2483. Minimum Penalty for a Shop
Description
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 theith
hour - whereas
'N'
indicates that no customers come at theith
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'
.
Solutions
Solution 1: Prefix Sum + Enumeration
First, we calculate how many customers arrive in the first $i$ hours and record it in the prefix sum array $s$.
Then we enumerate the closing time $j$ of the shop, calculate the cost, and take the earliest closing time with the smallest cost.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $customers$.
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