Average Waiting Time - Practice Coding Problems

Average Waiting Time - Problem

Array Simulation Medium

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

arrival_i is the arrival time of the i^th customer. The arrival times are sorted in non-decreasing order.
time_i is the time needed to prepare the order of the i^th 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.

Input & Output

Example 1 — Basic Case

$ Input: customers = [[1,2],[2,5],[4,3]]

› Output: 5.00000

💡 Note: Customer 1: arrives at 1, served 1→3, waits 2 time units. Customer 2: arrives at 2, must wait until chef free at 3, served 3→8, waits 6 time units. Customer 3: arrives at 4, must wait until chef free at 8, served 8→11, waits 7 time units. Average = (2+6+7)/3 = 5.0

Example 2 — No Waiting

$ Input: customers = [[5,2],[5,4],[6,3]]

› Output: 3.00000

💡 Note: Customer 1: arrives at 5, served immediately 5→7, waits 2 units. Customer 2: arrives at 5, waits until 7, served 7→11, waits 6 units. Customer 3: arrives at 6, waits until 11, served 11→14, waits 8 units. Average = (2+6+8)/3 = 16/3 ≈ 5.33

Example 3 — Single Customer

$ Input: customers = [[2,3]]

› Output: 3.00000

💡 Note: Only one customer arrives at time 2, takes 3 time units to prepare, so waiting time is exactly 3.0

Constraints

1 ≤ customers.length ≤ 10⁵
1 ≤ arrival_i, time_i ≤ 10⁴
0 ≤ arrival_i ≤ arrival_i+1

Visualization

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Understanding the Visualization

Input

Array of [arrival_time, cook_time] for each customer

Process

Track chef availability and calculate waiting times

Output

Average of all customer waiting times

Key Takeaway

🎯 Key Insight: Track chef availability time and calculate each customer's total waiting time from arrival to service completion

Asked in

a Amazon 15 G Google 12 M Microsoft 8

The key insight is to simulate the restaurant process by tracking when the chef becomes available and calculating each customer's waiting time as the difference between their completion time and arrival time. The optimal approach uses a single pass through all customers with O(n) time and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Simulation	O(n)	O(1)	Calculate waiting time for each customer by simulating the entire process
Optimized Simulation	O(n)	O(1)	Single pass tracking chef availability with optimized calculations

Brute Force Simulation — Algorithm Steps

Initialize chef's free time as 0
For each customer, determine start time as max(arrival, chef_free_time)
Calculate waiting time as (start_time + cook_time) - arrival_time
Update chef's free time and accumulate total waiting time

Visualization

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Step-by-Step Walkthrough

Initialize

Chef is free at time 0

Process Each Customer

Calculate start time, waiting time, and update chef availability

Calculate Average

Sum all waiting times and divide by number of customers

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

double solution(int customers[][2], int n) {
    int chefFreeTime = 0;
    double totalWaitingTime = 0;
    
    for (int i = 0; i < n; i++) {
        int arrival = customers[i][0];
        int cookTime = customers[i][1];
        
        // When does chef start cooking this order?
        int startTime = (arrival > chefFreeTime) ? arrival : chefFreeTime;
        
        // When does chef finish this order?
        int finishTime = startTime + cookTime;
        
        // Customer's waiting time is from arrival to finish
        int waitingTime = finishTime - arrival;
        totalWaitingTime += waitingTime;
        
        // Update when chef becomes free
        chefFreeTime = finishTime;
    }
    
    return totalWaitingTime / n;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse [[1,2],[3,4]] format
    int customers[100][2];
    int n = 0;
    
    char* ptr = line + 1; // Skip first [
    while (*ptr && *ptr != ']') {
        if (*ptr == '[') {
            ptr++;
            customers[n][0] = strtol(ptr, &ptr, 10);
            ptr++; // Skip comma
            customers[n][1] = strtol(ptr, &ptr, 10);
            n++;
            while (*ptr && *ptr != ']' && *ptr != '[') ptr++;
            if (*ptr == ']') ptr++;
        } else {
            ptr++;
        }
    }
    
    double result = solution(customers, n);
    printf("%f\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through all customers

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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