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I got the basic rotation right pretty fast, reversing the flattened list and reshaping.
Start by clarifying the problem and constraints, then present a clean solution for the 3x3 case using index mapping. Discuss in-place vs out-of-place trade-offs, and finally generalize to n by m grids, highlighting edge cases and complexity.
Pro tip: Mention that for a 180-degree rotation, you can swap elements in pairs (i, j) with (n-1-i, m-1-j) to achieve in-place rotation without extra space. Also, note that the same logic applies to any rectangular grid, but be careful with odd dimensions.
Ask if the grid is square, if it's mutable, and if in-place is required. Confirm the definition of 180-degree rotation (both rows and columns reversed).
Create a new grid and map each element (i, j) to (n-1-i, m-1-j). This is straightforward and easy to verify.
For in-place, iterate over half the elements and swap with their 180-degree counterpart. Handle odd dimensions by skipping the center element if n and m are both odd.
Explain that the same index mapping works for any n and m. The in-place swap still works by iterating over the first half of the flattened grid or using nested loops with careful bounds.
State time complexity O(n*m) and space complexity O(1) for in-place, O(n*m) for out-of-place. Mention edge cases: empty grid, 1x1, non-square, and odd dimensions.
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