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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int main(void){
    int n, k;
    char s[100001];
    scanf("%d %d", &n, &k);
    scanf("%s", &s);
    int s0 = s[0];
    int count = 1;
    char t[100001];
    int t_c = 0;
    for(int i = 1;i < n;i++){
        if(s[i] == s0){
            count++;
        }else{
            t[t_c] = count;
            count = 1;
            t_c++;
            s0 = s[i];
        }
    }
    t[t_c] = count;
    t[t_c+1] = 0;
    
    int max = 0;
    if(k >= (t_c+1)/2){
        for(int j = 0;j <= t_c;j++){
            max = max + t[j];
        }
        printf("%d\n", max);
    }else if(s[0] == 0){
        for(int j = 0;j < k;j++){
            max = max + t[j*2+0] + t[j*2+1];
        }
        for(int i = 1;i <= (t_c + 1)/2 - k;i++){
            int isum = t[i*2-1];
            for(int j = 0;j < k;j++){
                isum = isum + t[i*2+j*2+0] + t[i*2+j*2+1];
            }
            if(max < isum){
                max = isum;
            }
        }
        printf("%d\n", max);
    }else{
        max = max + t[0];
        for(int j = 0;j < k;j++){
            max = max + t[j*2+2] + t[j*2+1];
        }
        for(int i = 1;i <= (t_c + 1)/2 - k;i++){
            int isum = t[i*2];
            for(int j = 0;j < k;j++){
                isum = isum + t[i*2+j*2+2] + t[i*2+j*2+1];
            }
            if(max < isum){
                max = isum;
            }
        }
        printf("%d\n", max);
    }
    return 0;
}