PROBLEM DESCRIPTION
Given an undirected graph with V vertices labelled from 0 to V-1 and E edges, check whether it contains any cycle or not. Graph is in the form of adjacency list where adj[i] contains all the nodes ith node is having edge with.
SOLUTION
Good Explanation - geeksforgeeks
Let’s say we start from node 0, and node 1 is its neighbor. So, from node 0, we go to node 1. If we only track whether a node has been visited, this can cause a problem because node 0 will again appear as a neighbor of node 1, since the graph is undirected.
To handle this, we need to track the parent node. If a neighbor is not the parent, then we can conclude that a cycle has been found.
For every visited vertex v, if there is an adjacent u such that u is already visited and u is not parent of v, then there is a cycle in graph.
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class Solution {
public boolean isCycle(int V, ArrayList<ArrayList<Integer>> adj) {
boolean[] visited = new boolean[V];
// Traverse each node in the graph
// This is needed since the graph can be disconnected
for(int node = 0; node < V; node++) {
// If the node is not yet visited, initiate DFS from this node
if(!visited[node]) {
// If DFS finds a cycle, return true
if(dfs(adj, visited, node, -1))
return true;
}
}
// no cycle found
return false;
}
public boolean dfs(ArrayList<ArrayList<Integer>> adj, boolean[] visited, int node, int parent) {
// Mark the current node as visited
visited[node] = true;
// Explore all adjacent nodes (neighbors)
for(Integer neighbor : adj.get(node)) {
// If the neighbor hasn't been visited, perform DFS on the neighbor
if(!visited[neighbor]) {
// If DFS finds a cycle, return true
if(dfs(adj, visited, neighbor, node))
return true;
} else {
// If the neighbor is visited and is not the parent node, it's a cycle
if(neighbor != parent)
return true;
}
}
return false;
}
}