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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)


#include <iostream>
#include <string>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <array>
#include <cmath>

using namespace std;

using ll = long long int;

struct range_t {
    char ch;
    int begin, end;
};

int main() {
    int N, K;
    char str[(int)1e5] = {'\0'};

    cin >> N >> K;
    cin >> str;

    string S(str);

    std::vector<range_t> blockRange;

    int begin = 0;
    char prev = S.front();
    for (int i = 0; i < S.size(); i++) {
        if (prev != S.at(i)) {
            blockRange.push_back({ prev , begin, i });
            begin = i;
        } else if (i == S.size() - 1) {
            blockRange.push_back({ prev , begin, i + 1 });
        }
        prev = S.at(i);
    }


    int max = 0;
    if (blockRange.size() >= K * 2 + 1) {
        for (int rb = 0; true; rb++) {
            int size = (blockRange.at(rb).ch == '0') ? K * 2 : K * 2 + 1;
            if (rb + size >= blockRange.size()) break;

            int cnt = blockRange[rb + size].end - blockRange[rb].begin;
            if (cnt > max) max = cnt;
        }
    }
    else {
        for (int rb = 0; rb < blockRange.size(); rb++) {
            int cnt = 0;
            int flip = 0;
            for (int offset = 0; rb + offset < blockRange.size(); offset++) {
                const auto & target = blockRange.at(rb + offset);
                
                if (target.ch == '0') flip++;
                if (flip > K) break;

                cnt += target.end - target.begin;
            }
        
            if (cnt > max) max = cnt;
        }
    }
   
    cout << max;

    return 0;
}