Class Solution
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- g2401_2500.s2483_minimum_penalty_for_a_shop.Solution
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public class Solution extends Object
2483 - Minimum Penalty for a Shop.Medium
You are given the customer visit log of a shop represented by a 0-indexed string
customersconsisting only of characters'N'and'Y':- if the
ithcharacter is'Y', it means that customers come at theithhour - whereas
'N'indicates that no customers come at theithhour.
If the shop closes at the
jthhour (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
jthhour, it means the shop is closed at the hourj.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 <= 105customersconsists only of characters'Y'and'N'.
- if the
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Constructor Summary
Constructors Constructor Description Solution()
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Method Detail
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bestClosingTime
public int bestClosingTime(String customers)
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