<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>

#include <vector>

class DSU {
public:
    DSU(int n) {
        parent.resize(n);
        rank.resize(n, 0);
        for (int i = 0; i < n; ++i) {
            parent[i] = i;
        }
    }

    int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]); // Path compression
        }
        return parent[x];
    }

    void unite(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        
        if (rootX != rootY) {
            // Union by rank
            if (rank[rootX] > rank[rootY]) {
                parent[rootY] = rootX;
            } else if (rank[rootX] < rank[rootY]) {
                parent[rootX] = rootY;
            } else {
                parent[rootY] = rootX;
                rank[rootX]++;
            }
        }
    }

private:
    std::vector<int> parent;
    std::vector<int> rank;
};


class Solution {
public:


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

        int delrow[4] = {0,1};
        int delcol[4] = {1,0};
        DSU dsu(m*n);

        vector<vector<int>> edges;
        auto getIndex = [&](int row, int col) {
            return row * m + col;
        };
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int element = grid[i][j];

                for(int k=0;k<2;k++){
                    int nr = i+delrow[k];
                    int nc = j+delcol[k];

                    if(nr < n && nr>=0 && nc<m && nc>=0){
                        vector<int> edge = {max(element,grid[nr][nc]),getIndex(i,j),getIndex(nr,nc)};
                        edges.push_back(edge);
                    }
                }
            }
        }
        int ans=0;
        sort(edges.begin(),edges.end());
        for(auto edge : edges){
            int time = edge[0];
            int element1 = edge[1];
            int element2 = edge[2];

            dsu.unite(element1,element2);
            if(dsu.find(0)==dsu.find(m*n-1))return edge[0];
        }
    return ans;
    }
};
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