Problem Description

In this problem, we are operating a restaurant with a single chef and a list of customers who arrive and place orders. Each customer has two properties:

• arrival_i: The arrival time for the i-th customer.
• time_i: The time required to prepare the order for the i-th customer.

Customers arrive in a non-decreasing order of their arrival times. The chef starts working on each order when he is not busy with another, and only works on one order at a time. If the customer arrives while the chef is busy, they need to wait until the chef can start their order. Once the chef starts working on an order, they work on it until it's finished before moving onto the next customer's order. Our task is to calculate the average waiting time of all customers. The waiting time for a customer consists of the time they wait before the chef begins their order and the time it takes to prepare their order.

Intuition

The key to solving this problem is to track the time at which each customer's order will be completed, and then calculate the total wait-time. To do this, we maintain a variable t that represents the current time when the chef finishes preparing the previous order. We iterate through all the customers, and for each customer, we update t to be the larger of t or the customer's arrival time (since the chef can only start preparing the order after the customer has arrived), then add the preparation time b. This gives us the time when the chef will complete the current order. For each customer, the waiting time is the difference between the time of completion t and the arrival time a. We sum up all these waiting times to get the total waiting time tot.

Once we've processed all customers, we calculate the average waiting time by dividing the total wait-time by the total number of customers, which gives us the solution. The solution is efficient because it goes through the customers only once, updating the total wait-time on the fly.

Solution Approach

To implement the given solution, we are following a straightforward approach using a simple loop without the need for complex data structures or algorithms. Here's a step-by-step breakdown of the solution:

1. Initialize tot to 0, which will hold the total waiting time for all customers.

2. Initialize t to 0, which will keep track of the time when the chef completes an order. Think of t as the chef's available time to start a new order.

3. Loop over each customer in the list of customers. For each customer denoted as (a, b):

   • Update t to the maximum of t (the time when the chef will be free from the previous order) and arrival_i (the time when the current customer arrives). This is important because if a customer arrives before the chef is finished with the previous order, the chef can only start the next order after finishing the current one. However, if the chef is already free by the time the next customer arrives (t < arrival_i), they start the order at the customer's arrival time.
   • Add the preparation time time_i to t, i.e., t = max(t, a) + b. Now t represents the time when the order for the current customer will be completed.
   • Calculate the waiting time for the customer as the difference between the total time when their order is completed (t) and their arrival time (a), and add this to the total waiting time tot.

4. The average waiting time is computed by dividing tot by the number of customers len(customers).

5. Return the average waiting time as a float, which corresponds to the problem requirement of calculating the average.

This solution works in O(n) time because it employs a single for-loop that goes through the customers, and O(1) extra space as it uses only two variables that keep track of the total waiting time and the current time, regardless of the number of customers.

Example Walkthrough

Suppose we have the following list of customers, where each pair represents the arrival time and the time it takes to prepare their order, respectively:

Customers = [(1, 2), (2, 5), (4, 3)]

Here's how the solution approach would be applied:

1. We initialize tot to 0. This will accumulate the total waiting time for all customers. We also initialize t to 0, representing when the chef can start the next order.

2. We then start iterating over the customers list:

   • For the first customer (1, 2), since t (0) is less than arrival_1 (1), we update t to arrival_1 (1) and then add time_1 (2), so t becomes 3. The waiting time for this customer is t - arrival_1 = 3 - 1 = 2, so we update tot to 2.
   • The second customer (2, 5) arrives when the chef is busy, so the chef can only start at time t (3). We update t to 3 and add time_2 (5), which means t becomes 8. The waiting time for this customer is 8 - 2 = 6, and we update tot to 2 + 6 = 8.
   • The third customer (4, 3) arrives when the chef is free, as the chef finished the previous order at t (8), which is after the third customer's arrival. Therefore, t remains 8, and we add time_3 (3), making t become 11. The waiting time for this customer is 11 - 4 = 7, and the total waiting time tot is updated to 8 + 7 = 15.

3. After iterating through all customers, we divide the total waiting time tot (15) by the number of customers (3), which gives us an average waiting time of 15 / 3 = 5.0.

4. Finally, we return the average waiting time of 5.0 as the answer to the problem.

This example illustrates how the solution approach effectively calculates the average waiting time of customers in a restaurant using a simple loop and a straightforward update of the tot and t variables.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(n), where n is the length of the input list customers. This is because the code involves a single loop that iterates through each customer exactly once, performing a constant amount of work for each customer without any nested loops.

In the loop, two major operations are performed for each customer: calculating the time at which the chef starts preparing the customer's food (max(t, a) + b) and updating the total waiting time (tot += t - a). Since both of these operations are executed in constant time, the time complexity remains linear with respect to the number of customers.

Space Complexity

The space complexity of the given code is O(1). Aside from the input list customers, only a fixed number of integer variables (tot and t) are used for calculations. These variables do not depend on the size of the input and, as such, do not scale with the input. Regardless of the number of customers, the space used by the algorithm is constant.

Thus, the space required for the algorithm does not grow with the size of the input, which results in constant space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.