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 java.util.Scanner

import scala.collection.mutable.ArrayBuffer


object Main {
  def read(): (Int, Int, String) = {
    val sc = new Scanner(System.in)
    val n, k = sc.nextInt()
    val s = sc.next()
    (n, k, s)
  }

  def compressRunlength(s: String): IndexedSeq[(Int, Char)] = {

    var currentChar = s(0)
    var currentCount = 1
    var buf = ArrayBuffer.empty[(Int, Char)]
    for (i <- 1 until s.length) {
      if (currentChar == s(i)) {
        currentCount += 1
      } else {
        buf.append((currentCount, currentChar))
        currentChar = s(i)
        currentCount = 1
      }
    }
    buf.append((currentCount, currentChar))
    buf.toIndexedSeq
  }

  def solve(n: Int, k: Int, s: String): Int = {
    // s.length == n
    val runLength = compressRunlength(s)
    val cumSum = runLength.map(_._1).scanLeft(0)(_ + _)

    // (i, ..., j-1)の和: cumSum(j)-cumSum(i)
    def sum(min: Int, max: Int): Int = cumSum(max + 1) - cumSum(min)

    def cut(i: Int) = Math.min(i, runLength.length - 1)

    // i番目が0: i, i+1,...,i+2k-1
    // i番目が1: i, i+1,...,i+2k
    runLength.indices
      .map(i =>
        if (runLength(i)._2 == '0') (i, cut(i + 2 * k - 1)) else (i, cut(i + 2 * k)))
      .map { case (min, max) =>
        sum(min, max)
      }.max
  }


  def main(args: Array[String]): Unit = {
    val (n, k, s) = read()
    println(solve(n, k, s))
  }
}