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 <bits/stdc++.h>

using namespace std;

int main(){
	long N, K;
	string S;
    cin>>N>>K;
    cin>>S;
    int s_len = S.length();

    int zero_max=0, zero_idx=0;
    for(int i=0; i<s_len;){
        int c=0;
        int idx=i;
        while(S[i]=='0'){
            c++;
            ++i;
        }
        if(c>=zero_max){
            zero_max = c;
            zero_idx = idx;
        }
        ++i;
    }

    int one_max=0, one_idx=0;
    for(int i=0; i<s_len;){
        int c=0;
        int idx=i;
        while(S[i]=='1'){
            c++;
            ++i;
        }
        if(c>=one_max){
            one_max = c;
            one_idx = idx;
        }
        ++i;
    }

    int l, r;
    long ans=0;
    if(zero_max <= one_max){
        ans = one_max;
        l = one_idx + one_max - 1;
    }else{
        ans = zero_max;
        l = zero_idx + zero_max - 1;
    }

    for(int i=0; i<K; ++i){
        int rc = 0;
        int rr = r-1;
        while(rr>=0 && S[rr]=='0'){
            rc++;
            rr--;
        }
        while(rr>=0 && S[rr]=='1'){
            rc++;
            rr--;
        }

        int lc = 0;
        int ll = l+1;
        while(ll<s_len && S[ll]=='0'){
            lc++;
            ll++;
        }
        while(ll<s_len && S[ll]=='1'){
            lc++;
            ll++;
        }

        if(lc < rc){
            r = rr - 1;
            ans += rc;
        }else{
            l = ll + 1;
            ans += lc;
        }
    }

    cout<<ans<<endl;

    return 0;
}