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+1368\. Minimum Cost to Make at Least One Valid Path in a Grid
+
+**Difficulty:** Hard
+
+**Topics:** `Array`, `Breadth-First Search`, `Graph`, `Heap (Priority Queue)`, `Matrix`, `Shortest Path`
+
+Given an `m x n` grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of `grid[i][j]` can be:
+
+- `1` which means go to the cell to the right. (i.e go from `grid[i][j]` to `grid[i][j + 1]`)
+- `2` which means go to the cell to the left. (i.e go from `grid[i][j]` to `grid[i][j - 1]`)
+- `3` which means go to the lower cell. (i.e go from `grid[i][j]` to `grid[i + 1][j]`)
+- `4` which means go to the upper cell. (i.e go from `grid[i][j]` to `grid[i - 1][j]`)
+
+Notice that there could be some signs on the cells of the grid that point outside the grid.
+
+You will initially start at the upper left cell `(0, 0)`. A valid path in the grid is a path that starts from the upper left cell `(0, 0)` and ends at the bottom-right cell `(m - 1, n - 1)` following the signs on the grid. The valid path does not have to be the shortest.
+
+You can modify the sign on a cell with `cost = 1`. You can modify the sign on a cell one time only.
+
+Return _the minimum cost to make the grid have at least one valid path_.
+
+**Example 1:**
+
+![grid1](https://assets.leetcode.com/uploads/2020/02/13/grid1.png)
+
+- **Input:** grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
+- **Output:** 3
+- **Explanation:** You will start at point (0, 0).
+  The path to (3, 3) is as follows. (0, 0) --> (0, 1) --> (0, 2) --> (0, 3) change the arrow to down with cost = 1 --> (1, 3) --> (1, 2) --> (1, 1) --> (1, 0) change the arrow to down with cost = 1 --> (2, 0) --> (2, 1) --> (2, 2) --> (2, 3) change the arrow to down with cost = 1 --> (3, 3)
+  The total cost = 3.
+
+**Example 2:**
+
+![grid2](https://assets.leetcode.com/uploads/2020/02/13/grid2.png)
+
+- **Input:** grid = [[1,1,3],[3,2,2],[1,1,4]]
+- **Output:** 0
+- **Explanation:** You can follow the path from (0, 0) to (2, 2).
+
+
+**Example 3:**
+
+![grid3](https://assets.leetcode.com/uploads/2020/02/13/grid3.png)
+
+- **Input:** grid = [[1,2],[4,3]]
+- **Output:** 1
+
+
+**Constraints:**
+
+- `m == grid.length`
+- `n == grid[i].length`
+- `1 <= m, n <= 100`
+- `1 <= grid[i][j] <= 4`
+
+
+**Hint:**
+1. Build a graph where `grid[i][j]` is connected to all the four side-adjacent cells with weighted edge. the weight is `0` if the sign is pointing to the adjacent cell or `1` otherwise.
+2. Do BFS from `(0, 0)` visit all edges with `weight = 0` first. the answer is the distance to `(m -1, n - 1)`.
+
+
+
+**Solution:**
+
+We can use the **0-1 BFS** approach. The idea is to traverse the grid using a deque (double-ended queue) where the cost of modifying the direction determines whether a cell is added to the front or back of the deque. The grid is treated as a graph where each cell has weighted edges based on whether its current direction matches the movement to its neighbors.
+
+Let's implement this solution in PHP: **[1368. Minimum Cost to Make at Least One Valid Path in a Grid](https://github.com/mah-shamim/leet-code-in-php/tree/main/algorithms/001368-minimum-cost-to-make-at-least-one-valid-path-in-a-grid/solution.php)**
+
+```php
+<?php
+/**
+ * @param Integer[][] $grid
+ * @return Integer
+ */
+function minCost($grid) {
+    ...
+    ...
+    ...
+    /**
+     * go to ./solution.php
+     */
+}
+
+// Example Test Cases
+$grid1 = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]];
+echo minCost($grid1) . "\n"; // Output: 3
+
+$grid2 = [[1,1,3],[3,2,2],[1,1,4]];
+echo minCost($grid2) . "\n"; // Output: 0
+
+$grid3 = [[1,2],[4,3]];
+echo minCost($grid3) . "\n"; // Output: 1
+?>
+```
+
+### Explanation:
+
+1. **Direction Mapping:** Each direction (`1` for right, `2` for left, `3` for down, `4` for up) is mapped to an array of movement deltas `[dx, dy]`.
+
+2. **0-1 BFS:**
+   - A deque is used to prioritize cells with lower costs. Cells that do not require modifying the direction are added to the front (`unshift`), while those that require a modification are added to the back (`enqueue`).
+   - This ensures that cells are processed in increasing order of cost.
+
+3. **Distance Array:** A 2D array `$dist` keeps track of the minimum cost to reach each cell. It is initialized with `PHP_INT_MAX` for all cells except the starting cell `(0, 0)`.
+
+4. **Edge Weights:**
+   - If the current cell's sign matches the intended direction, the cost remains the same.
+   - Otherwise, modifying the direction incurs a cost of `1`.
+
+5. **Termination:** The loop terminates once all cells have been processed. The result is the value in `$dist[$m - 1][$n - 1]`, representing the minimum cost to reach the bottom-right corner.
+
+### Complexity:
+- **Time Complexity:** _**O(m × n)**_, since each cell is processed once.
+- **Space Complexity:** _**O(m × n)**_, for the distance array and deque.
+
+**Contact Links**
+
+If you found this series helpful, please consider giving the **[repository](https://github.com/mah-shamim/leet-code-in-php)** a star on GitHub or sharing the post on your favorite social networks 😍. Your support would mean a lot to me!
+
+If you want more helpful content like this, feel free to follow me:
+
+- **[LinkedIn](https://www.linkedin.com/in/arifulhaque/)**
+- **[GitHub](https://github.com/mah-shamim)**
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diff --git a/algorithms/001368-minimum-cost-to-make-at-least-one-valid-path-in-a-grid/solution.php b/algorithms/001368-minimum-cost-to-make-at-least-one-valid-path-in-a-grid/solution.php
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+<?php
+
+class Solution {
+
+    /**
+     * @param Integer[][] $grid
+     * @return Integer
+     */
+    function minCost($grid) {
+        $m = count($grid);
+        $n = count($grid[0]);
+
+        // Direction vectors for the signs (right, left, down, up)
+        $directions = [
+            [0, 1],  // 1: right
+            [0, -1], // 2: left
+            [1, 0],  // 3: down
+            [-1, 0]  // 4: up
+        ];
+
+        // Deque for 0-1 BFS
+        $deque = new SplQueue();
+        $deque->enqueue([0, 0, 0]); // [row, col, cost]
+
+        // Distance array to keep track of the minimum cost to reach each cell
+        $dist = array_fill(0, $m, array_fill(0, $n, PHP_INT_MAX));
+        $dist[0][0] = 0;
+
+        while (!$deque->isEmpty()) {
+            list($x, $y, $cost) = $deque->dequeue();
+
+            // If we've already found a better way, skip this cell
+            if ($dist[$x][$y] < $cost) {
+                continue;
+            }
+
+            // Traverse all 4 directions
+            foreach ($directions as $dirIndex => $dir) {
+                $nx = $x + $dir[0];
+                $ny = $y + $dir[1];
+                $newCost = $cost;
+
+                // Check if the new cell is within bounds
+                if ($nx >= 0 && $nx < $m && $ny >= 0 && $ny < $n) {
+                    // If the direction of the current cell matches the intended movement
+                    if ($grid[$x][$y] == $dirIndex + 1) {
+                        $newCost = $cost; // No additional cost
+                    } else {
+                        $newCost = $cost + 1; // Modify the direction with cost 1
+                    }
+
+                    // Update distance and add to deque
+                    if ($newCost < $dist[$nx][$ny]) {
+                        $dist[$nx][$ny] = $newCost;
+                        if ($newCost == $cost) {
+                            $deque->unshift([$nx, $ny, $newCost]); // Add to front
+                        } else {
+                            $deque->enqueue([$nx, $ny, $newCost]); // Add to back
+                        }
+                    }
+                }
+            }
+        }
+
+        // Return the minimum cost to reach the bottom-right corner
+        return $dist[$m - 1][$n - 1];
+    }
+}
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