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::Read;

pub fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();
    let k: usize = iter.next().unwrap().parse().unwrap();
    let s = iter.next().unwrap().as_bytes();
    let mut series_nums: Vec<usize> = Vec::new();
    if s[0] == ('0' as u8) {series_nums.push(0)}
    let mut current = s[0];
    let mut acc: usize = 1;
    for i in 1..n {
        if current != s[i] {
            series_nums.push(acc);
            acc = 0;
        }
        current = s[i];
        acc += 1;
    }
    series_nums.push(acc);
    if s[s.len() - 1] == ('0' as u8) {series_nums.push(0)};

    let mut cumsum: Vec<usize> = vec![0;series_nums.len() + 1];
    for i in 0..series_nums.len() {
        cumsum[i+1] = cumsum[i] + series_nums[i];
    }

    let mut ret: usize = 0;
    for left in 0..cumsum.len() {
        if left % 2 == 0 {
            let mut right = left + 2 * k + 1;
            if right >= cumsum.len() {
                right = cumsum.len() - 1;
            }
            ret = std::cmp::max(ret, cumsum[right] - cumsum[left]);
        }
    }
    println!("{}", ret);
}