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Average Waiting Time

There is a restaurant with a single chef. You are given an array customers, where customers[i] = [arrivali, timei]:

  • arrivali is the arrival time of the ith customer. The arrival times are sorted in non-decreasing order.
  • timei is the time needed to prepare the order of the ith customer.

When a customer arrives, he gives the chef his order, and the chef starts preparing it once he is idle. The customer waits till the chef finishes preparing his order. The chef does not prepare food for more than one customer at a time. The chef prepares food for customers in the order they were given in the input.

Return the average waiting time of all customers. Solutions within 10-5 from the actual answer are considered accepted.

 

Example 1:

Input: customers = [[1,2],[2,5],[4,3]]
Output: 5.00000
Explanation:
1) The first customer arrives at time 1, the chef takes his order and starts preparing it immediately at time 1, and finishes at time 3, so the waiting time of the first customer is 3 - 1 = 2.
2) The second customer arrives at time 2, the chef takes his order and starts preparing it at time 3, and finishes at time 8, so the waiting time of the second customer is 8 - 2 = 6.
3) The third customer arrives at time 4, the chef takes his order and starts preparing it at time 8, and finishes at time 11, so the waiting time of the third customer is 11 - 4 = 7.
So the average waiting time = (2 + 6 + 7) / 3 = 5.

Example 2:

Input: customers = [[5,2],[5,4],[10,3],[20,1]]
Output: 3.25000
Explanation:
1) The first customer arrives at time 5, the chef takes his order and starts preparing it immediately at time 5, and finishes at time 7, so the waiting time of the first customer is 7 - 5 = 2.
2) The second customer arrives at time 5, the chef takes his order and starts preparing it at time 7, and finishes at time 11, so the waiting time of the second customer is 11 - 5 = 6.
3) The third customer arrives at time 10, the chef takes his order and starts preparing it at time 11, and finishes at time 14, so the waiting time of the third customer is 14 - 10 = 4.
4) The fourth customer arrives at time 20, the chef takes his order and starts preparing it immediately at time 20, and finishes at time 21, so the waiting time of the fourth customer is 21 - 20 = 1.
So the average waiting time = (2 + 6 + 4 + 1) / 4 = 3.25.

 

Constraints:

  • 1 <= customers.length <= 105
  • 1 <= arrivali, timei <= 104
  • arrival<= arrivali+1

Solution Explanation for LeetCode 1701: Average Waiting Time

This problem involves calculating the average waiting time for customers at a restaurant with a single chef. The input is a list of customers, each represented as [arrival_time, preparation_time]. The chef prepares orders sequentially, starting work only when idle or when a customer arrives.

Approach: Simulation

The most efficient approach is a direct simulation of the chef's actions. We track the chef's current availability (t) and the total waiting time (tot).

  1. Initialization: t (chef's current time) and tot (total waiting time) are initialized to 0.

  2. Iteration: The code iterates through each customer in the customers array.

  3. Chef's Availability: For each customer:

    • max(t, a) determines the chef's start time. If the chef is already busy (t > a), the customer waits until the chef is free. Otherwise, the chef starts immediately.
    • t = max(t, a) + b updates the chef's completion time for the current order.
  4. Waiting Time Calculation: The customer's waiting time is t - a, which is added to tot.

  5. Average Calculation: Finally, the average waiting time is calculated by dividing tot by the number of customers.

Time and Space Complexity

  • Time Complexity: O(n), where n is the number of customers. The algorithm iterates through the customers once.
  • Space Complexity: O(1). The algorithm uses a constant amount of extra space to store variables like t and tot.

Code Examples

The following code examples demonstrate the solution in several programming languages:

Python:

class Solution:
    def averageWaitingTime(self, customers: List[List[int]]) -> float:
        tot = t = 0
        for a, b in customers:
            t = max(t, a) + b
            tot += t - a
        return tot / len(customers)

Java:

class Solution {
    public double averageWaitingTime(int[][] customers) {
        double tot = 0;
        int t = 0;
        for (int[] e : customers) {
            int a = e[0], b = e[1];
            t = Math.max(t, a) + b;
            tot += t - a;
        }
        return tot / customers.length;
    }
}

C++:

class Solution {
public:
    double averageWaitingTime(vector<vector<int>>& customers) {
        double tot = 0;
        int t = 0;
        for (auto& e : customers) {
            int a = e[0], b = e[1];
            t = max(t, a) + b;
            tot += t - a;
        }
        return tot / customers.size();
    }
};

JavaScript:

function averageWaitingTime(customers) {
    let tot = 0;
    let t = 0;
    for (let i = 0; i < customers.length; i++) {
        let a = customers[i][0];
        let b = customers[i][1];
        t = Math.max(t, a) + b;
        tot += t - a;
    }
    return tot / customers.length;
}

These code examples all follow the same logic, differing only in syntax. The efficiency remains the same across all languages.