2483. Minimum Penalty for a Shop

You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters ‘N’ and ‘Y’:

• if the ith character is ‘Y’, it means that customers come at the ith hour
• whereas ‘N’ indicates that no customers come at the ith hour.

If the shop closes at the jth hour (0 <= j <= n), the penalty is calculated as follows:

• For every hour when the shop is open and no customers come, the penalty increases by 1.
• For every hour when the shop is closed and customers come, the penalty increases by 1.

Return the earliest hour at which the shop must be closed to incur a minimum penalty.

Note that if a shop closes at the jth hour, it means the shop is closed at the hour j.

class Solution {
public:
    int bestClosingTime(string s) {
        int come = count(begin(s), end(s), 'Y'), none = s.length() - come;
        int pass = 0;
        int score = none, time = s.length();
        for(int i = 0; i <= s.length(); i++) {
            int now = come + pass;
            if(score == now) time = min(time, i);
            else if(score > now) {
                score = now;
                time = i;
            }
            if(i != s.length()) {
                if(s[i] == 'N') pass += 1;
                else come -= 1;
            }
        }
        return time;
    }
};