Average Waiting Time

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: 5.33333

💡 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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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

✓ Two Pointers

⏱️ Time: N/A Space: N/A

Brute Force Simulation

⏱️ Time: O(n) Space: O(1)

For each customer, calculate when they start being served based on arrival time and when chef becomes free, then compute their total waiting time.

Optimized Simulation

⏱️ Time: O(n) Space: O(1)

Track the chef's current availability time and process customers sequentially, calculating waiting time efficiently without redundant calculations.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

static int customers[1000][2];
static int customerCount = 0;

void parseInput(char* line) {
    customerCount = 0;
    int i = 0;
    int len = strlen(line);
    
    // Skip initial whitespace and opening bracket
    while (i < len && (line[i] == ' ' || line[i] == '[')) i++;
    
    while (i < len) {
        // Skip whitespace and opening bracket for customer
        while (i < len && (line[i] == ' ' || line[i] == '[' || line[i] == ',')) i++;
        
        if (i >= len || line[i] == ']') break;
        
        // Parse arrival time
        int arrival = 0;
        while (i < len && line[i] >= '0' && line[i] <= '9') {
            arrival = arrival * 10 + (line[i] - '0');
            i++;
        }
        
        // Skip comma
        while (i < len && (line[i] == ',' || line[i] == ' ')) i++;
        
        // Parse service time
        int serviceTime = 0;
        while (i < len && line[i] >= '0' && line[i] <= '9') {
            serviceTime = serviceTime * 10 + (line[i] - '0');
            i++;
        }
        
        customers[customerCount][0] = arrival;
        customers[customerCount][1] = serviceTime;
        customerCount++;
        
        // Skip closing bracket for this customer
        while (i < len && (line[i] == ']' || line[i] == ',' || line[i] == ' ')) i++;
    }
}

double calculateAverageWaitingTime() {
    if (customerCount == 0) return 0.0;
    
    int currentTime = 0;
    double totalWaitingTime = 0.0;
    
    for (int i = 0; i < customerCount; i++) {
        int arrivalTime = customers[i][0];
        int serviceTime = customers[i][1];
        
        // Chef becomes available at max(currentTime, arrivalTime)
        if (currentTime < arrivalTime) {
            currentTime = arrivalTime;
        }
        
        // Customer waits from arrival until service is complete
        int serviceCompleteTime = currentTime + serviceTime;
        int waitingTime = serviceCompleteTime - arrivalTime;
        
        totalWaitingTime += waitingTime;
        currentTime = serviceCompleteTime;
    }
    
    return totalWaitingTime / customerCount;
}

int main() {
    char line[10000];
    
    if (fgets(line, sizeof(line), stdin)) {
        line[strcspn(line, "\n")] = 0;
        parseInput(line);
        double result = calculateAverageWaitingTime();
        printf("%.10g\n", result);
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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