<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 n = grid.size();
        if(grid[0][0] != 0 || grid[n-1][n-1] != 0) return -1;

        vector<vector<int>> dist(n, vector<int>(n, INT_MAX));
        queue<pair<vector<int>, int>> que;
        que.push({{0, 0}, 1});
        dist[0][0] = 1;

        vector<int> x = {-1, -1, 0, 1, 1, 1, 0, -1};
        vector<int> y = {0, 1, 1, 1, 0, -1, -1, -1};

        while(!que.empty()) {
            int i = que.front().first[0];
            int j = que.front().first[1];
            int cost = que.front().second;
            que.pop();

            for(int k = 0; k<x.size(); k++) {
                if(i+x[k] < 0 || i+x[k] >= n || j+y[k] < 0 || j+y[k] >= n || dist[i+x[k]][j+y[k]] != INT_MAX) {
                    continue;
                }
                if(grid[i+x[k]][j+y[k]] == 1) dist[i+x[k]][j+y[k]] = -1;
                else {
                    que.push({{i+x[k], j+y[k]}, cost+1});
                    dist[i+x[k]][j+y[k]] = cost+1;
                }
            }
        }

        return dist[n-1][n-1] == INT_MAX ? -1 : dist[n-1][n-1];
    }
};
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