PPractick

The problem

A printer for a museum's "spiral mosaic" exhibit reads tiles from a rectangular grid in spiral order: across the top row, down the right column, back along the bottom row, up the left column, then around again one layer further in. The logic sounds simple, but spiral traversal is a notorious source of off-by-one bugs, especially on non-square grids and single rows or columns. Four shrinking boundaries (top, bottom, left, right) keep it under control.

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

Function signature

vector<int> spiralOrder(const vector<vector<int>>& m);  // m is non-empty, possibly non-square

Test cases

Input m	Output	Explanation
[[1,2,3],[4,5,6],[7,8,9]]	[1,2,3,6,9,8,7,4,5]
[[1,2,3,4],[5,6,7,8],[9,10,11,12]]	[1,2,3,4,8,12,11,10,9,5,6,7]	3 × 4: the inner layer is a single row
[[1,2],[3,4]]	[1,2,4,3]
[[1,2,3]]	[1,2,3]	A single row must not be read backwards again
[[1],[2],[3]]	[1,2,3]	A single column must not be read upwards again
[[7]]	[7]

(Confirm: the matrix may be rectangular; every element appears exactly once in the output.)

How interviewers phrase it

• "Return all elements of a matrix in spiral order" (LeetCode 54).
• Variants: generate an n × n matrix filled in spiral order (LeetCode 59), or spiral from a given starting cell.

Clarifying questions to ask

1. Is the matrix square or rectangular?
2. Clockwise, starting at the top-left?
3. Can it be empty?

Follow-ups to expect

1. "How do you avoid double-counting on a single row or column?"
2. "How many layers are there?"
3. "Is there another way?"
4. "Complexity?"

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