<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:

    bool boundary_conditions(int r, int c, int m, int n) {
        if(c < 0 || c >= n || r < 0 || r >= m)
            return false;
        return true;
    }

    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        if(grid[0][0] != 0 || grid[n-1][m-1] != 0) return -1;
        vector<vector<bool>> visited(m, vector<bool>(n,0));
        vector<vector<int>> distance(m, vector<int>(n,0));
        queue <pair<int,int>> Q;
        Q.push(make_pair(0,0));
        visited[0][0] = 1;
        distance[0][0] = 1;
        vector<pair<int,int>> directions = {{1,0},{0,1},{-1,0},{0,-1},{-1,-1},{1,1},{-1,1},{1,-1}};
        while(!Q.empty()) {
            int row = Q.front().first;
            int col = Q.front().second;
            Q.pop();
            if(row == m - 1 && col == n -1) {
                return distance[row][col];
            }
            for(int k = 0; k < directions.size(); k++) {
                int new_row = directions[k].first + row;
                int new_col = directions[k].second + col;
                if(boundary_conditions(new_row, new_col, m, n) && !visited[new_row][new_col] && grid[new_row][new_col] == 0) {
                    distance[new_row][new_col] = distance[row][col] + 1;
                    visited[new_row][new_col] = 1;
                    Q.push(make_pair(new_row, new_col));
                }
            }
        }
        return -1;
    }
};

//Shortest path can be found by BFS
//if 0, start looking in all 8 directions. 
//each element in the matrix should have a distance and visited variable associated with it
//if not visited and it is 0 and in bounds, add it to the Q, increment distance = curr_distance + 1 and visited is set to true, add it to the Q and move on to the next.
//if you reach the end of the matrix, return distance
//if you don't, keep going over the other elements in the matrix.
//if we reach the end without finding a path, return -1;

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