Number of operations to make a network connected

Question:

There are n computers numbered from 0 to n - 1 connected by ethernet cables connections forming a network where connections[i] = [ai, bi]represents a connection between computers ai and bi. Any computer can reach any other computer directly or indirectly through the network.

You are given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected.

Return the minimum number of times you nee d to do this in order to make all the computers connected. If it is not possible, return -1.

Example 1:

Input: n = 4, connections = [[0,1],[0,2],[1,2]] Output: 1 Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.

Example 2:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]] Output: 2

Example 3:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]] Output: -1 Explanation: There are not enough cables.

Solution:

For Disjoint Set class copy template from UNION BY RANK (LESS INTUITIVE).

public int makeConnected(int n, int[][] connections) {//connections or edges
        DisjointSet dsu = new DisjointSet(n);
        int m = connections.length;
        int extraEdges = 0; //count extra edges to connect to the ultimate parents
 
        for(int i = 0; i < m; i++){
            int u = connections[i][0];
            int v = connections[i][1];
            if(dsu.findUltimateParent(u) == dsu.findUltimateParent(v)){
                extraEdges++;
            } else{
                dsu.unionByRank(u,v);
            }
        }
 
        int uParents = 0;
        for(int i = 0; i < n; i++){
            if(dsu.parent[i] == i){
                uParents++;
            }
        }
	    
	    int totalEdgesNeeded = uParents - 1;
	    
        if(extraEdges >= totalEdgesNeeded){
            return totalEdgesNeeded;
        } else{
            return -1;
        }
    }