<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:
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        // Nodes flow to pacific ocean
        vector<vector<bool>> pacific = reachOcean(heights);

        // Flip the input heights to get nodes flow to atlantic ocean
        reverse(heights.begin(), heights.end());
        int M = heights.size(), N = heights[0].size();
        for (int i=0; i<M; i++) {
            reverse(heights[i].begin(), heights[i].end());
        }
        // // cout
        // for (int i=0; i<M; i++) {
        //     for (int j=0; j<N; j++) {
        //         cout << heights[i][j];
        //     }
        //     cout << endl;
        // }
        vector<vector<bool>> atlantic = reachOcean(heights);
        // Flip the output atlantic to get nodes flow to atlantic ocean
        reverse(atlantic.begin(), atlantic.end());
        for (int i=0; i<M; i++) {
            reverse(atlantic[i].begin(), atlantic[i].end());
        }
        
        vector<vector<int>> res;
        for (int i=0; i<M; i++) {
            for (int j=0; j<N; j++) {
                // cout << "(" << pacific[i][j] << " " << atlantic[i][j] << ")";
                if (pacific[i][j] && atlantic[i][j])
                    res.push_back({i, j});
            }
            // cout << endl;
        }
        return res;
    }

    vector<vector<bool>> reachOcean(vector<vector<int>>& heights) {
        if (heights.empty()) return {{}};

        // Initialization
        // int M, N;
        int M = heights.size(), N = heights[0].size();
        vector<vector<int>> dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        vector<vector<bool>> visited;
        for (int i=0; i<M; i++) {
            vector<bool> v_line(N, false);
            // vector<bool> g_line(N, false);
            visited.push_back(v_line);
            // good.push_back(g_line);
        }


        for (int i=0; i<M; i++) {
            for (int j=0; j<N; j++) {
                if (!visited[i][j] && (i == 0 || j == 0)) {
                    // DFS
                    stack<vector<int>> dfs;
                    vector<int> node = {i, j};
                    dfs.push(node);
                    visited[i][j] = true;
                    while(!dfs.empty()) {
                        node = dfs.top();
                        dfs.pop();
                        // // Mark reachable
                        // good[node[0]][node[1]] = true;
                        // Get neighbors (node with larger height)
                        for (int i_d=0; i_d<dirs.size(); i_d++) {
                            vector<int> nb = {node[0]+dirs[i_d][0], node[1]+dirs[i_d][1]};
                            if (nb[0] >= 0 && nb[0] < M && nb[1] >= 0 && nb[1] < N &&
                                heights[node[0]][node[1]] <= heights[nb[0]][nb[1]] &&
                                !visited[nb[0]][nb[1]]) {
                                dfs.push(nb);
                                visited[nb[0]][nb[1]] = true;
                            }
                        }
                    }
                }
            }
        }

        return visited;
    }

};
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