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 = {
    solve
  }

  def solve(): Unit = {
    val sc = new java.util.Scanner(System.in)
    val n, k = sc.nextInt()
    sc.nextLine()
    val s = sc.nextLine.toCharArray

    val nums = ArrayBuffer[Int]()
    var now = 1
    var cnt = 0
    for (i <- 0 until n) {
      if (s(i) == ('0' + now).toChar) {
        cnt += 1
      } else {
        nums.append(cnt)
        now = 1 - now
        cnt = 1
      }
    }
    if (cnt != 0) nums.append(cnt)

    // k=1の場合、1-0-1の1つ分の合算値をチェックしていく
    // 1-0-1-0 のように0の区間で終了していると、区間をチェックできない
    //  -> 0区間で終わる場合は、1の区間を末尾に付け足す（数値は0）
    if (nums.size % 2 == 0) nums.append(0)

    // 累積和をとる
    val sum = nums.scanLeft(0)(_ + _)

    // チェックする区間（k + 1個の1の区間 + k個の0の区間）
    val add = 2 * k + 1

    var ans = 0
    for (i <- 0 until nums.size by 2) {
      val left = i
      val right = math.min(i + add, nums.size)
      val tmp = sum(right) - sum(left)

      ans = math.max(ans, tmp)
    }

    println(ans)
  }
}