<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:
    int m;
    int n;

    bool flow(int cur_height, int next_height){
        return next_height >= cur_height;
    }

    void bfsCoastPac(queue<vector<int>>& q, vector<vector<int>>& heights, vector<vector<bool>>& t_pac){
        vector<vector<bool>> visit(this->m, vector<bool>(this->n, false));
        for(int i = 0; i < this->m; i++){
            visit[i][0] = true;
            q.push({i, 0});
        }

        for(int i = 0; i < this->n; i++){
            visit[0][i] = true;
            q.push({0, i});
        }
        while(!q.empty()){
            int row = q.front()[0];
            int col = q.front()[1];
            t_pac[row][col] = true;
            q.pop();

            vector<vector<int>> directions = {{1, 0}, {-1, 0}, {0, 1}, {0 , -1}};
            for(auto dir : directions){
                int newRow = row + dir[0]; 
                int newCol = col + dir[1]; 
                // condition check
                if(newRow >= 0 && newRow < this->m && newCol >= 0 && newCol < this->n && visit[newRow][newCol] == false && flow(heights[row][col], heights[newRow][newCol])){
                    visit[newRow][newCol] = true;
                    q.push({newRow, newCol});
                }
            }
        }
    }

    void bfsCoastAtl(queue<vector<int>>& q, vector<vector<int>>& heights, vector<vector<bool>>& t_atl){
        vector<vector<bool>> visit(this->m, vector<bool>(this->n, false));
        for(int i = 0; i < this->m; i++){
            visit[i][this->n - 1] = true;
            q.push({i, this->n - 1});
        }

        for(int i = 0; i < this->n; i++){
            visit[this->m - 1][i] = true;
            q.push({this->m - 1, i});
        }

        while(!q.empty()){
            int row = q.front()[0];
            int col = q.front()[1];
            t_atl[row][col] = true;
            q.pop();

            vector<vector<int>> directions = {{1, 0}, {-1, 0}, {0, 1}, {0 , -1}};
            for(auto dir : directions){
                int newRow = row + dir[0]; 
                int newCol = col + dir[1]; 
                // condition check
                if(newRow >= 0 && newRow < this->m && newCol >= 0 && newCol < this->n && visit[newRow][newCol] == false && flow(heights[row][col], heights[newRow][newCol])){
                    visit[newRow][newCol] = true;
                    q.push({newRow, newCol});
                }
            }
        }
    }

    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        this->m = heights.size();
        this->n = heights[0].size();
        vector<vector<int>> result;
        vector<vector<bool>> t_pac(m, vector<bool>(n, false));
        vector<vector<bool>> t_atl(m, vector<bool>(n, false));
        queue<vector<int>> q;
        // pac
        bfsCoastPac(q, heights, t_pac);

        //atl
        bfsCoastAtl(q, heights, t_atl);

        for(int i = 0; i < this->m; i ++){
            for(int j = 0; j < this->n; j++){
                if(t_pac[i][j] && t_atl[i][j]) result.push_back({i, j});
            }
        }

        return result;
    }
};
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