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)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn main() {
    let mut v: Vec<usize> = read_vec();
    let (n, k) = (v[0], v[1]);
    let mut s = read::<String>().chars().collect::<Vec<_>>();
    let mut one_starts = Vec::new();
    let mut zero_starts = Vec::new();
    let mut one_ends = Vec::new();
    let mut zero_ends = Vec::new();

    if s[0] == '0' {
        zero_starts.push(0i64);
    } else {
        one_starts.push(0i64);
    }
    for i in 1..n {
        if s[i] != s[i - 1] {
            if s[i] == '0' {
                zero_starts.push(i as i64);
                one_ends.push(i as i64);
            } else {
                one_starts.push(i as i64);
                zero_ends.push(i as i64);
            }
        }
    }
    if s[s.len() - 1] == '0' {
        zero_ends.push(s.len() as i64);
    } else {
        one_ends.push(s.len() as i64);
    }
    // println!("zero starts {:?}", &zero_starts);
    // println!("zero ends {:?}", &zero_ends);
    // println!("one starts {:?}", &one_starts);
    // println!("one ends {:?}", &one_ends);

    if zero_starts.len() <= k {
        println!("{}", s.len());
        return;
    }
    let inv_one_starts = one_starts.iter().rev().map(|x| -x).collect::<Vec<_>>();
    let mut ans = 0;
    for i in 0..zero_starts.len() - k + 1 {
        let mut start_pos;
        if zero_starts[i] == 0 {
            start_pos = 0;
        } else {
            start_pos = -inv_one_starts[inv_one_starts.lower_bound(&(-zero_starts[i]))];
        }
        let mut end_pos;
        if zero_ends[i + k - 1] == s.len() as i64 {
            end_pos = s.len() as i64;
        } else {
            end_pos = one_ends[one_ends.lower_bound(&(zero_ends[i + k - 1]))];
        }
        ans = max(ans, end_pos - start_pos);
    }
    println!("{}", ans);
}

use std::cmp::Ordering;

pub trait BinarySearch<T> {
    fn lower_bound(&self, &T) -> usize;
}

impl<T: Ord> BinarySearch<T> for [T] {
    fn lower_bound(&self, x: &T) -> usize {
        let mut low = 0;
        let mut high = self.len();

        while low != high {
            let mid = (low + high) / 2;
            match self[mid].cmp(x) {
                Ordering::Less => {
                    low = mid + 1;
                }
                Ordering::Equal | Ordering::Greater => {
                    high = mid;
                }
            }
        }
        low
    }
}