<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:
    void dfs(int i, int j,vector<vector<int>>& heights,unordered_set<int> &vis, int prevHeight)
        {
            int cellNumber = (i*heights[0].size()) + j;
            if(i < 0 || j < 0 || i >= heights.size() || j >= heights[0].size() || 
            heights[i][j] < prevHeight || vis.count(cellNumber))
            {
                return;
            }

            vis.insert(cellNumber);
            int dirx[4] = {-1,0,1,0};
            int diry[4] = {0,1,0,-1};

            for(int d = 0 ; d < 4; d++)
            {
                dfs(i+dirx[d],j+diry[d],heights,vis,heights[i][j]);  
            }

            return;
            
        }
        vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
            unordered_set<int> altaltic;
            unordered_set<int> pacific;

            int m = heights.size(), n = heights[0].size();

            for(int c = 0; c < n; c++ )
            {
                dfs(0,c,heights,pacific,heights[0][c]);    
                dfs(m-1,c,heights,altaltic,heights[m-1][c]);    
            }

            for(int r = 0; r < m; r++ )
            {
                dfs(r,0,heights,pacific,heights[r][0]);    
                dfs(r,n-1,heights,altaltic,heights[r][n-1]);    
            }

            vector<vector<int>> res;
            for(int i=0; i < m; i++)
            {
                for(int j = 0; j < n; j++)
                {
                    int cellNumber = (i*n)+j;
                    if(altaltic.count(cellNumber) && pacific.count(cellNumber))
                    {
                        res.push_back({i,j});
                    }
                }
            }

            return res;
        }
    };
24000