<p>There is an <code>m x n</code> rectangular island that borders both the <strong>Pacific Ocean</strong> and <strong>Atlantic Ocean</strong>. The <strong>Pacific Ocean</strong> touches the island&#39;s left and top edges, and the <strong>Atlantic Ocean</strong> touches the island&#39;s right and bottom edges.</p>

<p>The island is partitioned into a grid of square cells. You are given an <code>m x n</code> integer matrix <code>heights</code> where <code>heights[r][c]</code> represents the <strong>height above sea level</strong> of the cell at coordinate <code>(r, c)</code>.</p>

<p>The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell&#39;s height is <strong>less than or equal to</strong> the current cell&#39;s height. Water can flow from any cell adjacent to an ocean into the ocean.</p>

<p>Return <em>a <strong>2D list</strong> of grid coordinates </em><code>result</code><em> where </em><code>result[i] = [r<sub>i</sub>, c<sub>i</sub>]</code><em> denotes that rain water can flow from cell </em><code>(r<sub>i</sub>, c<sub>i</sub>)</code><em> to <strong>both</strong> the Pacific and Atlantic oceans</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/08/waterflow-grid.jpg" style="width: 400px; height: 400px;" />
<pre>
<strong>Input:</strong> heights = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]
<strong>Output:</strong> [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]
<strong>Explanation:</strong> The following cells can flow to the Pacific and Atlantic oceans, as shown below:
[0,4]: [0,4] -&gt; Pacific Ocean 
&nbsp;      [0,4] -&gt; Atlantic Ocean
[1,3]: [1,3] -&gt; [0,3] -&gt; Pacific Ocean 
&nbsp;      [1,3] -&gt; [1,4] -&gt; Atlantic Ocean
[1,4]: [1,4] -&gt; [1,3] -&gt; [0,3] -&gt; Pacific Ocean 
&nbsp;      [1,4] -&gt; Atlantic Ocean
[2,2]: [2,2] -&gt; [1,2] -&gt; [0,2] -&gt; Pacific Ocean 
&nbsp;      [2,2] -&gt; [2,3] -&gt; [2,4] -&gt; Atlantic Ocean
[3,0]: [3,0] -&gt; Pacific Ocean 
&nbsp;      [3,0] -&gt; [4,0] -&gt; Atlantic Ocean
[3,1]: [3,1] -&gt; [3,0] -&gt; Pacific Ocean 
&nbsp;      [3,1] -&gt; [4,1] -&gt; Atlantic Ocean
[4,0]: [4,0] -&gt; Pacific Ocean 
       [4,0] -&gt; Atlantic Ocean
Note that there are other possible paths for these cells to flow to the Pacific and Atlantic oceans.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> heights = [[1]]
<strong>Output:</strong> [[0,0]]
<strong>Explanation:</strong> The water can flow from the only cell to the Pacific and Atlantic oceans.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == heights.length</code></li>
	<li><code>n == heights[r].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= heights[r][c] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int m;
    int n;
    vector<vector<int>> heights;
    vector<vector<int>> directions = {{0,1},{0,-1},{1,0},{-1,0}};
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        if(heights.size()==0 || heights[0].size()==0) return {};

        m = heights.size();
        n = heights[0].size();
        this->heights=heights;

        queue<vector<int>> pacific;
        queue<vector<int>> atlantic;

        for(int row = 0;row<m;row++){
            for(int col = 0;col<n;col++){
                if(row==0 || col==0) pacific.push({row,col});
                if(row==m-1 || col==n-1) atlantic.push({row,col});
            }
        }

        vector<vector<int>> pacific_reach = bfs(pacific);
        vector<vector<int>> atlantic_reach = bfs(atlantic);

        vector<vector<int>> intersection;
        for(int row = 0;row<m;row++){
            for(int col = 0;col<n;col++){
                if(pacific_reach[row][col]==1 && atlantic_reach[row][col]==1){
                    intersection.push_back({row,col});
                    
                }

            }
        }
        return intersection;
    }
    vector<vector<int>> bfs(queue<vector<int>>& q){
        vector<vector<int>> reachable = vector(m,vector(n,0));
        while(!q.empty()){
            int size = q.size();
            for(int i = 0;i<size;i++){
                vector<int> curr = q.front();
                q.pop();
                int r = curr[0],c = curr[1];
                reachable[r][c] = 1;
                cout<<"hihi\n";
                for(vector<int>& direction: directions){
                    int nextR = r+direction[0],nextC = c+direction[1];

                    //check if the next coordinate is valid
                    if(nextR<0 || nextR >= m||nextC<0 || nextC>=n){
                        continue;
                    }

                    if(heights[nextR][nextC]<heights[r][c]) continue;

                    if(reachable[nextR][nextC]==1) continue;
                    
                    q.push({nextR,nextC});
                }
            }
        }
        return reachable;
    }
};
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