<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 isPossible(int sr, int sc, int dis, vector<vector<int>>& grid, vector<vector<int>>& visited){
        int n = grid.size();
        int m = grid[0].size();

        if( sr == n-1 && sc == m-1 ) return true;
        visited[sr][sc] = 1;
        vector<int> r = { -1, 0, 1, 0};
        vector<int> c = { 0, 1, 0, -1};
        int u1 = false;
        for( int i = 0; i < 4; ++i){
            int nsr = sr + r[i];
            int nsc = sc + c[i];
            if( 0 <= nsr && nsr < n && 0 <= nsc && nsc < m ){
                if( !visited[nsr][nsc] && grid[nsr][nsc] <= dis ){
                    u1 = u1 or isPossible( nsr, nsc, dis, grid, visited );
                }
            }
        }
    return u1;
    }

    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int mini = INT_MAX;
        int maxi = INT_MIN;

        for( int i = 0; i < n; ++i){
            for( int j = 0; j < m; ++j){
               mini = min( mini, grid[i][j]);
               maxi = max( maxi, grid[i][j]);
            } 
        }
        int left = mini;
        int right = maxi;
        int ans = -1;
        
        while( left <= right ){
            int mid = (left + right)/2;
            vector<vector<int>> visited( n, vector<int>(m,0)); 
            if( grid[0][0] <= mid && isPossible(0, 0, mid, grid, visited) ){
                ans = mid;
                right = mid - 1;
            }else{
                left = mid + 1;
            }
        }
    return ans;
    }
};
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