Read the grid in a spiral

Return all the values of a grid in spiral order: left along the top, down the right side, back along the bottom, up the left side, then inward and repeat.

Worked example#

For

1 2 3
4 5 6
7 8 9

the path is 1 2 3 across the top, 6 9 down the right, 8 7 back along the bottom, 4 up the left, and finally 5 in the middle.

The answer is 1 2 3 6 9 8 7 4 5.

Four walls closing in#

The clean way to write this is to hold four boundaries: top, bottom, left and right. Walk one edge, then move that boundary inward, and stop when the boundaries cross.

The case that catches everyone#

After the first two edges the remaining shape may be a single row or a single column. If you walk the bottom edge without checking top <= bottom first, you walk a row you have already read and emit its values twice.

So the last two of the four moves each need a guard:

if top <= bottom:  walk the bottom edge
if left <= right:  walk the left edge

The first two do not, because the loop condition has just checked them. A grid that is one row tall, or one column wide, is the shortest way to see this fail.

The cost#

Time O(rows x cols), since every cell is emitted exactly once. Memory O(1) beyond the answer itself.

Constraints#

• 1 <= rows, cols <= 100
• The grid need not be square
• -1000 <= value <= 1000