LeetCode: Number of Connected Components

/**
Number of Connected Components in an Undirected Graph
Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph.
Example 1:
     0          3
     |          |
     1 --- 2    4
Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.
Example 2:
     0           4
     |           |
     1 --- 2 --- 3
Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]], return 1.
Note: You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.
*/
import java.util.*;

public class NumberOfConnectedComponents{

/*
-similar to union find
-first each node is on its own component
-as we iterate through the edges and if we see the two nodes of the edge are in different component, we create a new entry to have them in a single component
*/
public static int countComponents(int n, int[][] edges) {
        if (n <= 1) {
            return n;
        }
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            map.put(i, i);
        }
        for (int[] edge : edges) {
            int fa_x = find(map, edge[0]);
            int fa_y = find(map, edge[1]);
System.out.println(fa_x +"..."+fa_y+"..."+n);
            if (fa_x != fa_y) {
                map.put(fa_x, fa_y);
n--;
            }
        }
        return n;
    }

    static int find(HashMap<Integer, Integer> map, int x) {
        int parent_comp = map.get(x);
        while (parent_comp != map.get(parent_comp)) {
            parent_comp = map.get(parent_comp);
        }
        return parent_comp;
    }

public static void main(String args[]){
int[][] edges = new int[4][2];
edges[0]= new int[]{0,1};
edges[1]= new int[]{1,2};
edges[2]= new int[]{3,4};
edges[3]= new int[]{0,5};
int comp = countComponents(6,  edges);
System.out.println(comp);

}
}