<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>

// unordered_set<pair<int, int>> is invalid!
// We can just store the visited information to vector<vector<bool>> pacific(m, vector<bool>(n));
//                                          and vector<vector<bool>> atlantic(m, vector<bool>(n));
// Ref: https://leetcode.com/problems/pacific-atlantic-water-flow/solutions/1743503/c-code-bfs-as-well-as-dfs-approach-simple-solution-runtime-20ms/
class Solution {
public:
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        int ROWS = heights.size(), COLS = heights[0].size();
        // Method 1: need to use size_t pos = string.find(',')
        // int r = std::stoi(string.substr(0, pos));
        // int c = std::stoi(string.substr(pos + 1));
        unordered_set<string> pac, atl;
        // Method 2:
        // unordered_set<pair<int, int>> pac, atl;
        vector<vector<int>> res;
        for (int c = 0; c < COLS; c++) {
            // From the upper Pacific Ocean
            dfs(0, c, pac, 0, heights, ROWS, COLS);
            // From the lower Atlantic Ocean
            dfs(ROWS - 1, c, atl, 0, heights, ROWS, COLS);
        }
        for (int r = 0; r < ROWS; r++) {
            // From the left Pacific Ocean
            dfs(r, 0, pac, 0, heights, ROWS, COLS);
            // From the right Atlantic Ocean
            dfs(r, COLS - 1, atl, 0, heights, ROWS, COLS);
        }
        for (auto pos : pac) {
            // if (atl.count(pos)) res.push_back({pos.first, pos.second});
            if (atl.count(pos)) {
                size_t p = pos.find(',');
                int r = std::stoi(pos.substr(0, p));
                int c = std::stoi(pos.substr(p + 1));
                res.push_back({r, c});
            }
        }
        return res;
    }
private:
    // void dfs(int r, int c, unordered_set<pair<int, int>>& seen, int prevH, const vector<vector<int>>& heights) {
    void dfs(int r, int c, unordered_set<string>& seen, int prevH, const vector<vector<int>>& heights,
             const int& ROWS, const int& COLS) {
        string temp = to_string(r) + "," + to_string(c);
        if (seen.count(temp) || r < 0 || r >= ROWS || c < 0 || c >= COLS || heights[r][c] < prevH) return;
        // seen.insert(make_pair(r, c));
        seen.insert(temp);
        dfs(r + 1, c, seen, heights[r][c], heights, ROWS, COLS);
        dfs(r - 1, c, seen, heights[r][c], heights, ROWS, COLS);
        dfs(r, c + 1, seen, heights[r][c], heights, ROWS, COLS);
        dfs(r, c - 1, seen, heights[r][c], heights, ROWS, COLS);
    }
};
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