# 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.

We can solve this problem using Union find method.

``` 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; } 1234567891011121314151617181920212223242526272829303132 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;} ```

Here are some good articles to learn union find data structure:

https://www.cs.princeton.edu/~rs/AlgsDS07/01UnionFind.pdf

http://algs4.cs.princeton.edu/15uf/