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
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 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.
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.
Input: customers = "NNNNN" Output: 0 Explanation: It is best to close the shop at the 0th hour as no customers arrive.
Input: customers = "YYYY" Output: 4 Explanation: It is best to close the shop at the 4th hour as customers arrive at each hour.
1 <= customers.length <= 105customersconsists only of characters'Y'and'N'.
implSolution{pubfnbest_closing_time(customers:String) -> i32{letmut count_n = 0;letmut count_y = customers.chars().filter(|&c| c == 'Y').count();letmut min_penalty = count_n + count_y;letmut ret = 0;for(i, c)in customers.chars().enumerate(){if c == 'N'{ count_n += 1;}elseif c == 'Y'{ count_y -= 1;}if min_penalty > count_n + count_y { min_penalty = count_n + count_y; ret = i + 1;}} ret asi32}}