Pacific Atlantic Water Flow
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0417.Pacific%20Atlantic%20Water%20Flow/images/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>
Depth-First Search; Breadth-First Search; Array; Matrix
typedef pair<int, int> pii;

class Solution {
public:
    vector<vector<int>> heights;
    int m;
    int n;

    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        m = heights.size();
        n = heights[0].size();
        this->heights = heights;
        queue<pii> q1;
        queue<pii> q2;
        unordered_set<int> vis1;
        unordered_set<int> vis2;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || j == 0) {
                    vis1.insert(i * n + j);
                    q1.emplace(i, j);
                }
                if (i == m - 1 || j == n - 1) {
                    vis2.insert(i * n + j);
                    q2.emplace(i, j);
                }
            }
        }
        bfs(q1, vis1);
        bfs(q2, vis2);
        vector<vector<int>> ans;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int x = i * n + j;
                if (vis1.count(x) && vis2.count(x)) {
                    ans.push_back({i, j});
                }
            }
        }
        return ans;
    }

    void bfs(queue<pii>& q, unordered_set<int>& vis) {
        vector<int> dirs = {-1, 0, 1, 0, -1};
        while (!q.empty()) {
            for (int k = q.size(); k > 0; --k) {
                auto p = q.front();
                q.pop();
                for (int i = 0; i < 4; ++i) {
                    int x = p.first + dirs[i];
                    int y = p.second + dirs[i + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n && !vis.count(x * n + y) && heights[x][y] >= heights[p.first][p.second]) {
                        vis.insert(x * n + y);
                        q.emplace(x, y);
                    }
                }
            }
        }
    }
};