<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 row = 0;
    int col = 0;
    vector<vector<int>> gridCopy;

    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if(grid[0][0] != 0){
            return -1;
        }
        queue<pair<int, int>> toVisit;
        vector<vector<int>> shortestDistance;
        gridCopy = grid;
        row = grid.size();
        col = grid[0].size();
        cout << row << endl;
        for(int i = 0; i < grid.size(); i++){
            vector<int> temp;
            for(int j = 0; j < grid[0].size(); j++){
                temp.push_back(0);
            }
            shortestDistance.push_back(temp);
        }
        shortestDistance[0][0] = 1;
        gridCopy[0][0] = 1;

        toVisit.push({0, 0});
        vector<pair<int, int>> directions = {
            {-1, -1},
            {-1, 0},
            {-1, 1},
            {0, 1},
            {1, 1},
            {1, 0},
            {1, -1},
            {0, -1},
        };
        cout << "HI";
        while(!toVisit.empty()){
            pair<int, int> current = toVisit.front();
            // cout << "ROW: " << current.first << "COL: " << current.second <<endl;
            toVisit.pop();
            // gridCopy[current.first][current.second] = 1;
            grid[current.first][current.second] = 1;
            for(pair<int, int> coords : directions){
                // cout << coords.first << coords.second << endl;
                if(inRange(current, coords)){
                                    cout << "ADDING ROW: " << current.first + coords.first << "" << "AND COL: " << current.second + coords.second <<endl;
                    toVisit.push({current.first + coords.first, current.second + coords.second});
                    gridCopy[current.first + coords.first][current.second + coords.second] = gridCopy[current.first][current.second] + 1;
                }
            }
            if(current.first == row - 1 && current.second == col - 1){
                return gridCopy[row - 1][col - 1];
            }
        }
        return -1;
    }
    bool inRange(pair<int, int> currLocation, pair<int, int> direction){
        return (currLocation.first + direction.first >= 0 && currLocation.first + direction.first < row &&
        currLocation.second + direction.second >= 0 && currLocation.second + direction.second < col) &&
        gridCopy[currLocation.first + direction.first][currLocation.second + direction.second] == 0;
    }
};
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