Welcome to Subscribe On Youtube
2483. Minimum Penalty for a Shop
Description
You are given the customer visit log of a shop represented by a 0indexed string customers
consisting only of characters 'N'
and 'Y'
:
 if the
i^{th}
character is'Y'
, it means that customers come at thei^{th}
hour  whereas
'N'
indicates that no customers come at thei^{th}
hour.
If the shop closes at the j^{th}
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 j^{th}
hour, it means the shop is closed at the hour j
.
Example 1:
Input: customers = "YYNY" Output: 2 Explanation:  Closing the shop at the 0^{th} hour incurs in 1+1+0+1 = 3 penalty.  Closing the shop at the 1^{st} hour incurs in 0+1+0+1 = 2 penalty.  Closing the shop at the 2^{nd} hour incurs in 0+0+0+1 = 1 penalty.  Closing the shop at the 3^{rd} hour incurs in 0+0+1+1 = 2 penalty.  Closing the shop at the 4^{th} hour incurs in 0+0+1+0 = 1 penalty. Closing the shop at 2^{nd} or 4^{th} 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 0^{th} hour as no customers arrive.
Example 3:
Input: customers = "YYYY" Output: 4 Explanation: It is best to close the shop at the 4^{th} hour as customers arrive at each hour.
Constraints:
1 <= customers.length <= 10^{5}
customers
consists only of characters'Y'
and'N'
.
Solutions

class Solution { public int bestClosingTime(String customers) { int n = customers.length(); int[] s = new int[n + 1]; for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (customers.charAt(i) == 'Y' ? 1 : 0); } int ans = 0, cost = 1 << 30; for (int j = 0; j <= n; ++j) { int t = j  s[j] + s[n]  s[j]; if (cost > t) { ans = j; cost = t; } } return ans; } }

class Solution { public: int bestClosingTime(string customers) { int n = customers.size(); vector<int> s(n + 1); for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (customers[i] == 'Y'); } int ans = 0, cost = 1 << 30; for (int j = 0; j <= n; ++j) { int t = j  s[j] + s[n]  s[j]; if (cost > t) { ans = j; cost = t; } } return ans; } };

class Solution: def bestClosingTime(self, customers: str) > int: n = len(customers) s = [0] * (n + 1) for i, c in enumerate(customers): s[i + 1] = s[i] + int(c == 'Y') ans, cost = 0, inf for j in range(n + 1): t = j  s[j] + s[1]  s[j] if cost > t: ans, cost = j, t return ans

func bestClosingTime(customers string) (ans int) { n := len(customers) s := make([]int, n+1) for i, c := range customers { s[i+1] = s[i] if c == 'Y' { s[i+1]++ } } cost := 1 << 30 for j := 0; j <= n; j++ { t := j  s[j] + s[n]  s[j] if cost > t { ans, cost = j, t } } return }

impl Solution { #[allow(dead_code)] pub fn best_closing_time(customers: String) > i32 { let n = customers.len(); let mut penalty = i32::MAX; let mut ret = 1; let mut prefix_sum = vec![0; n + 1]; // Initialize the vector for (i, c) in customers.chars().enumerate() { prefix_sum[i + 1] = prefix_sum[i] + (if c == 'Y' { 1 } else { 0 }); } // Calculate the answer for i in 0..=n { if penalty > ((prefix_sum[n]  prefix_sum[i]) as i32) + ((i  prefix_sum[i]) as i32) { penalty = ((prefix_sum[n]  prefix_sum[i]) as i32) + ((i  prefix_sum[i]) as i32); ret = i as i32; } } ret } }