<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]!=0 || grid[grid.size()-1][grid[0].size()-1]!=0) return -1;
        
        queue<pair<int,int>> q;
        vector<vector<bool>> vis(grid.size(), vector<bool>(grid.size(), false));
        q.push(make_pair(0, 0));
        vis[0][0] = true;
        
        int steps = 1;  // Start at step 1 from (0, 0)
        pair<int, int> node;
        vector<pair<int,int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
        
        while(!q.empty()) {
            int qsize = q.size();  // Get the number of nodes at the current level (step)
            for(int k = 0; k < qsize; k++) {  // Process all nodes at the current level
                node = q.front();
                q.pop();
                int i = node.first, j = node.second;
                
                // If we reached the bottom-right corner
                if(i == grid.size()-1 && j == grid[0].size()-1) {
                    return steps;
                }
                
                // Explore all 8 directions
                for(auto& d : directions) {
                    int ni = i + d.first, nj = j + d.second;
                    if(ni >= 0 && nj >= 0 && ni < grid.size() && nj < grid[0].size() && grid[ni][nj] == 0 && !vis[ni][nj]) {
                        q.push(make_pair(ni, nj));
                        vis[ni][nj] = true;
                    }
                }
            }
            steps++;  // Increment step after processing all nodes at the current level
        }
        
        return -1;  // Return -1 if there's no valid path
    }
};
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