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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"fmt"
)

func maxInt(a, b int) int {
	switch {
	case a >= b:
		return a
	default:
		return b
	}
}

func main() {
	var n, k int
	var str string
	fmt.Scan(&n, &k, &str)

	var partLength []int
	prevChar := str[0:1]
	cnt := 1
	for i := 1; i < n; i++ {
		if str[i:i+1] == prevChar {
			cnt++
		} else {
			partLength = append(partLength, cnt)
			prevChar = str[i : i+1]
			cnt = 1
		}
	}
	partLength = append(partLength, cnt)

	var cumSum []int
	cumSum = append(cumSum, 0)
	for i := range partLength {
		cumSum = append(cumSum, cumSum[i]+partLength[i])
	}
	if str[0:1] == "0" {
		cumSum = append([]int{0}, cumSum...)
	}
	if str[n-1:n] == "0" {
		cumSum = append(cumSum, n)
	}

	if len(cumSum) <= 2*k+2 {
		fmt.Println(n)
	} else {
		maxRet := 0
		for i := 0; i < len(cumSum)-(2*k+1); i += 2 {
			maxRet = maxInt(maxRet, cumSum[i+(2*k+1)]-cumSum[i])
		}
		fmt.Println(maxRet)
	}

}