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Problem Statement:(👈click here for GFG)
Given an undirected graph with V vertices 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.
Example 1:
Input: V = 5, E = 5 adj = {{1}, {0, 2, 4}, {1, 3}, {2, 4}, {1, 3}} Output: 1 Explanation:1->2->3->4->1 is a cycle.
Example 2:
Input: V = 4, E = 2 adj = {{}, {2}, {1, 3}, {2}} Output: 0 Explanation:No cycle in the graph.
Your Task:
You don't need to read or print anything. Your task is to complete the function isCycle() which takes V denoting the number of vertices and adjacency list as input parameters and returns a boolean value denoting if the undirected graph contains any cycle or not, return 1 if a cycle is present else return 0.
NOTE: The adjacency list denotes the edges of the graph where edges[i] stores all other vertices to which ith vertex is connected.
Expected Time Complexity: O(V + E)
Expected Space Complexity: O(V)
Constraints:
1 ≤ V, E ≤ 105
Solution:
class Solution {
// Function to detect cycle in an undirected graph.
isCycle(V, adj) {
// Create an array to keep track of visited vertices
let visited = new Array(V).fill(false);
// Perform DFS from each unvisited vertex
for (let i = 0; i < V; i++) {
if (!visited[i]) {
if (dfs(i, -1, visited, adj)) {
// If a cycle is detected, return true
return 1;
}
}
}
// If no cycle is detected, return false
return 0;
}
}
// Helper function to perform DFS
function dfs(vertex, parent, visited, adj) {
// Mark the current vertex as visited
visited[vertex] = true;
// Traverse all the adjacent vertices
for (let i = 0; i < adj[vertex].length; i++) {
let adjVertex = adj[vertex][i];
// If the adjacent vertex is not visited, visit it and continue DFS
if (!visited[adjVertex]) {
if (dfs(adjVertex, vertex, visited, adj)) {
return true;
}
}
// If the adjacent vertex is visited and not the parent of the current vertex, a cycle is detected
else if (adjVertex != parent) {
return true;
}
}
// If no cycle is detected, return false
return false;
}
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