Adobe coding & dsa interview questions

Sample questions with short answers

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

1. 1Rotate an N×N image matrix by 90 degrees in place.MediumAdobe-specificShow short answer

   Short answer

   Transpose the matrix (swap a[i][j] with a[j][i] for j > i), then reverse each row. O(N²) time and O(1) extra space. Adobe's image-heavy interviews often follow up with anti-clockwise rotation (reverse rows first, then transpose) or non-square matrices.

   Interviewers listen for

   • transpose then reverse each row
   • O(1) extra space
   • anti-clockwise follow-up
2. 2Implement a function to find the kth largest element in an unsorted array.MediumAdobe-specificShow short answer

   Short answer

   Use a min-heap of size k for O(n log k), or Quickselect for average O(n) by partitioning around a pivot and recursing into the side that contains the kth index. Mention Quickselect's O(n²) worst case and random pivots. Clarify duplicates and 1-based vs 0-based k.

   Interviewers listen for

   • min-heap of size k
   • Quickselect average O(n)
   • worst case and clarifications
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)