Amazon coding & dsa interview questions

Sample questions with short answers

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

1. 1Given a grid of 0s and 1s, count the number of islands.MediumAmazon-specificShow short answer

   Short answer

   Scan the grid; on each unvisited 1 start a DFS or BFS that marks every connected land cell, and increment the count. O(rows × cols) time; space is the recursion stack or queue. Amazon loves follow-ups: largest island, diagonal adjacency, or streaming rows — say how each changes the traversal.

   Interviewers listen for

   • flood fill with DFS or BFS
   • mark visited cells
   • O(rows × cols)
2. 2Find the top K most frequent words in a large document.MediumAmazon-specificShow short answer

   Short answer

   Count frequencies with a hash map, then keep a min-heap of size K ordered by frequency (and word for ties) while iterating the map; pop the heap at the end. O(n log K). For truly large inputs mention streaming counts across machines and merging per-shard top-K lists.

   Interviewers listen for

   • hash map frequency count
   • min-heap of size K
   • ties and merging per-shard top-K
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)