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>

int main() {

    int n, k, i, j=0, c_seg=1, c_seq=0, f01=-1, ans=0, max_ans=0;
    scanf("%d %d", &n, &k);

    char s[n];
    scanf("%s", s);
    // printf("%s\n", s);

    for(i=0; i<n; i++) {
        if(s[i]=='0') {
            if(f01==1) {
                c_seg += 1;
            }

            f01 = 0;

        } else {
            if(f01==0) {
                c_seg +=1;
            }

            f01 = 1;

        }
    }

    // printf("%d\n", c_seg);

    int a[c_seg][2];
    f01 = -1;

    for(i=0; i<n; i++) {
        if(s[i]=='0') {
            if(f01==1) {
                a[j][0] = c_seq;
                a[j][1] = 1;
                j += 1;
                c_seq = 0;
            }

            f01 = 0;
            c_seq += 1;

        } else {
            if(f01==0) {
                a[j][0] = c_seq;
                a[j][1] = 0;
                j += 1;
                c_seq = 0;
            }

            f01 = 1;
            c_seq += 1;

        }
    }
    a[j][0] = c_seq;
    if(f01==0) {
        a[j][1] = 0;
    } else {
        a[j][1] = 1;
    }

    // for(i=0; i<c_seg; i++) {
    //     printf("%d", a[i][0]);
    //     printf("%d", a[i][1]);
    // }
    // printf("\n");

    if(a[0][1]==0) {
        for(j=0; j<2*k; j++) {
            if(c_seg<=j) {
                break;
            }
            max_ans += a[j][0];
        }

        for(i=1; i<c_seg/2+1-k+1; i++) {
            for(j=0; j<2*k+1; j++) {
                if(j+i*2-1 >= c_seg) {
                    break;
                }
                ans += a[j+i*2-1][0];
            }
            if(ans > max_ans) {
                max_ans = ans;
            }
            ans = 0;
        }
    } else if(c_seg==1 && a[0][1]==1) {
        max_ans = a[0][0];
    } else {
        for(i=0; i<c_seg/2-k+1; i++) {
            for(j=0; j<2*k+1; j++) {
                if(j+i*2 >= c_seg) {
                    break;
                }
                ans += a[j+i*2][0];
            }
            if(ans > max_ans) {
                max_ans = ans;
            }
            ans = 0;
        }
    }

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

    return 0;
}