top, bottom, left and right mark the rectangle of cells that are not in result yet. Cell matrix[r][c] is unread exactly when top <= r <= bottom and left <= c <= right. This is the card's invariant.

Each sweep reads one side of that rectangle, and then moves the boundary on that side inward by 1. The side that was just read is no longer inside the rectangle. So the rectangle always shrinks by exactly the cells that were read.

In Spiral Matrix (LC 54) with a 3 by 4 matrix, the first four sweeps read 1, 2, 3, 4, then 8, 12, then 11, 10, 9, then 5. The rectangle is now one row and two columns: top = 1, bottom = 1, left = 1, right = 2. The second ring reads 6, 7. The final result is [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7].

The four sweeps of one ring read the top row, the right column, the bottom row and the left column of the rectangle. The top row is read first, and top += 1 removes it from the rectangle. So the right sweep starts at top, which is the row below the top-right corner. That corner is not read twice.

The same order removes the other corners. right -= 1 removes the right column before the bottom sweep, so the bottom sweep starts at the cell left of the bottom-right corner. bottom -= 1 removes the bottom row before the left sweep, so the left sweep stops above the bottom-left corner.

Every cell is in one ring, and every ring is read by one pass of the while loop. So result gets R * C values, one per cell, and no cell appears twice. The number of rings is the smaller side divided by 2, rounded up. A 3 by 4 matrix has 2 rings, a 5 by 5 matrix has 3 rings, and a matrix with one row has 1 ring.

A full ring has at least two rows and two columns. The last ring can be one row, one column or one cell. In that case the top sweep and the right sweep can already use up the rows or the columns.

The 3 by 4 example shows this. In the second ring, the top sweep reads 6, 7 and sets top = 2, while bottom = 1. The test top <= bottom is false, so the bottom sweep is skipped. Without the guard, it runs range(right, left - 1, -1) on row 1 and adds 6 a second time. The result then ends with ..., 5, 6, 7, 6.

The single row [[1, 2, 3]] gives [1, 2, 3, 2, 1] without the first guard. The single column [[1], [2], [3]] gives [1, 2, 3, 2] without the second guard, left <= right. With both guards, the results are [1, 2, 3] for both inputs.

The rule is: after a sweep changes a boundary, ask whether the next side still has cells. The top sweep and the right sweep do not need a guard, because the while condition already guarantees one row and one column at the start of each ring.

Every cell is read once, so the time is O(R * C). The only extra memory is four integers, so the auxiliary space is O(1). The output list holds R * C values, but it is the answer and does not count as extra memory.

A different method keeps a visited grid and a direction, and turns right when the next cell is outside the matrix or already visited. It produces the same order, but it needs O(R * C) extra memory for visited.

The same four boundaries solve Spiral Matrix II (LC 59), which writes the numbers 1 to n * n into an empty grid in spiral order: replace result.append(...) with an assignment from a counter. Spiral Matrix IV (LC 2326) does the same with the values of a linked list. Rotate Image (LC 48) and Cyclically Rotating a Grid (LC 1914) treat each ring as a closed loop of cells, and move the values along that loop.