<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) {
        if(grid[0][0] == 1) return -1;

        int n = grid.size();
        vector<vector<int>> dist(n, vector<int> (n));

        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                if(grid[i][j] == 0) dist[i][j] = INT_MAX;
                else dist[i][j] = -1;
            }
        }

        dist[0][0] = 1;
        // queue<vector<int>> q;
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        pq.push({0, 0, 1});

        int delX[8] = {0, 1, 1, 1, 0, -1, -1, -1};
        int delY[8] = {1, 1, 0, -1, -1, -1, 0, 1};

        while(!pq.empty()) {
            int x = pq.top()[0];
            int y = pq.top()[1];
            int dis = pq.top()[2];
            pq.pop();

            for(int i = 0; i < 8; i++) {
                int newX = x + delX[i];
                int newY = y + delY[i];

                if(newX >= 0 && newX < n && newY >= 0 && newY < n && grid[newX][newY] != 1) {
                    if(dis + 1 < dist[newX][newY]) {
                        dist[newX][newY] = dis + 1;
                        pq.push({newX, newY, dis + 1});
                    }
                }
            }
        }

        if(dist[n - 1][n - 1] == INT_MAX) return -1;

        return dist[n - 1][n - 1];
    }
};
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