<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:
  vector<pair<int,int>> pairs={
        {0,-1},{0,1},{1,0},{-1,0}
    };
    bool isok(int i,int j,int n,int m){
        if(i>=0 and i<n and j>=0 and j<m){
           return true;
        }
        return false;
    }
    bool isreachable(int i,int j,int mid,vector<vector<int>>& heights){
        int n=heights.size();
        int m=heights[0].size();
        queue<pair<pair<int,int>,int>>q;
        vector<vector<int>>vis(n,vector<int>(m,0)),dis(n,vector<int>(m,1e9));
        q.push({{0,0},0});
            vis[0][0]=1;
        while(!q.empty()){
            auto it=q.front();
            q.pop();
            int x=it.first.first;
            int y=it.first.second;
            // if(vis[x][y])continue;
            int d=it.second;
            for(auto itt:pairs){
                int xx=itt.first+x;
                int yy=itt.second+y;
                if(isok(xx,yy,n,m) && !vis[xx][yy] && heights[x][y]<=mid && heights[xx][yy]<=mid){
                    dis[xx][yy]=1+d;
                    vis[xx][yy]=1;
                    q.push({{xx,yy},dis[xx][yy]});
                }
            }
        }
        if(dis[n-1][m-1]!=1e9){
            return true;
        }
        return false;
    }
    int swimInWater(vector<vector<int>>& heights) {
         int n=heights.size();
        int m=heights[0].size();
        int mn=1e9,mx=0;
        for(auto it:heights){
           for(auto itt:it){
             mn=min(mn,itt);
            mx=max(mx,itt);
           }
        }
        int l=0,r=mx-mn,ans;
        
        while(r-l>1){
              int mid=(l+r)>>1;
              if(isreachable(0,0,mid,heights)){
                 r=mid;
              }
              else{
                l=mid;
              }
        }
        if(isreachable(0,0,l,heights))ans=l;
        else ans =r;
        // for(int i=max(0,ans-2);i<=ans;i++){
        //     if(isreachable(0,0,i,heights)){
        //         return i;
        //     }
        // }
        return ans;
    }
};
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