<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:
    bool dfs(int i, int j, vector<vector<int>>& grid, int a, 
    vector<int>& rows, vector<int>& colm,vector<int> &visit){
        visit[i*grid.size()+j]=true;
        if((i==(grid.size()-1)) && (j==(grid.size()-1))){
            return true;
        }
        for(int k=0;k<4;k++){
            int r = rows[k]+i;
            int c = colm[k]+j;
            if(r>=0 && r<grid.size() && c>=0 && c<grid.size() && grid[r][c]<=a &&
            visit[r*grid.size()+c]==false){
                if(dfs(r,c,grid,a,rows,colm,visit)) return true;
            }
        }
        return false;
    }
    bool check(vector<vector<int>>& grid, int a,vector<int>& rows, vector<int>& colm){
        int n = grid.size();
        vector<int> visit(n*n,false);
        if(grid[0][0]<=a)
        return dfs(0,0,grid,a,rows,colm,visit);
        else return false;
    }
    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        int left = 0;
        int right = n*n - 1;
        int mid;
        vector<int> rows = {-1,0,1,0};
        vector<int> colm = {0,-1,0,1};
        while(left <= right){
            mid = ((right-left)/2) + left;
            if(check(grid,mid,rows,colm)){
                right = mid - 1;
            }
            else{
                left = mid + 1;
            }
        }
        return left;
    }
};
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