<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) {
        queue<vector<int>> coords;
        if (grid[0][0] == 0) {
            coords.push(vector<int>{0, 0, 1});
        }

        vector<pair<int, int>> candidates;
        set<int> visited = {0};
        int x, y, p, q, iter;
        while (!coords.empty()) {
            x = coords.front()[0];
            y = coords.front()[1];
            iter = coords.front()[2];
            coords.pop();
            if (x == grid.size() - 1 && y == grid[0].size() - 1) {
                return iter;
            }
            candidates = {
                make_pair(x - 1, y - 1), make_pair(x - 1, y), make_pair(x - 1, y + 1),
                make_pair(x, y - 1), make_pair(x, y + 1),
                make_pair(x + 1, y - 1), make_pair(x + 1, y), make_pair(x + 1, y + 1)
            };
            for (auto &c : candidates) {
                p = c.first;
                q = c.second;
                if (0 <= p && p <= grid.size() - 1 && 0 <= q && q <= grid[0].size() - 1 && grid[p][q] == 0 &&
                    !visited.contains(p + 10*(grid.size() + 1) * q)) {
                    visited.insert(p + 10*(grid.size() + 1) *  q);
                    coords.push(vector<int>{p, q, iter + 1});
                }
                //000
                //110
                //111
            }
        }
        return -1;
    }
};
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