<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 getPoss(queue<vector<int>> q, vector<vector<int>> &ocean, vector<vector<int>> &vis, vector<vector<int>> &a, int n, int m) {
        vector<vector<int>> pos = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        while(!q.empty()) {
            vector<int> co = q.front();
            q.pop();
            for (int i=0;i<4;i++) {
                int x = co[0] + pos[i][0];
                int y = co[1] + pos[i][1];
                if (x < 0 || x >= n || y < 0 || y >= m || vis[x][y]) continue;
                if (a[x][y] < a[co[0]][co[1]]) continue;
                vis[x][y] = 1;
                q.push({x, y});
                ocean.push_back({x, y});
            }
        }
    }

    vector<vector<int>> pacificAtlantic(vector<vector<int>>& a) {
        int n = a.size();
        int m = a[0].size();
        vector<vector<int>> atlantic, pacific;
        vector<vector<int>> vis(n, vector<int>(m, 0));
        
        queue<vector<int>> q;
        for (int i=0;i<n;i++) {
            q.push({i, m-1});
            vis[i][m-1] = 1;
            atlantic.push_back({i, m-1});
        }
        for (int j=0;j<m;j++) {
            if (vis[n-1][j]) continue;
            q.push({n-1, j});
            vis[n-1][j] = 1;
            atlantic.push_back({n-1, j});
        }
        getPoss(q, atlantic, vis, a, n, m);
        q = queue<vector<int>>();
        vis = vector(n, vector<int>(m, 0));
        for (int i=0;i<n;i++) {
            q.push({i, 0});
            vis[i][0] = 1;
            pacific.push_back({i, 0});
        }
        for (int j=0;j<m;j++) {
            if (vis[0][j]) continue;
            q.push({0, j});
            vis[0][j] = 1;
            pacific.push_back({0, j});
        }
        getPoss(q, pacific, vis, a, n, m);
        sort(atlantic.begin(), atlantic.end());
        sort(pacific.begin(), pacific.end());
        int i = 0, j = 0;
        vector<vector<int>> res;
        while(i < atlantic.size() && j < pacific.size()) {
            if (atlantic[i] == pacific[j]) { 
                res.push_back(atlantic[i]);
                i++;
                j++;
            }
            else if (atlantic[i] < pacific[j]) i++;
            else j++;
        }
        return res;
    }
};
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