<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: 
    #define pi pair<int,pair<int,int>>
    bool f(int m,vector<vector<int>>&a,vector<vector<int>>&vis){

        priority_queue<pi,vector<pi>,greater<pi>>q;
        q.push({a[0][0],{0,0}});
        vis[0][0]=1;
        while(!q.empty()){
            auto it=q.top();q.pop();
            int r=it.second.first,c=it.second.second,dist=it.first;
            if(r==a.size()-1 and c==a[0].size()-1)return dist<=m;
            for(int x=-1;x<=1;x++){
                for(int y=-1;y<=1;y++){
                    if(x==0 or y==0){
                        int nr=r+x,nc=c+y;
                        if(nr>=0 and nc>=0 and nr<a.size() and nc<a[0].size() and !vis[nr][nc]){
                            q.push({max(dist,a[nr][nc]),{nr,nc}});
                            vis[nr][nc]=1;
                        }
                    }
                }
            }

        }
        return 0;
    }
    int swimInWater(vector<vector<int>>& grid) {
        int l=0,h=2500;
        int ans=-1;
        while(l<=h){
            int mid=(l+h)/2;
            vector<vector<int>>vis(grid.size(),vector<int>(grid[0].size(),0));
            if(f(mid,grid,vis)){
                ans=mid;
                h=mid-1;
            }
            else l=mid+1;
        }
        return ans;
    }
};
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