Average Waiting Time

Problem Description

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

• arrival[i] 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, they give the chef their order, and the chef starts preparing it once they are idle. The customer waits until the chef finishes preparing their 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.

Examples

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 their 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 their 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 their 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 their 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 their 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 their 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 their 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 <= 10^5
• 1 <= arrival[i], time[i] <= 10^4
• arrival[i] <= arrival[i + 1]

Solution for 1701. Average Waiting Time

Approach

cur means the current time. For each customer arriving at t with their order needing d, we start preparing their order at max(cur, t), we finish their order at max(cur, t) + d, and they wait for max(cur, t) + d - t.

We need to sum up the waiting time and also update the cur time after serving each customer.

Code in Different Languages

Python

Written by @agarwalhimanshugaya

    def averageWaitingTime(self, A):
     wait = cur = 0.
     for t, d in A:
         cur = max(cur, t) + d
         wait += cur - t
     return wait / len(A)

Java

Written by @agarwalhimanshugaya

    public double averageWaitingTime(int[][] A) {
     double wait = 0, cur = 0;
     for (int[] a: A) {
         cur = Math.max(cur, 1.0 * a[0]) + a[1];
         wait += cur - a[0];
     }
     return 1.0 * wait / A.length;
 }

C++

Written by @hiteshgahanolia

    double averageWaitingTime(vector<vector<int>>& A) {
     double wait = 0, cur = 0;
     for (auto& a: A) {
         cur = max(cur, 1.0 * a[0]) + a[1];
         wait += cur - a[0];
     }
     return 1.0 * wait / A.size();
 }

Complexity Analysis

• Time Complexity: $O(N)$
• Space Complexity: $O(1)$

References

• LeetCode Problem: Average Waiting Time
• Solution Link: LeetCode Solution