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 => in}
import scala.collection.{mutable => mu}
import java.lang.{Character => char}

object Main extends App {
  val Array(n,k) = in.readLine.split(" ").map(_.toInt)
  val s = in.readLine.map(char.getNumericValue)

  val table = mu.ArrayBuffer.empty[Int]
  s./:(-1)( (b, i) => {
    val v = if(i == 1) 1 else -1
    if(i != b) table.append(v)
    else table(table.size-1) += v
    i
  })

  val sums = table.scanLeft(0)((acc,i) => acc + i.abs)
  0 +=: table

  def rec(acc:Int, cnt:Int, left:Int, right:Int, now:Int):Int = {
    if(right == table.size) acc max (sums(right-1) - sums(left))
    else {
      val leftCnt = if(table(left) < 0) 1 else 0
      if(cnt-leftCnt == k && now == 1)
        rec(acc max (sums(right) - sums(left)), cnt-leftCnt, left+1, right, now)
      else
        rec(acc, cnt+(now^1), left, right+1, now^1)
    }
  }

  val ans = rec(0,0,0,1,s.head)
  println(ans)
}