<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 is_valid(int num, int n){
        return num >=0 && num < n;
    }
    void update_surround(vector<vector<int>>& grid, const pair<int,int> &point, queue<pair<int,int>> &cell_queue){
        vector<pair<int,int>> surround = {{-1,-1}, {0, -1}, {1, -1}, {-1, 0}, {1, 0}, {-1, 1}, {0, 1}, {1, 1}};
        int n = grid.size();
        int x = point.first;
        int y = point.second; 
        for(int i = 0; i < surround.size(); ++i){
            int new_x = point.first + surround[i].first;
            int new_y = point.second + surround[i].second;
            if(is_valid(new_x, n) && is_valid(new_y, n)){
                if(grid[new_x][new_y] == 0){
                    grid[new_x][new_y] = grid[x][y]+1;
                    cell_queue.push({new_x, new_y});
                }
            }
        }
    }
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if(grid[0][0] == 1)return -1;
        // convert all 1 into INT_MAX in every cells.
        for(vector<int>& row : grid){
            for(int& cell : row){
                if(cell == 1){
                    cell = INT_MAX;
                }
            }
        }

        // start from 0,0
        grid[0][0] = 1;
        int n = grid.size();
        queue<pair<int,int>> cell_queue; // the cells waiting for being checked.
        cell_queue.push({0,0});
        while(!cell_queue.empty()){
            pair<int,int> point = cell_queue.front();
            cell_queue.pop();
            if(point.first == n-1 && point.second == n-1){
                return grid[n-1][n-1];
            }

            update_surround(grid, point, cell_queue);
        }
        return -1;
    }
};
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