<p>Given an <code>n x n</code> binary matrix <code>grid</code>, return <em>the length of the shortest <strong>clear path</strong> in the matrix</em>. If there is no clear path, return <code>-1</code>.</p>

<p>A <strong>clear path</strong> in a binary matrix is a path from the <strong>top-left</strong> cell (i.e., <code>(0, 0)</code>) to the <strong>bottom-right</strong> cell (i.e., <code>(n - 1, n - 1)</code>) such that:</p>

<ul>
	<li>All the visited cells of the path are <code>0</code>.</li>
	<li>All the adjacent cells of the path are <strong>8-directionally</strong> connected (i.e., they are different and they share an edge or a corner).</li>
</ul>

<p>The <strong>length of a clear path</strong> is the number of visited cells of this path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example1_1.png" style="width: 500px; height: 234px;" />
<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example2_1.png" style="height: 216px; width: 500px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> -1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j] is 0 or 1</code></li>
</ul>

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int n = grid.size(); // sizee of grid
        if (grid[0][0] != 0 || grid[n-1][n-1] != 0) { // cheak 0,0 to  n-1,n-1 all are 0 if 1                                then      their was no valid path
            return -1; // cheaked all elemnt are not 1 if 1 no valid path return -1
        } 

        // Directions for 8 possible moves
        vector<vector<int>> directions = {
            {-1, -1}, {-1, 0}, {-1, 1},
            { 0, -1},          { 0, 1},
            { 1, -1}, { 1, 0}, { 1, 1}
        };
         // Min-heap to store cells with their current path length
        priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<tuple<int, int, int>>> pq;
        pq.push({1, 0, 0});  // {distance, r, c}
        grid[0][0] = 1;  // Mark starting cell as visited with distance 1

        while (!pq.empty()) {
            auto [dist, x, y] = pq.top();
            pq.pop(); 
             // If we reach the bottom-right corner, return the distance
            if (x == n-1 && y == n-1) {
                return dist;
            }

            // Explore all 8 possible directions
            for (auto dir : directions) {
                int nx = x + dir[0];
                int ny = y + dir[1];

                // Check if the new position is valid and unvisited
                if (nx >= 0 && ny >= 0 && nx < n && ny < n && grid[nx][ny] == 0) {
                    pq.push({dist + 1, nx, ny});
                    grid[nx][ny] = dist + 1;  // Mark as visited with the updated distance
                }
            }
        }

        return -1;  // No path found
    }
};
80865