<p>You are given an <code>n x n</code> integer matrix <code>grid</code> where each value <code>grid[i][j]</code> represents the elevation at that point <code>(i, j)</code>.</p>

<p>The rain starts to fall. At time <code>t</code>, the depth of the water everywhere is <code>t</code>. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most <code>t</code>. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.</p>

<p>Return <em>the least time until you can reach the bottom right square </em><code>(n - 1, n - 1)</code><em> if you start at the top left square </em><code>(0, 0)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim1-grid.jpg" style="width: 164px; height: 165px;" />
<pre>
<strong>Input:</strong> grid = [[0,2],[1,3]]
<strong>Output:</strong> 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/29/swim2-grid-1.jpg" style="width: 404px; height: 405px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
<strong>Output:</strong> 16
<strong>Explanation:</strong> The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;&nbsp;n<sup>2</sup></code></li>
	<li>Each value <code>grid[i][j]</code> is <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    int swimInWater(vector<vector<int>>& grid) {
        int len=grid.size();
        priority_queue< pair< int , pair<int , int> > , vector<pair< int , pair<int , int> >> , greater<pair< int , pair<int , int> >> >pq;
        vector<vector<int>>arr(len , vector<int>(len , 0)) , dist(len , vector<int>(len , 0)) , visited(len , vector<int>(len , 0));
        dist[0][0]=grid[0][0];
        pq.push({ grid[0][0] , { 0 , 0 }});
        vector<int>dirr={0, -1 , 0 , 1  },dirc={ 1 , 0 , -1 , 0 };
        int r , c , rw , cl , curval;

        for(int i=0;i<len;i++){
            for(int j=0;j<len;j++){
                dist[i][j]=grid[i][j];
            }
        }

        while(!pq.empty()){
            curval=pq.top().first , rw=pq.top().second.first , cl=pq.top().second.second;
            pq.pop();
            cout<<"row is "<<rw<<" col is "<<cl<<endl;
            if(visited[rw][cl]==1)continue;
            
            for(int i=0;i<4;i++){
                r=rw+dirr[i] , c=cl+dirc[i];
                if( r>=0 && r<len && c<len && c>=0){
                    cout<<" r is "<<r<<" c ic "<<c<<endl;
                    if(visited[r][c]==0){

                        if(curval<=grid[r][c]){
                            cout<<"normal"<<endl;
                            pq.push({ grid[r][c] , { r , c }});
                            dist[r][c]=grid[r][c];
                        }
                        else{
                            if(dist[r][c]==grid[r][c]){
                                dist[r][c]=curval;
                                pq.push({ curval , { r , c }});
                            }
                            else{
                                if(curval<dist[r][c]){
                                    cout<<"newval"<<endl;
                                    pq.push({ curval , { r , c }});
                                    dist[r][c]=curval;
                                }
                            }
                        }

                    }
                }
            }
            visited[rw][cl]=1;
        }
        return dist[len-1][len-1];
    }
};
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