<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 bfs(vector<vector<int>>& heights, vector<vector<bool>>& visited, queue<pair<int, int>>& q) {
        vector<vector<int>> dirs = {{1,0}, {-1,0}, {0,1}, {0,-1}};
        while(!q.empty()) {
            pair<int, int> coords = q.front();
            q.pop();
            int i = coords.first;
            int j = coords.second;

            for(vector<int> dir:dirs) {
                int di = i + dir[0];
                int dj = j + dir[1];
                if(0 <= di && di < heights.size() && 0 <= dj && dj < heights[0].size() && !visited[di][dj] && heights[di][dj] >= heights[i][j]) {
                    visited[di][dj] = true;
                    q.push({di, dj});
                }
            }

        }
    }
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        int m = heights.size();
        int n = heights[0].size();

        vector<vector<bool>> pacific(m, vector<bool>(n));
        vector<vector<bool>> atlantic(m, vector<bool>(n));

        queue<pair<int, int>> pq;
        queue<pair<int, int>> aq;

        for(int i = 0; i < m; ++i) {
            pq.push({i, 0});
            pacific[i][0] = true;
            aq.push({i, n-1});
            atlantic[i][n-1] = true;
        }
        for(int j = 0; j < n; ++j) {
            pq.push({0, j});
            pacific[0][j] = true;
            aq.push({m-1, j});
            atlantic[m-1][j] = true;
        }
        bfs(heights, pacific, pq);
        bfs(heights, atlantic, aq);

        vector<vector<int>> valid_coords;
        for(int i = 0; i < m; ++i) {
            for(int j = 0; j < n; ++j) {
                if(pacific[i][j] && atlantic[i][j]) {
                    valid_coords.push_back({i,j});
                }
            }
        }
        return valid_coords;
    }
};
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