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"
	"math"
)

type Status struct {
	val string
	len int
}

func main() {
	n := SingleInt()
	k := SingleInt()
	s := SingleStr()
	ss := make([]int, 0, n)

	now := "1"
	cnt := 0
	for i := 0; i < n; i++ {
		if string(s[i]) == now {
			cnt++
		} else {
			ss = append(ss, cnt)
			now = reverse(now)
			cnt = 1
		}
	}
	if cnt != 0 {
		ss = append(ss, cnt)
	}
	if len(ss)%2 == 0 {
		ss = append(ss, 0)
	}

	add := 2*k + 1
	ans := 0

	for i := 0; i < n; i += 2 {
		var tmp int

		left := i
		right := Min(i+add, len(ss))

		for j := left; j < right; j++ {
			tmp += ss[j]
		}
		ans = Max(tmp, ans)
	}
	fmt.Printf("%#v\n", ans)
}

func Max(nums ...int) int {
	if len(nums) == 0 {
		panic("function max() requires at least one argument.")
	}
	res := nums[0]
	for i := 0; i < len(nums); i++ {
		res = int(math.Max(float64(res), float64(nums[i])))
	}
	return res
}

func Min(nums ...int) int {
	if len(nums) == 0 {
		panic("function min() requires at least one argument.")
	}
	res := nums[0]
	for i := 0; i < len(nums); i++ {
		res = int(math.Min(float64(res), float64(nums[i])))
	}
	return res
}

func reverse(v string) string {
	if v == "0" {
		return "1"
	}
	return "0"
}

func SingleStr() string {
	var s string
	fmt.Scan(&s)
	return s
}

func SingleInt() int {
	var n int
	fmt.Scan(&n)
	return n
}