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.io.StdIn
import scala.math._
import scala.collection.mutable.ArrayBuffer


object Main {

  class Zero(var l: Int, var r: Int) {
    override def toString: String = s"[$l,$r]"
  }

  def main(args: Array[String]): Unit = {
    val Array(n: Int, k: Int) = StdIn.readLine.split(" ").map(_.toInt)
    val as = StdIn.readLine.map(_ == '1')

    val B = ArrayBuffer[Zero]()
    B += new Zero(-1, -1)  // Sentinel
    B += new Zero(-1, -1)

    REP(n) { i =>
      if (!as(i)) {
        if (B.last.l == -1) B.last.l = i
        B.last.r = i
      } else if (B.last.l != -1) {
        B += new Zero(-1, -1)
      }
    }
    if (B.last.l != -1) B += new Zero(-1, -1)
    B.last.l = n
    B.last.r = n

    debug(B.mkString(" "))

    var ans = 0
    REP(B.length - 1) { l =>
      val r = min(l + k + 1, B.length - 1)
      val len = B(r).l - 1 - (B(l).r + 1) + 1
      ans = max(ans, len)
    }
    println(ans)

    debug(n, k)
    as.foreach {i => debugNoNewline(i, "")}
  }

  def DEBUG(f: => Unit): Unit = {
    if (false) { f }
  }

  def debug(something: Any): Unit = DEBUG {
    System.err.println(something)
  }
  def debugNoNewline(something: Any): Unit = DEBUG {
    System.err.print(something)
  }

  def REP(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = offset
    val N = n + offset
    while (i < N) { f(i); i += 1 }
  }

}