2483.Minimum Penalty for a Shop
===

###### tags: `Medium` `String` `Prefix Sum`

[2483. Minimum Penalty for a Shop](https://leetcode.com/problems/minimum-penalty-for-a-shop/)

### 題目描述

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 i^th^ character is `'Y'`, it means that customers come at the i^th^ hour
* whereas `'N'` indicates that no customers come at the i^th^ hour.

If the shop closes at the j^th^ 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 j^th^ hour, it means the shop is closed at the hour `j`.

### 範例

**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` <= 10^5^
* `customers` consists only of characters `'Y'` and `'N'`.

### 解答

#### C++

class Solution {
public:
    const int INF = 1e5 + 10;

    int bestClosingTime(string customers) {
        ios_base::sync_with_stdio(0); cin.tie(0);
        
        int numCustomers = customers.size();
        vector<int> prefixN(numCustomers + 1, 0);

        for (int i = 1; i <= numCustomers; i ++) {
            prefixN[i] = prefixN[i - 1] + static_cast<int>(customers[i - 1] == 'N');
        }

        int suffixY = 0;
        int minPenalty = INF, minPenaltyHour = 0;

        for (int ithHour = numCustomers; ithHour >= 0; ithHour --) {
            suffixY += static_cast<int>(customers[ithHour] == 'Y');
            if (prefixN[ithHour] + suffixY <= minPenalty) {
                minPenalty = prefixN[ithHour] + suffixY;
                minPenaltyHour = ithHour;
            }
        }
        return minPenaltyHour;
    }
};

> [name=Jerry Wu]

#### C#

public class Solution {
    public int BestClosingTime(string customers) {
        int score = 0;
        int max = 0;
        int ans = 0;

        for (int i = 0; i < customers.Length; i++) {
            score = customers[i] == 'Y' ? score + 1 : score - 1;
            if (score > max) {
                max = score;
                ans = i + 1;
            }
        }

        return ans;
    }
}

>[name=Jim][time=Aug 29, 2023]
### Reference

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