<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) {
        
        const int ROWS = grid.size();
        const int COLS = grid[0].size();
        
        if(grid[0][0] == 1 || grid[ROWS-1][COLS-1] == 1) return -1;

        queue<int> nodes; //idx = row * COLS + col
        nodes.push(0);
        
        unordered_set<int> visited;
        visited.insert(0);
        int pathLength = 1;
        vector<pair<int,int>> moves = {{1,0},{0,1},{-1,0},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}};

        while(!nodes.empty()) {
            
            int N = nodes.size();

            for(int i = 0; i < N; ++i) {

                int currIdx = nodes.front();
                nodes.pop();
                int currRow = currIdx / COLS;
                int currCol = currIdx % COLS;
                if(currRow == ROWS - 1 && currCol == COLS - 1) {
                    return pathLength;
                }

                for(int j = 0; j < moves.size(); ++j) {
                    
                    int newRow = currRow + moves[j].first;
                    int newCol = currCol + moves[j].second;
                    int newIdx = newRow * COLS + newCol;

                    if(newRow < 0 || newCol < 0 || newRow >= ROWS || newCol >= COLS ||
                    grid[newRow][newCol] == 1 || visited.count(newIdx)) {
                        continue;
                    }

                    nodes.push(newIdx);
                    visited.insert(newIdx);
                }
            }
            ++pathLength;
        }
        return -1;
    }
};
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