<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:
    
    // Binary search on ansswer
    
    int dx[4]={0,0,1,-1};
    int dy[4]={1,-1,0,0};
    int n;

    bool dfs(int x,int y, int t,vector<vector<int>> &grid, vector<vector<int>> &visit){
        visit[x][y]=1;

        if(t< grid[x][y]) return false;
        
        if(x== n-1 && y== n-1){
            return true;
        }

        bool ans=false;

        for(int i=0;i<4;i+=1){
            int nx= x + dx[i];
            int ny= y + dy[i];

            if(nx>=0 && ny>=0 && nx<n && ny<n && grid[nx][ny]<=t && !visit[nx][ny]){
                if(nx== n-1 && ny == n-1){
                    ans=true;
                }else{
                    ans = ans | dfs(nx,ny,t,grid, visit);
                }
            }
        }
        return ans;
    }

    int swimInWater(vector<vector<int>>& grid) {
        n= grid.size();
        
        int low=0,high=1e5;
        while(low<=high){
            int mid= low + (high-low)/2;
            vector<vector<int>> visit(n+1, vector<int> (n+1,0));
            if(dfs(0,0,mid,grid,visit)){
                high=mid-1;
            }else{
                low=mid+1;
            }

        }
        return low;
    }
};
15326