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)

//
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
//
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

// solution 
use std::cmp;
#[allow(unused_mut)]
#[allow(non_snake_case)]
fn main() {
    input!{
        N: usize,
        K: usize,
        S: chars,
    }

    let mut interval_length = Vec::new(); 
    let mut ind = 0;
    while ind < S.len() {
        let mut count = 1;
        if ind+1 >= S.len() {
            interval_length.push(count);
            break
        }
        while S[ind] == S[ind+1] {
            count += 1;
            ind += 1;
            if ind >= S.len()-1 {
                break
            }
        }
        interval_length.push(count);
        ind += 1;
    }

    if S.last().unwrap() == &'0' {
        interval_length.push(0);
    }
    if S[0] == '0' {
        interval_length.insert(0, 0);
    }

    let max_op = interval_length.len() / 2;
    if K >= max_op {
        println!("{}", interval_length.iter().sum::<usize>());
        return
    }

    let mut sum = interval_length.iter().take(2*K+1).sum::<usize>();
    let mut prev = sum;
    let mut max = sum;
    for kstart in 0..(max_op-K) {
        sum = prev + interval_length[2*(K+kstart)+1] + interval_length[2*(K+kstart)+2] - interval_length[2*kstart] - interval_length[2*kstart+1];
        max = cmp::max(max, sum);
        prev = sum;
    }
    println!("{}", max);
}