2483.Minimum Penalty for a Shop

tags: Medium String Prefix Sum

2483. 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 <= 10⁵
• 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;
    }
};

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;
    }
}

JimAug 29, 2023

Reference

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