<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(vector<vector<int>> &grid , int x , int y , vector<vector<int>>&vis , int t){
        int n = grid.size();
        vis[x][y] = 1;
        int dx[4] = {-1 , 0 , 1 , 0};
        int dy[4] = {0 , 1 , 0 , -1};
        if(x  == n-1 and y == n-1)return true;
        for(int i= 0 ; i < 4 ; i++){
            int nx = x + dx[i];
            int ny = y + dy[i];
            if(nx >=0  and ny >=0 and nx < n and ny < n ){
                // Important Condition grid[nx][ny] <= t
                if( grid[nx][ny] <= t and vis[nx][ny] == 0){
                     if(dfs(grid, nx , ny , vis , t) == true)return true;
                }
            }
        }
        return false;
    }
    bool check(vector<vector<int>>& grid , int t){
        bool ans = false;
        int n = grid.size();
        vector<vector<int>> vis(n+1 , vector<int>(n+1  , 0));
        if(grid[0][0] <= t){
           ans =  dfs(grid ,0 , 0 , vis , t );
        }
        return ans;
    }
    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        int low = 0 , high = n*n;
        int ans = 0;
        while(low <= high){
            int mid = (low + high)/2;
            if(check(grid , mid )){
                ans = mid;
                high = mid - 1;
            }else {
                low = mid + 1;
            }
        }
        return ans;
    }
};
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