } else {
          if (i - K + 1 >= 0)
            ends(i) - starts(i - K + 1)
          else
            0
        }
      if (result < count) {
        result = count
      }
    }
    if (result == 0) {
      println(S.length)
    } else {
      println(result)
    }

  }
}