Total Traveled Distance - Practice Coding Problems

Total Traveled Distance - Problem

Database Easy

You're building a ride-sharing analytics dashboard and need to calculate the total distance traveled by each user in your system.

You have two tables:

Users table: Contains user information (user_id, name)
Rides table: Contains ride records (ride_id, user_id, distance)

Your task is to write a SQL query that returns each user's user_id, name, and total_traveled_distance. Here's the catch: some users might not have taken any rides yet, so their total distance should be 0.

The results should be ordered by user_id in ascending order.

Example:

If Alice (user_id=1) took 3 rides covering 10, 15, and 20 miles, her total should be 45 miles. If Bob (user_id=2) hasn't taken any rides, his total should be 0.

Input & Output

example_1.sql — Basic Case

$ Input: Users: +----------+-------+ | user_id | name | +----------+-------+ | 1 | Alice | | 2 | Bob | | 3 | Alex | +----------+-------+ Rides: +----------+----------+----------+ | ride_id | user_id | distance | +----------+----------+----------+ | 1 | 1 | 120 | | 2 | 2 | 317 | | 3 | 3 | 222 | | 4 | 1 | 44 | | 5 | 2 | 1 | +----------+----------+----------+

› Output: +----------+-------+------------------------+ | user_id | name | total_traveled_distance| +----------+-------+------------------------+ | 1 | Alice | 164 | | 2 | Bob | 318 | | 3 | Alex | 222 | +----------+-------+------------------------+

💡 Note: Alice (user_id=1) has 2 rides: 120 + 44 = 164 miles. Bob (user_id=2) has 2 rides: 317 + 1 = 318 miles. Alex (user_id=3) has 1 ride: 222 miles.

example_2.sql — User with No Rides

$ Input: Users: +----------+--------+ | user_id | name | +----------+--------+ | 1 | Alice | | 2 | Bob | | 3 | Charlie| +----------+--------+ Rides: +----------+----------+----------+ | ride_id | user_id | distance | +----------+----------+----------+ | 1 | 1 | 100 | | 2 | 1 | 200 | +----------+----------+----------+

› Output: +----------+--------+------------------------+ | user_id | name | total_traveled_distance| +----------+--------+------------------------+ | 1 | Alice | 300 | | 2 | Bob | 0 | | 3 | Charlie| 0 | +----------+--------+------------------------+

💡 Note: Alice has traveled 300 miles total, while Bob and Charlie haven't taken any rides yet, so their distances are 0.

example_3.sql — Single User Case

$ Input: Users: +----------+------+ | user_id | name | +----------+------+ | 1 | John | +----------+------+ Rides: +----------+----------+----------+ | ride_id | user_id | distance | +----------+----------+----------+ | 1 | 1 | 50 | +----------+----------+----------+

› Output: +----------+------+------------------------+ | user_id | name | total_traveled_distance| +----------+------+------------------------+ | 1 | John | 50 | +----------+------+------------------------+

💡 Note: Edge case with single user who has taken one ride of 50 miles.

Visualization

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Understanding the Visualization

Start with All Users

LEFT JOIN ensures every user from Users table appears in results

Match Ride Data

Find corresponding rides for each user (NULL if no rides)

Group and Sum

GROUP BY user to aggregate multiple rides per user

Handle Missing Data

COALESCE converts NULL sums to 0 for users with no rides

Key Takeaway

🎯 Key Insight: LEFT JOIN is perfect for "show all users, even inactive ones" scenarios. It's the SQL equivalent of a comprehensive report that doesn't leave anyone out!

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

Single pass through users (n) and rides (m) tables, plus sorting

✓ Linear Growth

Space Complexity

O(n)

Space for the result set containing all users

⚡ Linearithmic Space

Constraints

1 ≤ Users.user_id ≤ 10⁶
1 ≤ Rides.ride_id ≤ 10⁶
1 ≤ Rides.distance ≤ 1000
All user_id values in Users table are unique
All ride_id values in Rides table are unique
user_id in Rides table references valid users in Users table

Asked in

U Uber 45 L Lyft 32 D DoorDash 28 A Amazon 25

The optimal solution uses a LEFT JOIN between Users and Rides tables combined with GROUP BY and SUM aggregation. The key insight is using LEFT JOIN to preserve all users (even those without rides) and COALESCE to handle NULL values gracefully, converting them to 0.

Common Approaches

Approach	Time	Space	Notes
✓ Subquery for Each User (Brute Force)	O(n²)	O(1)	Use correlated subquery to calculate distance for each user individually
LEFT JOIN with GROUP BY (Optimal)	O(n + m)	O(n)	Join tables and aggregate in a single efficient query

Subquery for Each User (Brute Force) — Algorithm Steps

Select all users from Users table
For each user, run a subquery to sum their distances from Rides table
Use COALESCE to handle NULL values (users with no rides)
Order results by user_id

Visualization

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Step-by-Step Walkthrough

Select User 1

Pick first user from Users table

Scan All Rides

Search entire Rides table for user 1's trips

Sum Distances

Calculate total distance for user 1

Repeat Process

Repeat steps 1-3 for each remaining user

Code -

solution.c — C

// SQL solution in C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int user_id;
    char name[50];
} User;

typedef struct {
    int ride_id;
    int user_id;
    int distance;
} Ride;

typedef struct {
    int user_id;
    char name[50];
    int total_traveled_distance;
} Result;

int compare(const void *a, const void *b) {
    return ((Result*)a)->user_id - ((Result*)b)->user_id;
}

Result* solution(User* users, int user_count, Ride* rides, int ride_count) {
    Result* result = malloc(user_count * sizeof(Result));
    
    for (int i = 0; i < user_count; i++) {
        result[i].user_id = users[i].user_id;
        strcpy(result[i].name, users[i].name);
        result[i].total_traveled_distance = 0;
        
        for (int j = 0; j < ride_count; j++) {
            if (rides[j].user_id == users[i].user_id) {
                result[i].total_traveled_distance += rides[j].distance;
            }
        }
    }
    
    qsort(result, user_count, sizeof(Result), compare);
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each user (n), we scan all rides (m), resulting in n*m operations

⚠ Quadratic Growth

Space Complexity

O(1)

No additional space needed beyond the result set

✓ Linear Space

Constraints

1 ≤ Users.user_id ≤ 10⁶
1 ≤ Rides.ride_id ≤ 10⁶
1 ≤ Rides.distance ≤ 1000
All user_id values in Users table are unique
All ride_id values in Rides table are unique
user_id in Rides table references valid users in Users table

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Subquery for Each User (Brute Force) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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