<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]==1) return -1;

        int n = grid.size();
        if(n==1) return 1;
        if(grid[n-1][n-1]==1) return -1;

        unordered_map<int, pair<int, int>> dirmap;
        dirmap[1] = make_pair(-1, 0);        dirmap[5] = make_pair(-1, -1);
        dirmap[2] = make_pair(1, 0);         dirmap[6] = make_pair(-1, 1);
        dirmap[3] = make_pair(0, -1);        dirmap[7] = make_pair(1, -1);
        dirmap[4] = make_pair(0, 1);         dirmap[8] = make_pair(1, 1);

        
        int dir, r,c;
        queue<pair<int,pair<int, int>>> st;
        vector<vector<int>> dist(n, vector<int>(n,-1));

        dist[0][0] = 1;
        st.push({1, {0,0}});
        grid[0][0] = 1;

        while (!st.empty())
        {
            auto it = st.front();
            int dis = it.first;
            auto p = it.second;
            st.pop();

            dir = 0;
            if(p.first==n-1 && p.second==n-1)
                return dis;
            
            while (dir++ < 8){
                r = p.first + dirmap[dir].first;
                c = p.second + dirmap[dir].second;
                if (r >= 0 && r < n && c >= 0 && c< n && grid[r][c]==0 && (dist[r][c]==-1 || 1+dis<dist[r][c]))
                {
                    dist[r][c] = dis + 1;
                    st.push({dist[r][c], {r,c}});
                    grid[r][c] = 1;
                }
            }
                
        }

        return -1;
    }
};
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