Atlassian coding & dsa interview questions

Sample questions with short answers

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

1. 1Design a simple rate limiter class: allow N requests per user per minute.MediumAtlassian-specificShow short answer

   Short answer

   Keep a per-user deque of timestamps; on each request pop timestamps older than 60 seconds, then allow if the deque has fewer than N entries. O(1) amortised per request, O(N) memory per user. Atlassian's coding rounds are practical — discuss thread safety and a fixed-window counter alternative.

   Interviewers listen for

   • per-user sliding window of timestamps
   • pop timestamps older than 60 seconds then compare to N
   • thread safety and alternatives
2. 2Given tasks with dependencies, return a valid execution order or report a cycle.MediumAtlassian-specificShow short answer

   Short answer

   Build a directed graph and run Kahn's algorithm: start with nodes of in-degree zero, repeatedly pop one, append it to the order, and decrement its neighbours' in-degrees. If the order has fewer nodes than the graph, there is a cycle. O(V + E). DFS with colours is the alternative.

   Interviewers listen for

   • topological sort with Kahn's algorithm
   • fewer nodes in the order than the graph means a cycle
   • O(V + E)
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)