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2483. Minimum Penalty for a Shop

Description

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'.

Solutions

Solution 1: Prefix Sum + Enumeration

First, we calculate how many customers arrive in the first \(i\) hours and record it in the prefix sum array \(s\).

Then we enumerate the closing time \(j\) of the shop, calculate the cost, and take the earliest closing time with the smallest cost.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Where \(n\) is the length of the string \(customers\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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
    }
}

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