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: Time Limit Exceeded)

package main
import (
    "fmt"
    "math"
)

var zero = make(map[int]int)
var one  = make(map[int]int)

func main(){
    var N, K int
    var s string
    fmt.Scan(&N, &K, &s)
    fmt.Println(search(s, N, K))
}

func search(s string,N,  K int) float64{
    padding(s, N, K)
    ans, count := 0.0, 0
    memoLength := int(math.Max(float64(len(one)), float64(len(zero))))
    for i:=0;i< memoLength;i++{
        if s[0] == 49 && i == 0{
            count += one[i]
            i++
        }else if i != 0{
            count += one[i-1]
        }
        for j:=0; j<K && i+j<memoLength;j++{
            count += zero[i+j] + one[i+j]
        }
        ans = math.Max(ans, float64(count))
        // fmt.Println(count, ans)
        count = 0
    } 
    return ans
}

func padding(s string,N, K int){
    for i, now:= 0, 0; now < len(s);i++{
        for now < len(s) && s[now] == 48{
            zero[i]++
            now ++
        }
        for now < len(s) && s[now] == 49{
            one[i]++
            now ++
        }
    }
}