Rotate Image

You are given an n × n integer matrix representing an image. Rotate the image by 90 degrees clockwise.
Perform the rotation in place by modifying matrix directly. Return that same matrix after the mutation so the runner can compare its final values; do not allocate another n × n matrix in the final approach.

Function
rotateImage(matrix: int[][]) → int[][]

Examples
Example 1
matrix = [[1,2,3],[4,5,6],[7,8,9]]
return = [[7,4,1],[8,5,2],[9,6,3]]
The former bottom-left element becomes the new top-left element, and every position moves through one clockwise quarter-turn.
Example 2
matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
return = [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
Each original column becomes a row in reverse vertical order.

Constraints
n == matrix.length == matrix[i].length.
1 <= n <= 20.
-1000 <= matrix[i][j] <= 1000.

The trick: a clockwise quarter-turn equals a transpose (swap matrix[i][j] with matrix[j][i] for j > i) followed by reversing every row. That's O(n^2) time and O(1) extra space, which satisfies the in-place rule. The common pitfall is transposing the full grid, which swaps every pair twice and returns the original. Start the inner loop at i+1. The second pitfall is mixing up directions. Reverse rows for clockwise. For counterclockwise you'd reverse columns instead, or reverse rows first then transpose. Check it on Example 1: transpose gives [[1,4,7],[2,5,8],[3,6,9]], reversing rows gives [[7,4,1],[8,5,2],[9,6,3]]. With n up to 20, performance is never the issue. Correctness is. If you blank during the live OA, StealthCoder can hand you the two-step solution, but you should be able to write it from memory.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.