<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 dr[4]={0,1,0,-1};
    int dc[4]={1,0,-1,0};

    bool check(int val, vector<vector<int>>& grid){
        int m= grid.size(), n= grid[0].size();

        queue<vector<int>> q;
        q.push({0,0});
        int vis[m][n];
        memset(vis,0,sizeof(vis));
        vis[0][0]= true;
        if(grid[0][0]> val){
            return false;
        }

        while(!q.empty()){
            int r= q.front()[0];
            int c= q.front()[1];
            q.pop();
            if(r==m-1 and c==n-1){
                if(grid[r][c]>val){
                    return false;
                }
                return true;
            }

            for(int i=0; i<4; i++){
                int nr= r+dr[i];
                int nc= c+dc[i];
                if(nr>=0 and nr<m and nc>=0 and nc<n){
                    if(grid[nr][nc]<=val and !vis[nr][nc]){
                        vis[nr][nc]= 1;
                        q.push({nr,nc});
                    }
                }
            }
        }
        return false;
    }

    int swimInWater(vector<vector<int>>& grid) {
        int m= grid.size(), n= grid[0].size();
        int lo=0, hi= m*n;
        int ans= INT_MAX;

        while(lo<=hi){
            int mid= (lo+hi)/2;
            if(check(mid, grid)){
                ans= mid;
                hi=mid-1;
            }else{
                lo=mid+1;
            }
        }
        return ans;
    }
};
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