<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 {
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
public:
    bool bfs(int i,int j,vector<vector<int>> &grid,vector<vector<int>> vis,int mid){
        int n=grid.size();
        int m=grid[0].size();
        queue<pair<int,int>> q;
        q.push({i,j});
        vis[i][j]=1;
        while(!q.empty()){
            auto [x,y]=q.front();
            q.pop();
            if(x==n-1 and y==m-1)return true;
            for(int k=0;k<4;k++){
                int nx=x+dx[k];
                int ny=y+dy[k];
                if(nx<0 || ny<0 || nx>=n || ny>=m || vis[nx][ny]==1 || grid[nx][ny]>mid)continue;
                vis[nx][ny]=1;
                q.push({nx,ny}); 
            }
        }
        return false;
    }
    int swimInWater(vector<vector<int>>& grid) {
        int l=0,r=1e9;
        int ans=0;
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>> vis(n,vector<int>(m,0));
        while(l<=r){
            int mid=(l+r)/2;
            if (grid[0][0]<=mid and bfs(0,0,grid,vis,mid)){
                ans=mid;
                r=mid-1;
            }
            else{
                l=mid+1;
            }
        }
        return ans;
    }
};
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