<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>

vector<int> par, sz;
class DSU{
public:

    DSU(int n){
        par.resize(n+1);
        sz.resize(n+1);
        for(int i=0;i<=n;i++){
            par[i] = i;
            sz[i] = 1;
        }
    }
    int findpar(int x){
        if(x == par[x]) return x;
        else return par[x] = findpar(par[x]);
    }
    void merge(int x, int y){
        int ulpx = findpar(x), ulpy = findpar(y);
        if(ulpx == ulpy) return;
        if(sz[ulpx] >= sz[ulpy]){
            par[ulpy] = ulpx;
            sz[ulpx] += sz[ulpy];
        }
        else{
            par[ulpx] = ulpy;
            sz[ulpy] += sz[ulpx];
        }
        return;
    }
};
class Solution {
public:
    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        DSU dsu(n*n);
        vector<int> direcs{1, 0, -1, 0, 1};
        for(int t=0;t<n*n;t++){
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    if(grid[i][j] <= t){
                        for(int k=0;k<4;k++){
                            int r = i + direcs[k], c = j+direcs[k+1];
                            if(min(r,c) >= 0 && max(r, c) < n && grid[r][c] <= t){
                                dsu.merge(i*n+j, r*n+c);
                            }
                            if(dsu.findpar(0) == dsu.findpar((n-1)*n+(n-1))) return t;
                        }
                    }
                }
            }
        }
        return n*n-1;
    }
};
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