<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>> traverse(int n, pair<int, int> node){
        vector<pair<int, int>> ans;
        
        int dx[8] = {1, -1, 0, 0, 1, -1, -1, 1};
        int dy[8] = {0, 0, -1, 1, 1, -1, 1, -1};


        int i = node.first;
        int j = node.second;

        for(int k=0; k<8; k++){
            int x = i + dx[k];
            int y = j + dy[k];

            if(x >= 0 && y >= 0 && x < n && y < n){
                ans.push_back({x, y});
            }
        }

        return ans;
    }


    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int n = grid.size();

        if(grid[0][0] != 0 || grid[n-1][n-1] != 0) {
            return -1; // Start or end blocked
        }

        queue<pair<pair<int, int>, int>> q;
        vector<vector<int>> vis(n, vector<int>(n, 0));

        q.push({{0, 0}, 1});
        vis[0][0] = 1;

        while(!q.empty()){
            int r = q.front().first.first;
            int c = q.front().first.second;

            int steps = q.front().second;

            q.pop();

            if(r == n-1 && c == n-1){
                return steps;
            }

            if(grid[r][c] == 0){
                for(auto co: traverse(n, {r, c})){
                    if(grid[co.first][co.second] == 0 && vis[co.first][co.second] == 0){
                        q.push({co, steps+1});
                        vis[co.first][co.second] = 1;
                    }
                }
            }


        }

        return -1;

    }
};
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