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Uber·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Uber SWE interview, got a tree construction problem that looked straightforward until I started thinking about the recursion. Nothing too wild but it required some careful thinking about base cases.

Questions Asked (1)

Q1

Given an n x n binary matrix where n is a power of 2, build a Quad-Tree representation of it. Each node is either a leaf (all values in its region are the same) or an internal node with four children corresponding to the four quadrants.

Algorithms & Data Structures
Author's notes

The recursion itself isn't bad once you see it, but I fumbled the stopping condition at first.

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AI HintsAI Generated

Suggested Approach

Use a recursive divide-and-conquer strategy: for each region, check if all values are the same; if so, create a leaf node, otherwise split into four quadrants and recurse. This naturally builds the quad-tree in O(n^2) time by visiting each cell once.

Pro tip: Mention that you can optimize the uniformity check by using prefix sums to query any submatrix in O(1), reducing the overall time complexity to O(n^2) while avoiding redundant scans.

1. Define the Node structure

Clearly define the QuadTreeNode with attributes: val (boolean for leaf), isLeaf (boolean), and topLeft, topRight, bottomLeft, bottomRight pointers.

2. Implement recursive helper

Write a function that takes the matrix and the top-left coordinates (row, col) and size of the current region. It returns the root of the quad-tree for that region.

3. Check uniformity

For the current region, check if all values are the same. If yes, return a leaf node with that value. If no, proceed to split.

4. Split and recurse

Divide the region into four equal quadrants (size/2) and recursively build the quad-tree for each. Create an internal node with these four children.

5. Analyze complexity

Explain that the time complexity is O(n^2) because each cell is visited once, and space complexity is O(n^2) in the worst case (e.g., checkerboard pattern).

Key Points to Mention

  • Base case: region size 1 is always a leaf.
  • Uniformity check can be optimized with prefix sums for O(1) submatrix queries.
  • Recursive division into four quadrants of size n/2 x n/2.
  • Node structure: isLeaf, val, and four child pointers.
  • Time complexity O(n^2) and space complexity O(n^2) worst-case.
  • Handling of edge cases like n=1 or all same values.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.