<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 bfs(unordered_map<int,unordered_map<int,int>>& ocean,queue<pair<int,int>>&q, int n, int m, unordered_map<int,unordered_map<int,bool>>&vis,vector<vector<int>>& heights)
    {
        while(!q.empty())
    {
        pair<int,int> x = q.front();
        q.pop();

        if(vis[x.first][x.second])
        continue;
        vis[x.first][x.second] = true;

        ocean[x.first][x.second]++;
        vector<pair<int,int>> directions = {{0,1},{1,0},{-1,0},{0,-1}};

        for(pair<int,int> dir : directions)
        {
            int new_x = x.first + dir.first;
            int new_y = x.second + dir.second;

            if(new_x<0 || new_x >=n || new_y < 0 || new_y >=m || heights[x.first][x.second] >heights[new_x][new_y] || vis[new_x][new_y])
            continue;

            q.push({new_x,new_y});


        }

    } 

    }
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
    vector<vector<int>> res;

    unordered_map<int,unordered_map<int,int>> pacific,atlantic;

    queue<pair<int,int>> q;
    int n = heights.size();
    int m = heights[0].size();

    for(int i=0;i<n;i++)
    q.push({i,0});

    for(int j=1;j<m;j++)
    q.push({0,j});

    unordered_map<int,unordered_map<int,bool>> vis;
    bfs(pacific,q,n,m,vis,heights);

    // while(!q.empty())
    // {
    //     pair<int,int> x = q.front();
    //     q.pop();

    //     if(vis[x.first][x.second])
    //     continue;
    //     vis[x.first][x.second] = true;

    //     pacific[x.first][x.second]++;
    //     vector<pair<int,int>> directions = {{0,1},{1,0},{-1,0},{0,-1}};

    //     for(pair<int,int> dir : directions)
    //     {
    //         int new_x = x.first + dir.first;
    //         int new_y = x.second + dir.second;

    //         if(new_x<0 || new_x >=n || new_y < 0 || new_y >=m || heights[x.first][x.second] >heights[new_x][new_y] || vis[new_x][new_y])
    //         continue;

    //         q.push({new_x,new_y});


    //     }

    // } 

    vis.clear();

    for(int i=0;i<n;i++)
    q.push({i,m-1});

    for(int j=0;j<m-1;j++)
    q.push({n-1,j});
    bfs(atlantic,q,n,m,vis,heights);

    // while(!q.empty())
    // {
    //     pair<int,int> x = q.front();
    //     q.pop();

    //     if(vis[x.first][x.second])
    //     continue;
    //     vis[x.first][x.second] = true;

    //     atlantic[x.first][x.second]++;
    //     vector<pair<int,int>> directions = {{0,1},{1,0},{-1,0},{0,-1}};

    //     for(pair<int,int> dir : directions)
    //     {
    //         int new_x = x.first + dir.first;
    //         int new_y = x.second + dir.second;

    //         if(new_x<0 || new_x >=n || new_y < 0 || new_y >=m || heights[x.first][x.second] >heights[new_x][new_y] || vis[new_x][new_y])
    //         continue;

    //         q.push({new_x,new_y});


    //     }

    // } 

    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(pacific[i][j] && atlantic[i][j])
            res.push_back({i,j});
        }
    }


    return res;
    


    }
};
44400