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: Accepted)

#include <stdio.h>
#include <stdlib.h>

int *check(char *str, int len, int *cnt) {
    int i, j = 0;
    int *list = (int*)malloc(len * sizeof(int));
    list[j] = 0; ++j;
    for(i = 1; i < len; ++i) {
        if(str[i-1] != str[i]) { list[j] = i; ++j; }
    }
    list[j] = i; ++j;
    *cnt = j;
    return list;
}

void solve(char *str, int len, int try) {
    int i, j, k, max = 1, tmp, cnt;
    int *list = check(str, len, &cnt);
    //for(i = 0; i < cnt; ++i) { printf("%d ", list[i]); }
    //printf("\n");
    for(i = 0; i < cnt-1; ++i) {
        j = list[i];
        if(str[j] == '0') {
            if( i + 2*try < cnt) {
                k = (list[i+2*try] < len) ? list[i+2*try] : len;
            }
            else { k = len; }
        } else {
            if( i + 2*try + 1 < cnt) {
                k = (list[i+2*try+1] < len) ? list[i+2*try+1] : len;
            }
            else { k = len; }
        }
        tmp = k - j;
        max = (max < tmp) ? tmp : max;
    }
    printf("%d\n", max);
    free(list);
}

int main()
{
    int n, k;
    scanf("%d %d", &n, &k);
    char *str = (char*)malloc((n+4)*sizeof(char));
    scanf("%s", str);
    solve(str, n, k);
    free(str);
    return 0;
}