<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();
        int m = grid[0].size();

        if (grid[0][0] == 1) return -1;

        vector<vector<int>> dist(n, vector<int> (m, INT_MAX));
        vector<vector<int>> vis(n, vector<int> (m, 0));

        set<pair<int, pair<int, int>>> st;

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

        while (!st.empty()) {
            auto val = *(st.begin());
            int dis = val.first;
            int i = val.second.first;
            int j = val.second.second;
            vis[i][j] = 1;

            cout << dis << " " << i << " "<< j << endl;

            st.erase(val);

            int dir[8][2] = {
                {0, 1}, {0, -1},
                {1, 1}, {-1, -1},
                {1, -1}, {-1, 1},
                {1, 0}, {-1, 0}
            };

            for (int a = 0; a < 8; a++) {
                int newI = i+dir[a][0];
                int newJ = j+dir[a][1];

                if (newI >= 0 && newI < n && newJ >= 0 && newJ < m && grid[newI][newJ] == 0 && !vis[newI][newJ]) {
                    int newDist = dis+1;

                    if (newDist < dist[newI][newJ]) {
                        if (dist[newI][newJ] != INT_MAX) st.erase({dist[newI][newJ], {newI, newJ}});
                        dist[newI][newJ] = newDist;
                        st.insert({newDist, {newI, newJ}});
                    }
                }
            }

        }

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

    }
};
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