<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>

vector<int> dx = {1,-1,0,0};
vector<int> dy = {0,0,1,-1};

void bfs(vector<vector<int>> & heights, int x, int y, vector<vector<int>> &vis, set<pair<int,int>> &ans)
{
    int m = heights.size();
    int n = heights[0].size();
    queue<pair<int,int>> q;
    vis[x][y]=1;
    q.push({x,y});

    while(!q.empty())
    {
        auto it = q.front();
        q.pop();
        int curr_x = it.first;
        int curr_y = it.second;
        pair<int,int> p = {curr_x, curr_y};
        ans.insert(p);


        for(int i = 0 ; i < 4; i++)
        {
            int nx = curr_x + dx[i];
            int ny = curr_y + dy[i];

            if(nx>=0 && nx<m && ny>=0 && ny<n && !vis[nx][ny] && heights[nx][ny]>=heights[curr_x][curr_y])
            {
                vis[nx][ny] = 1;
                q.push({nx,ny});
                //validNeigh++;
            }

        }

    }

}

class Solution {
public:
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        int m = heights.size();
        int n = heights[0].size();

        vector<vector<int>> vis(m, vector<int>(n, 0));
        set<pair<int,int>> ansAtlantic;
        set<pair<int,int>> other;

        //top, left
        for(int i = 0 ; i < n; i++)
        {

            bfs(heights, 0, i, vis, ansAtlantic);
        }
        for(int  j = 0 ; j < m ; j++)
        {
            bfs(heights, j, 0, vis, ansAtlantic);
        }

        vis = vector<vector<int>>(m, vector<int>(n, 0));


        //bottom right
        for(int  j = 0 ; j < m ; j++)
        {
            bfs(heights, j, n-1, vis, other);
        }
        for(int i = 0 ; i < n; i++)
        {
            bfs(heights, m-1, i, vis, other);
        }

        vector<vector<int>> res;
        for(auto it: ansAtlantic)
        {
            if(other.find(it)!= other.end())
            {
                int cx = it.first;
                int cy = it.second;
                res.push_back({it.first, it.second});
                }
        }

        return res;

    }
};
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