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>
#include <string.h>
#include <stdbool.h>

int main() {
    int n = 0, k = 0;
    scanf("%d %d\n", &n, &k);
    int raw[n+1][2];
    int len_raw = 0;
    int count_0 = 0;
    for (int i=0; i<n; i++) {
        raw[i][0] = raw[i][1] = -1;
        char c = getchar();
        if (i == 0) {
            raw[0][0] = c - '0';
            raw[0][1] = 1;
            if (raw[0][0] == 0) count_0++;
        }
        else {
            if ((c - '0') == raw[len_raw][0]) raw[len_raw][1]++;
            else {
                len_raw++;
                raw[len_raw][0] = 1 - raw[len_raw-1][0];
                raw[len_raw][1] = 1;
                if (raw[len_raw][0] == 0) count_0++;
            }
        }
    }
    if (k >= count_0) {
        printf("%d\n", n);
        return 0;
    }
    int max = 0;
    int count = 0;
    int buf = 0;
    for (int i=0; i<=len_raw; i++) {
        count = 0;
        buf = 0;
        if (raw[i][0] == 0) {
            if (i == 0) {
                for (int j=0; j<=len_raw; j++) {
                    if (raw[j][0] == 0) count++;
                    if (count > k) break;
                    buf += raw[j][1];
                }
                if (buf > max) max = buf;
            }
            else {
                for (int j=i-1; j<=len_raw; j++) {
                    if (raw[j][0] == 0) count++;
                    if (count > k) break;
                    buf += raw[j][1];
                }
                if (buf > max) max = buf;
            }
        }
    }
    printf("%d\n", max);
    return 0;
}