<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:
void do_it(int t, vector<vector<int>> &res, queue<pair<int, int>> &q, vector<vector<int>> &visit, int n, int m, vector<vector<int>>& heights){
    while(!q.empty()){
        int i = q.front().first, j = q.front().second;
        q.pop();
        if(i-1 >= 0 && !visit[i-1][j] && heights[i-1][j] >= heights[i][j]){
            visit[i-1][j] = 1;
            res[i-1][j] |= t;
            q.push({i-1, j});
        }
        if(i+1 < m && !visit[i+1][j] && heights[i+1][j] >= heights[i][j]){
            visit[i+1][j] = 1;
            res[i+1][j]|= t;
            q.push({i+1, j});
        }
        if(j-1 >= 0 && !visit[i][j-1] && heights[i][j-1] >= heights[i][j]){
            visit[i][j-1] = 1;
            res[i][j-1]|= t;
            q.push({i, j-1});
        }
        if(j+1 < n && !visit[i][j+1] && heights[i][j+1] >= heights[i][j]){
            visit[i][j+1] = 1;
            res[i][j+1]|= t;
            q.push({i, j+1});
        }
    }
}
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        int m = heights.size(), n = heights[0].size();
        vector<vector<int>> res(m, vector<int>(n, 0)), v1(m, vector<int>(n, 0)), v2(m, vector<int>(n, 0));
        queue<pair<int, int>> q1, q2;
        for(int i = 0; i < m; i++){
            res[i][0] |= 1;
            v1[i][0] = 1;
            q1.push({i, 0});
            res[i][n-1] |= 2;
            v2[i][n-1] = 1;
            q2.push({i, n-1});
        }
        for(int j = 0; j < n; j++){
            res[0][j] |= 1;
            v1[0][j] = 1;
            q1.push({0, j});
            res[m-1][j] |= 2;
            v2[m-1][j] = 1;
            q2.push({m-1, j});
        }
        // cout << res[0][0] << " " << res[m-1][]
        // if(m == 1)
        //     res[0][0]--, res[m-1][n-1]--;
        // if(m != 1 && n != 1)
        //     res[0][n-1] = 2, res[m-1][0] = 2;
        do_it(1, res, q1, v1, n, m, heights);
        do_it(2, res, q2, v2, n, m, heights);
        vector<vector<int>> ans;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                // cout << res[i][j] & 3 << " ";
                if((res[i][j] & 3) == 3){
                    vector<int> temp(2);
                    temp[0] = i, temp[1] = j;
                    ans.push_back(temp);
                }
            }
        }
        return ans;
    }
};
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