<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:
    vector<pair<int, int>> GetNeighbors(vector<vector<int>> &grid, int sr, int sc) {
        vector<pair<int, int>> neighbors;
        int delta_r[] = {-1, -1, 0, 1, 1, 1, 0, -1};
        int delta_c[] = {0, 1, 1, 1, 0, -1, -1, -1};

        for (int i = 0; i < 8; i++) {
            int nr = sr + delta_r[i];
            int nc = sc + delta_c[i];
            if (nr >= 0 && nr < grid.size() && nc >= 0 && nc < grid[0].size()) {
                neighbors.emplace_back(make_pair(nr, nc));
            }
        }

        return neighbors;
    }

    int MyBFS(vector<vector<int>> &grid) {
        if (grid[0][0] == 1) {
            return -1;
        }
        if (grid.size() == 1 && grid[0].size() == 1) {
            return 1;
        }

        int res = 1;
        deque<pair<int, int>> q;

        grid[0][0] = 1;
        q.push_back({0, 0});

        while (!q.empty()) {
            res++;
            int n = q.size();
            for (int i = 0; i < n; i++) {
                for (auto nb : GetNeighbors(grid, q.front().first, q.front().second)) {
                    if (grid[nb.first][nb.second] != 0) {
                        continue;
                    }
                    if (nb.first == grid.size() - 1 && nb.second == grid[0].size() - 1) {
                        return res;
                    }
                    grid[nb.first][nb.second] = 1;
                    q.push_back({nb.first, nb.second});
                }
                q.pop_front();
            }
        }

        return -1;
    }

    int shortestPathBinaryMatrix(vector<vector<int>> &grid) {
        return MyBFS(grid);
    }
};

// Topics: Graph, Matrix, BFS

// Time Complexity:
67711