<p>Given an <code>m x n</code> integer matrix <code>heightMap</code> representing the height of each unit cell in a 2D elevation map, return <em>the volume of water it can trap after raining</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/08/trap1-3d.jpg" style="width: 361px; height: 321px;" />
<pre>
<strong>Input:</strong> heightMap = [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> After the rain, water is trapped between the blocks.
We have two small ponds 1 and 3 units trapped.
The total volume of water trapped is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/08/trap2-3d.jpg" style="width: 401px; height: 321px;" />
<pre>
<strong>Input:</strong> heightMap = [[3,3,3,3,3],[3,2,2,2,3],[3,2,1,2,3],[3,2,2,2,3],[3,3,3,3,3]]
<strong>Output:</strong> 10
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == heightMap.length</code></li>
	<li><code>n == heightMap[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= heightMap[i][j] &lt;= 2 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int row[4] = {-1,0,1,0};
    int col[4] = {0,1,0,-1};
    int trapRainWater(vector<vector<int>>& hmap) {
        // priority queue se hoga
        // boundary pe to stoe nahi kar sakte to unhe pehle hi vis mark kardo and add kardo saare boundary elements ko heap mai
        // fir usme se weakest nikalo and neighbours check karo
        // if neighbour value < weakest value -> paani store hosakta hai +delegate your work to neighbour
        // if neighbour value >= weakest value => no paani store + delegate your work(means pop kardo or use add karo heap mai)

        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        int n = hmap.size();
        int m = hmap[0].size();
        vector<vector<int>> vis(n, vector<int>(m, 0));
        for(int i=0;i<m;i++){
            vis[0][i] = vis[n-1][i] = 1;
            pq.push({hmap[0][i], 0, i});
            pq.push({hmap[n-1][i], n-1, i});
        }
        for(int i=0;i<n;i++){
            vis[i][0] = vis[i][m-1] = 1;
            pq.push({hmap[i][0], i, 0});
            pq.push({hmap[i][m-1], i, m-1});
        }

        int cnt = 0;
        while(!pq.empty()){
            auto top = pq.top();
            pq.pop();
            int oh = top[0];
            int r = top[1];
            int c = top[2];

            for(int i=0;i<4;i++){
                int newr = r + row[i];
                int newc = c + col[i];

                if(newr < 0 || newc < 0 || newr >= n || newc >= m || vis[newr][newc]) continue;
                vis[newr][newc] = 1;
                //agar bada or barabar hai
                if(hmap[newr][newc] >= oh){
                    pq.push({hmap[newr][newc], newr, newc});
                }else{
                    cnt += (oh - hmap[newr][newc]);
                    pq.push({oh, newr, newc});
                }
            }
        }

        return cnt;
    }
};
29506