PPractick

The problem

A photo editor stores images as n × n grids of pixel values, and the "rotate right" button must turn the image 90 degrees clockwise in place, since phones don't have memory to spare for a second copy of a large image. Mapping each pixel directly to its new position and writing it in place would overwrite pixels you still need. The elegant solution is two simple in-place steps: transpose the matrix (swap across the main diagonal), then reverse each row.

Example: [[1, 2, 3], [4, 5, 6], [7, 8, 9]] → [[7, 4, 1], [8, 5, 2], [9, 6, 3]].

Function signature

void rotate(vector<vector<int>>& m);  // n x n matrix, rotated 90 degrees clockwise in place

Test cases

Input m	Output	Explanation
[[1,2,3],[4,5,6],[7,8,9]]	[[7,4,1],[8,5,2],[9,6,3]]	The first row becomes the last column
[[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]	[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]	4 × 4
[[1,2],[3,4]]	[[3,1],[4,2]]
[[1]]	[[1]]

(Confirm: the matrix is square; the rotation is clockwise and done in place.)

How interviewers phrase it

• "Rotate an n × n image by 90 degrees clockwise, in place" (LeetCode 48).
• Variants: counter-clockwise, 180 degrees, or a non-square matrix (which can't be done in place).

Clarifying questions to ask

1. Clockwise or counter-clockwise?
2. Is the matrix square?
3. Must it be in place?

Follow-ups to expect

1. "Where does element (i, j) go?"
2. "Why does transpose-then-reverse work?"
3. "How do you rotate counter-clockwise?"
4. "Is there a one-pass alternative?"
5. "Complexity?"

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