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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.

For example:

```
0 3
| |
1 --- 2 4
```

Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], return 2.

**Java Solution – Union-find**

This problem can be solved by using union-find beautifully. Initially, there are n nodes. The nodes that are involved in each edge are merged.

```
public int countComponents(int n, int[][] edges) {
int count = n;
int[] root = new int[n];
// initialize each node is an island
for(int i=0; i<n; i++){
root[i]=i;
}
for(int i=0; i<edges.length; i++){
int x = edges[i][0];
int y = edges[i][1];
int xRoot = getRoot(root, x);
int yRoot = getRoot(root, y);
if(xRoot!=yRoot){
count--;
root[xRoot]=yRoot;
}
}
return count;
}
public int getRoot(int[] arr, int i){
while(arr[i]!=i){
arr[i]= arr[arr[i]];
i=arr[i];
}
return i;
}
```

There are k loops and each loop processing the root array costs log(n). Therefore, the time complexity is O(k*log(n)).