<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) {
        if (heights.size() == 0 || heights[0].size() == 0){
            return {};
        }
        m_rows = heights.size();
        m_cols = heights[0].size();
        queue<vector<int>> pacific;
        queue<vector<int>> atlantic;
        for(int row = 0; row < m_rows; ++row){
            pacific.push({row, 0});
            atlantic.push({row, m_cols -1});
        }
        for (int col = 0; col < m_cols; ++col){
            pacific.push({0, col});
            atlantic.push({m_rows-1, col});
        }
        auto pacificReachable = bfs(pacific, heights);
        auto atlanticReachable = bfs(atlantic, heights);
        vector<vector<int>> sol;
        sol.reserve(m_rows*m_cols);
        for (int row = 0; row < m_rows; ++row){
            for (int col = 0; col < m_cols; ++col){
                if (pacificReachable[row][col] && atlanticReachable[row][col]){
                    sol.push_back({row, col});
                }
            }
        }
        return sol;
    }
private:
    vector<vector<bool>> bfs(queue<vector<int>> &q, const vector<vector<int>>& heights){
        vector<vector<bool>> visited(m_rows, vector<bool>(m_cols, false));
        vector<vector<int>> dirs = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
        while(!q.empty()){
            const auto cell = q.front();
            q.pop();
            visited[cell[0]][cell[1]] = true;
            for (auto dir : dirs){
                int newRow = cell[0] + dir[0];
                int newCol = cell[1] + dir[1];
                if (newRow < 0 || newRow >= m_rows || newCol < 0 || newCol >= m_cols){
                    continue;
                }
                if (visited[newRow][newCol]){
                    continue;
                }
                if (heights[cell[0]][cell[1]] > heights[newRow][newCol]){
                    continue;
                }
                q.push({newRow, newCol});
                visited[newRow][newCol] = true;
            }
        }
        return visited;
    }

    vector<vector<int>> getNeighbors(int row, int col, const vector<vector<int>>& heights){
        vector<vector<int>> neighbors;
        if (row > 0){
            if (heights[row-1][col] >= heights[row][col]){
                neighbors.push_back({row-1, col});
            }
        }
        if (row < heights.size() - 1){
            if (heights[row+1][col] >= heights[row][col]){
                neighbors.push_back({row+1, col});
            }
        }
        if (col > 0){
            if (heights[row][col-1] >= heights[row][col]){
                neighbors.push_back({row, col-1});
            }
        }
        if (col < heights[0].size() - 1){
            if (heights[row][col+1] >= heights[row][col]){
                neighbors.push_back({row, col+1});
            }
        }
        return neighbors;
    }
private:
    int m_rows;
    int m_cols;
};
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