Longest Decreasing Matrix Path with Limited Relaxations

Given an integer matrix and a nonnegative integer k, return a longest orthogonal path of cell coordinates.
A move to an up, down, left, or right neighbor is:
free when the next value is strictly smaller than the current value; or
a relaxation when the next value is greater than or equal to the current value.
The path may use at most k relaxed moves. Because every cycle contains at least one non-decreasing move, the relaxation budget keeps every valid walk finite; a coordinate may therefore be revisited only after spending additional budget.
Return the coordinates as [row, column] rows. If several paths have maximum length, return the lexicographically smallest coordinate sequence. Setting k = 0 is the original longest-decreasing-path task, while returning coordinates and accepting positive k preserve both reported follow-ups.

Function
longestRelaxedDecreasingPath(matrix: int[][], k: int) → int[][]

Examples
Example 1
matrix = [[9,8,7],[2,3,6],[1,4,5]]
k = 0
return = [[0,0],[0,1],[0,2],[1,2],[2,2],[2,1],[1,1],[1,0],[2,0]]
The spiral visits values 9, 8, 7, 6, 5, 4, 3, 2, 1. Every move is strictly decreasing, so no relaxation is used.
Example 2
matrix = [[1,2]]
k = 1
return = [[0,1],[0,0],[0,1],[0,0]]
Start at value 2, move down to 1 for free, spend the one relaxation to return to 2, then move down to 1 again. The repeated coordinates have different remaining-budget states.
Example 3
matrix = [[4,3],[2,1]]
k = 0
return = [[0,0],[0,1],[1,1]]
There are multiple decreasing paths of length three. The returned one is lexicographically smallest.

Constraints
1 <= matrix.length, matrix[0].length <= 20.
All rows have the same length.
-10^9 <= matrix[r][c] <= 10^9.
0 <= k <= 5.

The trick is that the state isn't a cell, it's (r, c, budget left). With k at most 5 and a grid at most 20x20, that's about 2400 states, small enough to memoize. Define best(r, c, j) as the longest path starting at that cell with j relaxations left. For each neighbor, a strictly smaller value costs 0 and a value greater than or equal costs 1, which needs j >= 1. Finiteness holds because every cycle burns budget, so the recursion terminates. The pitfall is tie-breaking. Compare by length first, then compare coordinate sequences, and try all four neighbors rather than greedily picking the smallest. Also rebuild the path from stored choices, not just lengths. Starting cells need the same comparison. A plain BFS over paths blows up, so use DFS with memoization over the layered states, or DP ordered by budget.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.