Problem 1701 Average Waiting Time
Table of Contents
Problem Statement
Link - Problem 1701
Question
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.
timeiis the time needed to prepare the order of theithcustomer.
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 <= 10^5`
- `1 <= arrivali, timei <= 10^4`
- `arrival(i) <= arrival(i+1)`
Solution
class Solution {
public:
double averageWaitingTime(vector<vector<int>>& customers) {
std::ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
long long int endTime = customers[0][0] + customers[0][1];
double waitTime = customers[0][1];
int arrivalTime, time;
vector<int> it;
for(int i = 1; i<customers.size();i++)
{ it = customers[i];
arrivalTime = it[0];
time = it[1];
if(arrivalTime>endTime)
{
endTime = arrivalTime;
}
endTime += time;
waitTime += endTime - arrivalTime;
}
return waitTime/customers.size();
}
};
Complexity Analysis
Time Complexity: O(N)Space Complexity: O(1)
Explanation
1. Intuition
- This just a calculation of FCFS scheduling.
- Technically we are calculating the average turnaround time.
- We know that `turn around time = end time - arrival time`.
- `end time = service time + max(arrival time, previous end time)`.
- So we need accumulate the turn around time and divide it by the number of customers.
2. Implementation
- Initialize the `endTime` and `waitTime` with the first customer's total time and service time respectively.
- Iterate over the customers from the second customer.
- Calculate the `arrivalTime` and `time` of the current customer.
- If the `arrivalTime` is greater than the `endTime`, update the `endTime` with the `arrivalTime`.
- This shows that chef is idle and had to wait for the customer to arrive.
- Update the `endTime` with the `time` of the current customer.
- Add the `endTime - arrivalTime` to the `waitTime`.
- Return the `waitTime` divided by the number of customers.