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ConnectedComponentsUndirectedGraph.java
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ConnectedComponentsUndirectedGraph.java
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package Leetcode;
import java.util.*;
/**
* @author kalpak
*
* 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:
* Input: n = 5 and edges = [[0, 1], [1, 2], [3, 4]]
*
* 0 3
* | |
* 1 --- 2 4
*
* Output: 2
*
*
* Example 2:
* Input: n = 5 and edges = [[0, 1], [1, 2], [2, 3], [3, 4]]
*
* 0 4
* | |
* 1 --- 2 --- 3
*
* Output: 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.
*/
public class ConnectedComponentsUndirectedGraph {
public static int countComponents(int n, int[][] edges) {
if(n <= 1)
return n;
Map<Integer, List<Integer>> graph = new HashMap<>();
// Initialize the graph
for(int i = 0; i < n; i++) {
graph.put(i, new ArrayList<>());
}
// Put the edges into the graph.
// Put the edges both ways since the graph is undirected
for(int[] edge : edges) {
graph.get(edge[0]).add(edge[1]);
graph.get(edge[1]).add(edge[0]);
}
Set<Integer> visited = new HashSet<>();
int numberOfComponents = 0;
for(int i = 0; i < n; i++) {
if(!visited.contains(i)) {
dfsRecursive(i, visited, graph);
numberOfComponents++;
}
}
return numberOfComponents;
}
private static void dfsRecursive(int startNode, Set<Integer> visited, Map<Integer, List<Integer>> graph) {
visited.add(startNode);
List<Integer> adjacencyList = graph.get(startNode);
if(adjacencyList != null) {
for(int edge : adjacencyList) {
if(!visited.contains(edge))
dfsRecursive(edge, visited, graph);
}
}
return;
}
public static void main(String[] args) {
int[][] edges = new int[][]{{0, 1}, {1, 2}, {3, 4}};
System.out.println("The number of connected components are : " + countComponents(5, edges));
}
}