Minimum Penalty for a Shop
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 shop closes at the jth hour (0 <= j <=...
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 shop closes at the jth hour (0 <= j <= n), the penalty is calculated as follows:
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 <= 10^5
- customers consists only of characters 'Y' and 'N'.
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