<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 isInside(int x, int y, int rows, int cols)
    {
        if (x>=0 && x<rows && y>=0 && y<cols) return true;
        return false;
    }
    bool dfs(vector<vector<int>>& grid, 
             int x,
             int y,
             int timeT, 
             vector<vector<bool>>& visited)
    {
        visited[x][y]=1;
        if(x== grid.size()-1 && y==grid[0].size()-1) return true;
        
        int dirX[] = {-1,0,1,0};
        int dirY[] = {0,1,0,-1};
        
        bool currAns = false;
        
        for(int i=0; i<4; i++)
        {
            int newX = x + dirX[i];
            int newY = y + dirY[i];
            if(isInside(newX, newY, grid.size(), grid[0].size()))
            {
                if(!visited[newX][newY] && grid[newX][newY] <= timeT)
                {
                    currAns =  currAns | dfs(grid, newX, newY, timeT, visited);
                }
                    
            }
        }
        
        return currAns;
    }
    int swimInWater(vector<vector<int>>& grid) 
    {
        if(grid.empty()) return 0;
        
        int high = (int)grid.size() * (int)grid.size();
        int low = 0, ans = high;

        while(low <= high)
        {
            int mid = low + (high-low)/2;
            vector<vector<bool > > visited(grid.size(), vector<bool >(grid[0].size(),0));
            // cout<<mid<<endl;
            if( grid[0][0]<= mid && dfs(grid, 0, 0, mid, visited))
            {
                high = mid-1;
                ans = mid;
                // cout<<ans<<endl;
            }
            else
            {
                low = mid+1;
            }
        }
        
        return ans;
    }
};
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