<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 {
    int M;
    int N;
public:
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        M = heights.size();
        if(M == 0) return {};
        N = heights[0].size();
        vector<vector<int>> pac = vector<vector<int>>(M, vector<int>(N, 0));
        vector<vector<int>> atl = vector<vector<int>>(M, vector<int>(N, 0));
        vector<vector<int>> starts;
        for(int i = 0; i < M; i++)
            starts.push_back({i, -1});
        for(int j = 0; j < N; j++)
            starts.push_back({-1, j});
        bfs(heights, starts, pac);
        starts.clear();
        for(int i = 0; i < M; i++)
            starts.push_back({i, N});
        for(int j = 0; j < N; j++)
            starts.push_back({M, j});
        bfs(heights, starts, atl);
        vector<vector<int>> solution;
        for(int i = 0; i < M; i++)
            for(int j = 0; j < N; j++)
                if(atl[i][j] == 1 && pac[i][j] == 1)
                    solution.push_back({i, j});
        return solution;
    }
    void bfs(vector<vector<int>>& heights, vector<vector<int>>& starts, vector<vector<int>>& visited){
        vector<vector<int>> dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        queue<vector<int>> q;
        for(vector<int> x: starts) q.push(x);
        while(!q.empty()){
            vector<int> temp = q.front();
            q.pop();
            for(int k = 0; k < 4; k++){
                int m = temp[0] + dir[k][0];
                int n = temp[1] + dir[k][1];
                if(m < 0 || m >= M || n < 0 || n >= N) continue;
                if(visited[m][n] == 1) continue;
                if(temp[0] >= 0 && temp[0] < M && temp[1] >= 0 && temp[1] < N && heights[m][n] < heights[temp[0]][temp[1]]) continue;
                visited[m][n] = 1;
                q.push({m, n});
            }
        }
    }
};
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