Sort Matrix Borders

Given matrix, an n x m rectangular matrix of integers, let's define its 0-border as the union of its leftmost and rightmost columns, as well as its top and bottom rows. A vector's 0-border is the vector itself.
If we were to remove the matrix's 0-border, then the 0-border of the resulting matrix can be defined as the 1-border of the original matrix. We can continue this way to define the 2-border, 3-border, etc, until we reach the center of the matrix.
For each valid k, your task is to sort the elements in each k-border and place them clockwise in ascending order, starting from the top-left corner.
Note: You are not expected to provide the most optimal solution, but a solution with time complexity not worse than O(n * m * (n + m)) will fit within the execution time limit.

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

Examples
Example 1
matrix = [[9,7,-4,5],[1,6,2,-6],[12,20,2,0]]
return = [[-6,-4,0,1],[20,2,6,2],[12,9,7,5]]
For
matrix = [[9, 7, -4, 5],
[1, 6, 2, -6],
[12, 20, 2, 0]]
FastPrep-authored deterministic derivation: The source image shows this example input but is cropped before its output. Applying the visible border-sorting rule gives the executable output shown here.

Constraints
FastPrep execution-adapter constraints (not shown in the source image):
matrix is non-empty.
Every row has the same positive length.
Every entry is an integer.

The trick is treating each k-border as a ring. For k from 0 while the ring still exists, walk the perimeter clockwise: top row left to right, right column down, bottom row right to left, left column up. Collect the values into a list, sort ascending, then walk the exact same path and write them back in order. Because the traversal order is shared, the write-back is trivial. Total work is about n * m * log, well under the stated bound. The pitfalls are all edge cases: a ring that collapses to a single row, a single column, or a single cell. If you naively run all four sides you'll double-count cells and corrupt the output. Guard the bottom row with top != bottom and the left column with left != right. If you blank on those guards mid-assessment, StealthCoder is the hedge running invisibly while you keep typing.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.