<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();
        int m = grid[0].size();

        
        if (grid[0][0] == 1 || grid[n - 1][m - 1] == 1) return -1;

        vector<vector<int>> vis(n, vector<int>(m, 0));
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> q; // Min-heap
        int drow[] = {-1, -1, 0, 0, 1, 1, 1, -1};
        int dcol[] = {-1, 1, -1, 1, -1, 1, 0, 0};
        int ans = INT_MAX;

        
        q.push({1, 0, 0}); 
        vis[0][0] = 1;

        while (!q.empty()) {
            int dist = q.top()[0];
            int row = q.top()[1];
            int col = q.top()[2];
            
            q.pop();

            
            if (row == n - 1 && col == m - 1) return dist;

            
            for (int i = 0; i < 8; i++) {
                int nrow = row + drow[i];
                int ncol = col + dcol[i];

                
                if (nrow >= 0 && nrow < n && ncol >= 0 && ncol < m &&
                    grid[nrow][ncol] == 0 && vis[nrow][ncol] == 0) {
                    vis[nrow][ncol] = 1;
                    q.push({dist + 1, nrow, ncol});
                }
            }
        }

       
        return -1;
    }
};

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