It would be something like the following

1. store current value (var_next)

2. check adjacent cells for equal to the (var_next) or less while skipping over the (var_last) cell

- true, store value to (var_next)

3. once completed all checks:

- store the current value to (var_last)

- move to (var_next)

4. repeat steps 1. to 3. until there are no more cells to move to.

If you are looking for a different type of answer, then please update your question.

**UPDATE**

There was a little more to this than I thought. Needed to also track the path taken.

Here is the solution that I came up with. It's written in C#:

1. Tests used:

C#

int[,] cells1 = { { 1, 2, 3 }, { 6, 5, 4 }, { 7, 8, 9 } }; int[]? result1 = ProcessCells(cells1); DisplayResults(cells1, result1); int[,] cells2 = { { 0, 1, 2 }, { 2, 2, 2 }, { 3, 4, 5 } }; int[]? result2 = ProcessCells(cells2); DisplayResults(cells2, result2); int[,] cells3 = { { 6, 6, 6 }, { 5, 8, 7 }, { 4, 8, 9 } }; int[]? result3 = ProcessCells(cells3); DisplayResults(cells3, result3); // test a single cell matrix int[,] singleCells = { { 1 } }; int[]? singleResult = ProcessCells(singleCells); DisplayResults(singleCells, singleResult); // test a large matrices int[,] largeMatrix1 = { { 8, 9, 10, 11, 12, 13, }, { 7, 16, 15, 14, 14, 14, }, { 6, 17, 30, 31, 32, 33, }, { 5, 18, 29, 26, 25, 34, }, { 4, 19, 28, 27, 24, 35, }, { 3, 20, 21, 22, 23, 36, }, }; int[]? largeResult1 = ProcessCells(largeMatrix1); DisplayResults(largeMatrix1, largeResult1); int[,] largeMatrix2 = { { 1, 1, 1, 1, 1, 2, }, { 1, 3, 2, 2, 2, 2, }, { 1, 3, 8, 8, 8, 8, }, { 1, 4, 7, 6, 6, 9, }, { 1, 4, 7, 6, 6, 9, }, { 0, 4, 5, 5, 5, 9, }, }; int[]? largeResult2 = ProcessCells(largeMatrix2); DisplayResults(largeMatrix2, largeResult2); // test a solution that can not be completed int[,] badCells = { { 0, 0, 3 }, { 6, 0, 4 }, { 7, 8, 9 } }; int[]? badResult = ProcessCells(badCells); DisplayResults(badCells, badResult);

And here is the code:

C#

static int[]? ProcessCells(int[,] cells) { // get the bounds of the array int rows = cells.GetLength(0); int cols = cells.GetLength(1); // is the array valid? if (rows < 1) return default; if (cols < 1) return default; // remember where we have been... int[,] path = new int[rows, cols]; // track the results int[] results = new int[rows * cols]; // initialize results to failed for (int i = 0; i < results.Length; i++) results[i] = -1; // initialize the starting value and cell int startValue = cells[rows - 1, cols - 1]; int currentValue = startValue; int[] currentCell = {rows - 1, cols - 1}; path[rows - 1, cols - 1] = -1; // set pointer to start of results int index = 0; // log the start cell results[index] = cells[currentCell[0], currentCell[1]]; // now walk the 2D array while (currentCell[0] >= 0 && currentCell[1] >= 0) { // track valid cell int[]? nextCell = default; int nextValue = -1; // look around for (int direction = 0; direction < 4; direction++) { // get the adjacent cell that we want to look at int[]? cell = GetValidAdjacentCell(direction, currentCell, rows, cols); // check if not a valid direction if(cell is null || path[cell[0], cell[1]] != 0) continue; int testValue = cells[cell[0], cell[1]]; // do we have a valid value? if ((testValue == currentValue || testValue == currentValue - 1) && Math.Max(nextValue, testValue) == testValue) { nextValue = testValue; nextCell = new[] { cell[0], cell[1] }; } } // no more to do if (nextCell is null) break; // track value results[++index] = cells[nextCell[0], nextCell[1]]; // track where we are currentCell = new[] { nextCell[0], nextCell[1] }; currentValue = nextValue; // track where we have been path[nextCell[0], nextCell[1]] = -1; } // we are all done return results; } static int[]? GetValidAdjacentCell(int direction, int[] currentCell, int rows, int cols) { int cr = currentCell[0]; int cc = currentCell[1]; // check based on importance of order if (direction == 0) // left return cc == 0 ? default : new[] { cr, --cc }; if (direction == 1) // right return cc == cols - 1 ? default : new[] { cr, ++cc }; if (direction == 2) // down return cr == 0 ? default : new[] { --cr, cc }; if (direction == 3) // up return cr == rows - 1 ? default : new[] { ++cr, cc }; // invalid direct value return default; } static void DisplayResults(int[,] cells, int[]? path) { int rows = cells.GetLength(0); int cols = cells.GetLength(1); int padLeft = cells[rows - 1, cols - 1].ToString().Length + 1; Console.WriteLine("Matrix Traversed:"); for (int row = 0; row < rows; row++) { for (int col = 0; col < cols; col++) Console.Write(cells[row, col].ToString().PadLeft(padLeft)); Console.WriteLine(); } Console.WriteLine(); int count = path.Count(value => value != -1); if (path.Length != count) Console.WriteLine($"Path cannot be completed. Only traversed {count} cells of {path.Length}"); Console.WriteLine($"Path found: {string.Join(", ", path)}"); Console.WriteLine(); }

3. And the output:

Matrix Traversed: 1 2 3 6 5 4 7 8 9 Path found: 9, 8, 7, 6, 5, 4, 3, 2, 1 Matrix Traversed: 0 1 2 2 2 2 3 4 5 Path found: 5, 4, 3, 2, 2, 2, 2, 1, 0 Matrix Traversed: 6 6 6 5 8 7 4 8 9 Path found: 9, 8, 8, 7, 6, 6, 6, 5, 4 Matrix Traversed: 1 Path found: 1 Matrix Traversed: 8 9 10 11 12 13 7 16 15 14 14 14 6 17 30 31 32 33 5 18 29 26 25 34 4 19 28 27 24 35 3 20 21 22 23 36 Path found: 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 14, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3 Matrix Traversed: 1 1 1 1 1 2 1 3 2 2 2 2 1 3 8 8 8 8 1 4 7 6 6 9 1 4 7 6 6 9 0 4 5 5 5 9 Path found: 9, 9, 9, 8, 8, 8, 8, 7, 7, 6, 6, 6, 6, 5, 5, 5, 4, 4, 4, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0 Matrix Traversed: 0 0 3 6 0 4 7 8 9 Path cannot be completed. Only traversed 4 cells of 9 Path found: 9, 8, 7, 6, -1, -1, -1, -1, -1

The last test will fail as it has no complete path.

All code is fully commented.

Enjoy!

You can use a bottom-up approach to fill in the values of this array by iterating through the matrix and for each element, checking its neighbors (up, down, left, right) to determine if it is possible to reach them while satisfying your criteria. You can then use the values stored in the array to find the longest path.

Here's an outline of the algorithm:

Create a 2D array to store the maximum number of segments of decreasing values and minimum number of segments of the same value that can be obtained from each element in the matrix.

Initialize the array with the values from the input matrix.

Iterate through the matrix and for each element, check its neighbors (up, down, left, right) to determine if it is possible to reach them while satisfying your criteria.

If it is possible to reach a neighbor, update the values in the array for that neighbor by adding 1 to the number of segments of decreasing values or leaving it the same for the number of segments of the same value.

Once you have filled in the entire array, you can use it to find the longest path by starting at the element with the highest value of maximum number of segments of decreasing values and minimum number of segments of the same value.

Follow the path by checking the values stored in the array and moving to the neighbor with the highest value.

It's worth noting that this algorithm will have a time complexity of O(n^2) since you are iterating through the matrix twice.

You can also try a Depth-First Search (DFS) with a memory table, where the DFS starts from each element of the matrix and stores the information of the maximum number of segments of decreasing values and minimum number of segments of the same value that can be obtained from each element in the matrix, this will allow you to avoid repeating the same calculation and speeds up your algorithm.

In addition, you can try to optimize your DFS algorithm by using a priority queue to sort the possible next cells based on their decreasing value, then the cells that have the same value, this way you will not have to explore all the cells, but only the ones that are most likely to be part of the solution, this can help you improve the performance of your algorithm.