How do you solve Top Travellers on LeetCode?

Top Travellers (LeetCode #1407) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using aggregation functions like SUM() can significantly reduce computation time.

What algorithm or data structure is used to solve Top Travellers?

Top Travellers uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation Functions.

What are the interview tips for Top Travellers?

Always clarify the requirements before starting to code. Think about edge cases, such as users with no rides. Practice writing SQL queries to get comfortable with JOINs and GROUP BY.

What are common mistakes when solving Top Travellers?

Forgetting to handle users with no rides, which can lead to incorrect results. Not ordering the results correctly as per the requirements.

Top Travellers — LeetCode #1407 (Easy)

Tags: Database

Related patterns: Hash Map, Aggregation Functions

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

In this approach, we will calculate the total distance traveled by each user by iterating through the Rides table for each user in the Users table. This is straightforward but inefficient as it involves nested loops.

This complexity arises because for each user, we are iterating through all rides, leading to a nested loop structure.

Step 1: Initialize a dictionary to store the total distance for each user.
Step 2: For each user in the Users table, iterate through the Rides table to sum their distances.
Step 3: Store the total distance in the dictionary with the user's name as the key.
Step 4: Convert the dictionary to a list of tuples for sorting.
Step 5: Sort the list first by distance (descending) and then by name (ascending).
Step 6: Return the sorted list.

1. Users: [{1, 'Alice'}, {2, 'Bob'}, {3, 'Alex'}, {4, 'Donald'}, {7, 'Lee'}, {13, 'Jonathan'}, {19, 'Elvis'}] 2. Rides: [{1, 1, 120}, {2, 2, 317}, {3, 3, 222}, {4, 7, 100}, {5, 13, 312}, {6, 19, 50}, {7, 7, 120}, {8, 19, 400}, {9, 7, 230}] 3. Total distances: {1: 120, 2: 317, 3: 222, 4: 0, 7: 450, 13: 312, 19: 450} 4. Convert to list: [('Alice', 120), ('Bob', 317), ('Alex', 222), ('Donald', 0), ('Lee', 450), ('Jonathan', 312), ('Elvis', 450)] 5. Sort: [('Elvis', 450), ('Lee', 450), ('Bob', 317), ('Jonathan', 312), ('Alex', 222), ('Alice', 120), ('Donald', 0)] 6. Final output: [('Elvis', 450), ('Lee', 450), ('Bob', 317), ('Jonathan', 312)]

Optimal Solution approach

Time complexity: O(n). Space complexity: O(n).

By using SQL's aggregation functions effectively, we can compute the total distance for each user in a single pass through the Rides table, which is much more efficient than the brute-force method.

This complexity is due to a single pass through the Rides table to compute the total distances, making it much more efficient than the brute-force approach.

Step 1: Use a JOIN operation to combine Users and Rides tables on user_id.
Step 2: Use SUM() to calculate the total distance for each user.
Step 3: Use GROUP BY to group the results by user id.
Step 4: Use ORDER BY to sort the results first by total distance (descending) and then by name (ascending).
Step 5: Return the final result set.

1. Users: [{1, 'Alice'}, {2, 'Bob'}, {3, 'Alex'}, {4, 'Donald'}, {7, 'Lee'}, {13, 'Jonathan'}, {19, 'Elvis'}] 2. Rides: [{1, 1, 120}, {2, 2, 317}, {3, 3, 222}, {4, 7, 100}, {5, 13, 312}, {6, 19, 50}, {7, 7, 120}, {8, 19, 400}, {9, 7, 230}] 3. Total distances calculated: {1: 120, 2: 317, 3: 222, 4: 0, 7: 450, 13: 312, 19: 450} 4. Convert to list: [('Alice', 120), ('Bob', 317), ('Alex', 222), ('Donald', 0), ('Lee', 450), ('Jonathan', 312), ('Elvis', 450)] 5. Sort: [('Elvis', 450), ('Lee', 450), ('Bob', 317), ('Jonathan', 312), ('Alex', 222), ('Alice', 120), ('Donald', 0)] 6. Final output: [('Elvis', 450), ('Lee', 450), ('Bob', 317), ('Jonathan', 312)]

Key Insights

Using aggregation functions like SUM() can significantly reduce computation time.
LEFT JOIN ensures all users are included even if they have no rides.

Common Mistakes

Forgetting to handle users with no rides, which can lead to incorrect results.
Not ordering the results correctly as per the requirements.

Interview Tips

Always clarify the requirements before starting to code.
Think about edge cases, such as users with no rides.
Practice writing SQL queries to get comfortable with JOINs and GROUP BY.

#1407

Top Travellers

Easy

Database↗Hash Map↗Aggregation Functions↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

By using SQL's aggregation functions effectively, we can compute the total distance for each user in a single pass through the Rides table, which is much more efficient than the brute-force method.

⚙️

Algorithm

5 steps

1Step 1: Use a JOIN operation to combine Users and Rides tables on user_id.
2Step 2: Use SUM() to calculate the total distance for each user.
3Step 3: Use GROUP BY to group the results by user id.
4Step 4: Use ORDER BY to sort the results first by total distance (descending) and then by name (ascending).
5Step 5: Return the final result set.

solution.py6 lines

1# Full working Python code
2SELECT u.name, COALESCE(SUM(r.distance), 0) AS travelled_distance
3FROM Users u
4LEFT JOIN Rides r ON u.id = r.user_id
5GROUP BY u.id
6ORDER BY travelled_distance DESC, u.name ASC;

ℹ

Complexity note: This complexity is due to a single pass through the Rides table to compute the total distances, making it much more efficient than the brute-force approach.

1Using aggregation functions like SUM() can significantly reduce computation time.
2LEFT JOIN ensures all users are included even if they have no rides.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.