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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let k: usize = input();
    let s: Vec<char> = input::<String>().chars().collect();

    let mut group = vec![];
    if s[0] == '0' {
        group.push(('1', 0));
    }
    let mut i = 0;
    while i < n {
        let cur = s[i];
        let mut cnt = 0;
        while i < n && s[i] == cur {
            i += 1;
            cnt += 1;
        }
        group.push((cur, cnt));
    }

    let z = group.iter().filter(|&&(c, _)| c == '0').count();
    if z <= k {
        println!("{}", n);
        return;
    }

    let mut gn = group.len();
    if group[group.len() - 1].0 == '0' {
        group.push(('1', 0));
        gn += 1;
    }

    let mut sum = vec![0; group.len() + 1];
    for i in 0..gn {
        sum[i + 1] = sum[i] + group[i].1;
    }

    let mut ans = 0;
    for i in 0..gn - (2 * k + 1) + 1 {
        if i & 1 == 1 {
            continue;
        }
        let temp = sum[i + 2 * k + 1] - sum[i];
        ans = std::cmp::max(ans, temp);
    }
    println!("{}", ans);
}