The removal condition tripped me up at first.
First, clarify the problem constraints and edge cases, then propose a two-phase solution: marking cells for removal based on neighbor counts, followed by gravity-drop per column. Discuss time/space complexity and potential optimizations, and be prepared to code the solution.
Pro tip: Mention that you would use a copy of the matrix for marking to avoid cascading removals, and that gravity can be implemented efficiently with a two-pointer approach per column.
Ask about matrix size limits, whether diagonal neighbors count, and if multiple rounds are needed. Confirm that only one round of elimination is performed.
Iterate through each cell, count how many of its 4 neighbors (up, down, left, right) have the same value. If count >= 2, mark the cell for removal (e.g., in a separate boolean matrix or by setting to a sentinel).
Set all marked cells to 0. Ensure that marking is based on the original matrix to avoid cascading effects within the same round.
For each column, use a two-pointer approach: one pointer from the bottom to place non-zero values, another to scan upwards. This compacts non-zero values to the bottom, filling the top with zeros.
State time complexity O(m*n) and space O(m*n) for the marking matrix (or O(1) if modifying in place with care). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.