<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;

        unordered_map<int, unordered_map<int, int>> visited;
        queue<vector<int>> q;
        q.push({0, 0});
        visited[0][0] = 1;
        int count = 1;

        while (!q.empty()) {
            int qn = q.size();
            while (qn--) {
                auto v = q.front();
                q.pop();

                if (v[0] == grid.size()-1 && v[1] == grid[0].size()-1)
                    return count;
                
                for (int i = -1; i <= 1; i++) {
                    for (int j = -1; j <= 1; j++) {
                        int x = v[0] + i, y = v[1] + j;
                        if (x < 0 || x >= grid.size() || y < 0 || y >= grid[x].size()
                            || grid[x][y] == 1 || visited[x][y])
                            continue;
                        
                        // if (x == grid.size()-1 && y == grid[x].size()-1)
                        //     return count + 1;

                        q.push({x, y});
                        visited[x][y] = 1;
                    }
                }
            }

            count++;
        }

        return -1;
    }
};
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