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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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