My first instinct was to just scan each subregion naively every time, which works but gets slow on bigger grids.
Use a recursive divide-and-conquer strategy: for each submatrix, check if all values are the same; if so, return a leaf node with that value, otherwise split into four quadrants and recurse. Emphasize that the recursion depth is O(log n) and each cell is visited once, leading to O(n^2) time and O(log n) space for the recursion stack.
Pro tip: Mention that you can optimize the uniformity check by using prefix sums to query any submatrix in O(1), but note that the naive recursive check is still O(n^2) overall because each cell is visited once. Also, clarify that the space complexity is O(log n) for the recursion stack, not O(n^2), since the tree depth is logarithmic.
Confirm that n is a power of 2, the matrix is binary, and that uniform regions become leaf nodes. Ask if the tree should be built in-place or if additional space is allowed.
Design a function that takes the top-left coordinates and size of a submatrix. It checks if all values in the submatrix are the same; if so, returns a leaf node; otherwise, splits into four quadrants and recurses.
For the current submatrix, compare all values to the first element. If any differ, it's non-uniform. Optionally, use prefix sums to optimize repeated checks, but note the trade-off.
Explain that each cell is visited once, so time is O(n^2). The recursion depth is O(log n), so space is O(log n) for the stack, plus O(number of nodes) for the tree.
Mention that using prefix sums can reduce the uniformity check to O(1) per submatrix, but the overall time remains O(n^2) because of the number of submatrices. Also, discuss iterative vs recursive approaches.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.