<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 DisjointSet
{
    public:
    vector<int> parent , size , rank;

    DisjointSet(int n ){
        parent.resize(n+1);
        rank.resize(n+1 , 0);
        size.resize(n+1 , 1);
        for (int i = 0 ; i < n ; i++){
            parent[i] = i;
        }
    }

    int findUPar(int x){
        if (x == parent[x]) return x;
        return parent[x] = findUPar(parent[x]);
    }

    void UnionbySize(int x , int y){
        int ultparent1 = findUPar(x);
        int ultparent2 = findUPar(y);
        if (ultparent1 == ultparent2) return;
        if (size[ultparent1] > size[ultparent2]){
            parent[ultparent2] = ultparent1;
            size[ultparent1] += size[ultparent2];
        }
        else{
            parent[ultparent1] = ultparent2;
            size[ultparent2] += size[ultparent1];
        }
    }

    void UnionbyRank(int x , int y){
            int ultparent1 = findUPar(x);
            int ultparent2 = findUPar(y);
            if (ultparent1 == ultparent2) return;
            if (rank[ultparent1] > rank[ultparent2]){
                parent[ultparent2] = ultparent1;
                
            }
            else if (rank[ultparent2] > rank[ultparent1]){
                parent[ultparent1] = ultparent2;
            }

            else{
                parent[ultparent1] = ultparent2;
                rank[ultparent2] ++;
            }
        }
};

class Solution {
public:
    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        if (n==1){
            return 0;
        }
        DisjointSet ds(n*n);
        int t = 0;
        int lastnode = ((n-1) * n) + n-1;
        int dx[4] = {0,0,-1,1};
        int dy[4] = {1,-1,0,0};
        cout << n << endl;
        cout << lastnode;
        while (ds.findUPar(0) != ds.findUPar(lastnode)){
            for (int i = 0 ; i <  n ; i ++){
                for (int j = 0 ; j < n ; j++){
                    int node = i*n+j;
                    if (grid[i][j] <= t){
                        for (int k = 0 ; k < 4 ; k++){
                            int row = i + dx[k];
                            int col = j + dy[k];
                            int newnode = row * n + col; 
                            if (row < 0 || col < 0 || row>= n || col >= n) continue;
                            if (grid[row][col] <= t && ds.findUPar(newnode) != ds.findUPar(node)){
                                ds.UnionbyRank(newnode , node);
                            }
                        }
                    }
                }
            }
            t ++;
        }
        return t-1;
    }
};
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