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
.
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 <= 105
customers
consists only of characters'Y'
and'N'
.
impl Solution {
pub fn best_closing_time(customers: String) -> i32 {
let mut count_n = 0;
let mut count_y = customers.chars().filter(|&c| c == 'Y').count();
let mut min_penalty = count_n + count_y;
let mut ret = 0;
for (i, c) in customers.chars().enumerate() {
if c == 'N' {
count_n += 1;
} else if c == 'Y' {
count_y -= 1;
}
if min_penalty > count_n + count_y {
min_penalty = count_n + count_y;
ret = i + 1;
}
}
ret as i32
}
}