LeetCode 323: Number of Connected Components in an Undirected Graph Solution

Master LeetCode problem 323 (Number of Connected Components in an Undirected Graph), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

323. Number of Connected Components in an Undirected Graph

Medium

Depth-First Search Breadth-First Search Union Find Graph

Problem Explanation

Explanation:

To solve this problem, we can use a graph traversal technique such as Depth First Search (DFS) or Breadth First Search (BFS) to find connected components in the given undirected graph. We will start the traversal from each unvisited node and mark all the nodes reachable from it. Each connected component will be identified by a separate traversal from an unvisited node.

Here are the steps for the algorithm:

Create an adjacency list representation of the graph.
Initialize a set to keep track of visited nodes.
Initialize a variable to count the number of connected components.
Iterate through all the nodes in the graph.
For each unvisited node, perform a DFS or BFS traversal to mark all nodes reachable from it as visited. Increment the count of connected components.
Return the count of connected components.

Time Complexity: O(V + E), where V is the number of vertices and E is the number of edges in the graph. Space Complexity: O(V), where V is the number of vertices in the graph.

Solution Code

class Solution {
    public int countComponents(int n, int[][] edges) {
        List<List<Integer>> adjList = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adjList.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            adjList.get(edge[0]).add(edge[1]);
            adjList.get(edge[1]).add(edge[0]);
        }
        
        Set<Integer> visited = new HashSet<>();
        int count = 0;
        
        for (int i = 0; i < n; i++) {
            if (!visited.contains(i)) {
                dfs(i, adjList, visited);
                count++;
            }
        }
        return count;
    }
    
    private void dfs(int node, List<List<Integer>> adjList, Set<Integer> visited) {
        visited.add(node);
        for (int neighbor : adjList.get(node)) {
            if (!visited.contains(neighbor)) {
                dfs(neighbor, adjList, visited);
            }
        }
    }
}

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