Uber coding & dsa interview questions

Sample questions with short answers

Tap a question to reveal the short answer and key points.

1. 1Given driver locations, find the K nearest drivers to a rider.MediumUber-specificShow short answer

   Short answer

   Compute distances and keep a max-heap of size K, giving O(n log K). In practice drivers are bucketed by geohash or H3 cell so you only scan nearby cells, then sort the small candidate set. Mention Haversine vs planar approximation and how to handle ties and moving drivers.

   Interviewers listen for

   • max-heap of size K
   • geohash or H3 bucketing to limit candidates
   • Haversine vs planar approximation
2. 2Merge overlapping time intervals (driver shifts).MediumUber-specificShow short answer

   Short answer

   Sort intervals by start, then sweep: if the current interval starts before the last merged one ends, extend the end; otherwise push a new interval. O(n log n) due to the sort. Clarify whether touching intervals count as overlapping and handle empty input.

   Interviewers listen for

   • sort by start
   • single sweep extending the end
   • O(n log n) and touching intervals edge case
3. 3Given an array of integers and a target, return the indices of the two numbers that add up to the target. How would you approach it and what is the complexity?EasyAsked everywhereShow short answer

   Short answer

   Walk the array once keeping a hash map of value → index. For each element check whether target − value is already in the map; if so return both indices. This is O(n) time and O(n) space, versus the O(n²) brute-force double loop. Clarify whether exactly one solution exists and whether the same element can be reused.

   Interviewers listen for

   • hash map of value to index
   • walk the array once, O(n) time
   • clarify duplicates and reuse of an element
4. 4How do you detect a cycle in a linked list without using extra memory?MediumAsked everywhereShow short answer

   Short answer

   Use Floyd's tortoise-and-hare: advance a slow pointer by one node and a fast pointer by two. If the fast pointer reaches null there is no cycle; if the pointers ever meet there is one. It runs in O(n) time with O(1) extra space. To find the cycle's start, reset one pointer to the head and move both one step at a time until they meet again.

   Interviewers listen for

   • slow and fast pointers
   • O(1) extra space
   • pointers meeting proves a cycle
5. 5Explain the difference between BFS and DFS and when you would choose each.MediumAsked everywhereShow short answer

   Short answer

   BFS explores level by level using a queue, so it finds the shortest path in an unweighted graph and suits 'fewest steps' problems. DFS goes deep first using a stack or recursion, uses less memory on wide graphs, and suits connectivity, cycle detection, topological sort and backtracking. Both are O(V + E); pick by what the problem asks for and the graph's shape.

   Interviewers listen for

   • BFS uses a queue and finds shortest paths
   • DFS uses a stack or recursion
   • both O(V + E)