I jumped straight to the main logic and forgot to ask about negatives until they brought it up themselves, which was a bit embarrassing.
Start by clarifying the problem and edge cases, then propose a single-pass O(mn) solution that compares each cell (except those in the first row and first column) with its top-left neighbor. Discuss space trade-offs: O(1) extra space for in-place comparison versus O(m+n) for storing diagonal representatives, and mention how to handle empty matrices and negative values.
Pro tip: Emphasize that the O(1) space solution is optimal and that you can early-exit upon finding a mismatch, which is crucial for large matrices. Also, mention that negative values are handled naturally since we only compare equality, not magnitude.
Ask about matrix dimensions, empty input, single row/column, and whether negative values are allowed. Confirm that diagonals are defined from top-left to bottom-right.
Iterate through each cell except those in the first row and first column, and check if matrix[i][j] equals matrix[i-1][j-1]. If any mismatch, return false; otherwise, return true.
Time complexity is O(mn) since each cell is visited once. Space complexity is O(1) extra space, as we only use a few variables. Discuss alternative O(m+n) space approach if needed.
For empty matrix (m=0 or n=0), return true. For single row or column, automatically true since each diagonal has length 1. Negative values are handled by direct equality comparison.
Walk through a small example, including a mismatch case. Compare the O(1) space solution with a hashmap-based O(m+n) space solution, highlighting when the latter might be preferable (e.g., if matrix is read-only or we need to process diagonals independently).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.