2483

2483. Minimum Penalty for a Shop (Medium)

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.

 

Example 1:

Input: customers = "YYNY"
Output: 2
Explanation: 
- Closing the shop at the 0th hour incurs in 1+1+0+1 = 3 penalty.
- Closing the shop at the 1st hour incurs in 0+1+0+1 = 2 penalty.
- Closing the shop at the 2nd hour incurs in 0+0+0+1 = 1 penalty.
- Closing the shop at the 3rd hour incurs in 0+0+1+1 = 2 penalty.
- Closing the shop at the 4th hour incurs in 0+0+1+0 = 1 penalty.
Closing the shop at 2nd or 4th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.

Example 2:

Input: customers = "NNNNN"
Output: 0
Explanation: It is best to close the shop at the 0th hour as no customers arrive.

Example 3:

Input: customers = "YYYY"
Output: 4
Explanation: It is best to close the shop at the 4th hour as customers arrive at each hour.

 

Constraints:

• 1 <= customers.length <= 105
• customers consists only of characters 'Y' and 'N'.

Companies: Stripe

Related Topics:
String, Prefix Sum

Similar Questions:

• Grid Game (Medium)

Solution 1. Left-to-right state transition

The penalty at s[i] is the count of Y's to the right (excluding s[i]) plus the count of N's to the left (including s[i]).

// OJ: https://leetcode.com/problems/minimum-penalty-for-a-shop
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int bestClosingTime(string s) {
        int N = s.size(), y = 0, n = 0;
        for (char c : s) y += c == 'Y';
        int minPenalty = y, ans = 0;
        for (int i = 0; i < N; ++i) {
            y -= s[i] == 'Y';
            n += s[i] == 'N';
            if (y + n < minPenalty) {
                minPenalty = y + n;
                ans = i + 1;
            }
        }
        return ans;
    }
};