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Formatted question description: https://leetcode.ca/all/2483.html

# 2483. Minimum Penalty for a Shop

• Difficulty: Medium.
• Related Topics: String, Prefix Sum.
• Similar Questions: Grid Game.

## Problem

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 the ith hour

• whereas 'N' indicates that no customers come at the ith 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'.

## Solution (Java, C++, Python)

• 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 {
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 };
}

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
}
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).