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The key to solving this problem is to join the Users and Rides tables on the user's ID, group the results by user, and compute the sum of the distances. Once the sum is calculated for each user, sort the results primarily by the total distance traveled in descending order and secondarily by the user's name in ascending order.
Time Complexity: O(n log n), where n is the number of rides due to aggregation and sorting
Space Complexity: O(1), excluding the result space.
1SELECT u.name, COALESCE(SUM(r.distance), 0) AS travelled_distance FROM Users u LEFT JOIN Rides r ON u.id = r.user_id GROUP BY u.id, u.name ORDER BY travelled_distance DESC, u.name ASC;Use a LEFT JOIN to ensure that every user is included in the result, even users without rides. COALESCE is used to handle null sums for users without rides by replacing null with 0. Finally, the results are grouped by user name and ID, and results are sorted by the travelled distance from highest to lowest, with secondary sorting alphabetically by name.
Another approach involves using arrays, hashes, or dictionaries in high-level programming languages to manually join, group, and sort data. First, construct a mapping of users, then iterate over the ride data to accumulate distances for each user id. After summation, sort the data based on distance and name accordingly.
Time Complexity: O(n log n) due to sorting operation, where n is the number of users + rides.
Space Complexity: O(n) to store travelled distances and sorted result.
1#include <stdio.h>
2#include <string.h>
3#include <stdlib.h>
4
5typedef struct {
This solution utilizes standard arrays and the qsort library function for sorting. The approach involves iterating through the rides, summing the distances for each user and making use of the qsort function to order the results first by distance (descending) and then by name (ascending).