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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <limits.h>
#include <math.h>

int main(int argc, char *argv[])
{
    int n, k;
    char s[100001];
    int cnt[100005];
    int ptr;
    char last = ' '; // not '0' and not '1'
    int lastpos;
    int sum;
    int i;
    int max;

    scanf("%d %d", &n, &k);
    scanf("%s", s);

    for (i = 0; i < 100005; i++) {
        cnt[i] = 0;
    }

    lastpos = 0;
    last = s[lastpos];
    ptr = 0;
    i = 0;
    do {
        i++;
        if (last != s[i]) {
            int len = i - lastpos;
            int dest = ptr + (last - '0');
            cnt[dest] = len;

            if (last == '1') {ptr += 2;}
            lastpos = i;
            last = s[lastpos];
        }
    } while(s[i] != '\0');

    sum = 0;
    for (i = 0; i < k * 2; i++) {
        sum += cnt[i];
    }
    max = sum;

    for (i = k * 2; cnt[i] != 0; i++) {
        if (i % 2 == 0) {
            sum += cnt[i];
            sum -= cnt[i - (k * 2)];
            if (i > (k * 2 + 1)) {
                sum -= cnt[i - (k * 2 + 1)];
            }
        } else {
            sum += cnt[i];
        }
        if (sum > max) {max = sum;}
    }

    printf("%d\n", max);

    return 0;
}