<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 res = INT_MAX;
        int n = grid.size();
        int reachEnd = false;
        set<pair<int, int>> visited;
        queue<pair<pair<int, int>, int>> q; // {{x, y}, dist};
        vector<pair<int, int>> directions = 
            {{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {-1, -1}, {1, 1}, {-1, 1}, {1, -1}};

        // First or last node is not 0
        if (grid[0][0] != 0 || grid[n-1][n-1] != 0) return -1;

        q.push({{0, 0}, 1});

        while (!q.empty()) {
            int dist = q.front().second;
            int x = q.front().first.first;
            int y = q.front().first.second;
            q.pop();

            // Reach end 
            if (x==n-1 && y==n-1) {
                reachEnd = true;
                res = min(res, dist);
            }

            if (visited.find({x, y}) != visited.end()) {
                continue;
            }
            visited.insert({x, y});

            for (auto [dr, dc] : directions) {
                int r = dr + x;
                int c = dc + y;
                if (r>=0 && r<n && c>=0 && c<n) {
                    // Add to queue
                    if (grid[r][c] == 0) {
                        q.push({{r, c}, dist+1});
                    }
                }
            }
        }

        return reachEnd ? res : -1;
    }
};
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