<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 dp[205][205]; //to check visited or not
    void dfs(int i, int j,int prevVal, vector<vector<int>>& heights, vector<vector<int>>& path){
        vector<int>v;
        if(i<0 || j<0 || i>=heights.size() || j>=heights[0].size()) return;
        if(heights[i][j] < prevVal) return ;
        if(dp[i][j] != 0) return;
        path[i][j]++;
        dp[i][j]=1;

        dfs(i, j+1, heights[i][j], heights, path);
        dfs(i, j-1, heights[i][j], heights, path);
        dfs(i+1, j, heights[i][j], heights, path);
        dfs(i-1, j, heights[i][j], heights, path);


    }

    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        vector<vector<int>> path(205, vector<int>(205, 0)); //permanantly storing the visiting cell
        vector<vector<int>> ans;
        memset(dp, 0, sizeof(dp));
        for(int i=0;i<heights.size(); i++){
            for(int j=0; j<heights[i].size(); j++){
                if(i==0 || j==0){
                    dfs(i, j, -1, heights, path);
                }         
            }
        }
        memset(dp, 0, sizeof(dp)); //initializing the dp array with 0 again
        for(int i=0;i<heights.size(); i++){
            for(int j=0; j<heights[i].size(); j++){
                if(i==heights.size()-1 || j==heights[0].size()-1){
                    dfs(i, j, -1, heights, path);
                }

            }
        }
        //storing the indices to a 2D vector
        for(int i=0;i<heights.size();i++){
            for(int j=0;j<heights[0].size();j++){
                vector<int>v;
                if(path[i][j]==2){
                    v.push_back(i);
                    v.push_back(j);
                    ans.push_back(v); //if we put this line out of the if condition, there'll be stored some unnecessary null vector?
                }
                v.clear();

            }

        }

        return ans;

    }
};
48000