Print Nodes which are not part of any cycle in a Directed Graph
Given a directed graph G N nodes and E Edges consisting of nodes valued [0, N – 1] and a 2D array Edges[][2] of type {u, v} that denotes a directed edge between vertices u and v. The task is to find the nodes that are not part of any cycle in the given graph G.
Examples:
Input: N = 4, E = 4, Edges[][2] = { {0, 2}, {0, 1}, {2, 3}, {3, 0} }
Output: 1
Explanation:
From the given graph above there exists a cycle between the nodes 0 -> 2 -> 3 -> 0.
Node which doesn’t occurs in any cycle is 1.
Hence, print 1.Input: N = 6, E = 7, Edges[][2] = { {0, 1}, {0, 2}, {1, 3}, {2, 1}, {2, 5}, {3, 0}, {4, 5}}
Output: 4 5
Explanation:
From the given graph above there exists a cycle between the nodes:
1) 0 -> 1 -> 3 -> 0.
2) 0 -> 2 -> 1 -> 3 -> 0.
Nodes which doesn’t occurs in any cycle are 4 and 5.
Hence, print 4 and 5.
Naive Approach: The simplest approach is to detect a cycle in a directed graph for each node in the given graph and print only those nodes that are not a part of any cycle in the given graph.
Time Complexity: O(V * (V + E)), where V is the number of vertices and E is the number of edges.
Auxiliary Space: O(V)
Efficient Approach: To optimize the above approach, the idea is to store the intermediate node as a visited cycle node whenever any cycle in the given graph. To implement this part use an auxiliary array cyclePart[] that will store the intermediate cycle node while performing the DFS Traversal. Below are the steps:
- Initialize an auxiliary array cyclePart[] of size N, such that if cyclePart[i] = 0, then ith node doesn’t exist in any cycle.
- Initialize an auxiliary array recStack[] of size N, such that it will store the visited node in the recursion stack by marking that node as true.
- Perform the DFS Traversal on the given graph for each unvisited node and do the following:
- Now find a cycle in the given graph, whenever a cycle is found, mark the node in cyclePart[] as true as this node is a part of the cycle.
- If any node is visited in the recursive call and is recStack[node] is also true then that node is a part of the cycle then mark that node as true.
- After performing the DFS Traversal, traverse the array cyclePart[] and print all those nodes that are marked as false as these nodes are not the part of any cycle.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;class Graph { // No. of vertices int V; // Stores the Adjacency List list<int>* adj; bool printNodesNotInCycleUtil( int v, bool visited[], bool* rs, bool* cyclePart);public: // Constructor Graph(int V); // Member Functions void addEdge(int v, int w); void printNodesNotInCycle();};// Function to initialize the graphGraph::Graph(int V){ this->V = V; adj = new list<int>[V];}// Function that adds directed edges// between node v with node wvoid Graph::addEdge(int v, int w){ adj[v].push_back(w);}// Function to perform DFS Traversal// and return true if current node v// forms cyclebool Graph::printNodesNotInCycleUtil( int v, bool visited[], bool* recStack, bool* cyclePart){ // If node v is unvisited if (visited[v] == false) { // Mark the current node as // visited and part of // recursion stack visited[v] = true; recStack[v] = true; // Traverse the Adjacency // List of current node v for (auto& child : adj[v]) { // If child node is unvisited if (!visited[child] && printNodesNotInCycleUtil( child, visited, recStack, cyclePart)) { // If child node is a part // of cycle node cyclePart[child] = 1; return true; } // If child node is visited else if (recStack[child]) { cyclePart[child] = 1; return true; } } } // Remove vertex from recursion stack recStack[v] = false; return false;}// Function that print the nodes for// the given directed graph that are// not present in any cyclevoid Graph::printNodesNotInCycle(){ // Stores the visited node bool* visited = new bool[V]; // Stores nodes in recursion stack bool* recStack = new bool[V]; // Stores the nodes that are // part of any cycle bool* cyclePart = new bool[V]; for (int i = 0; i < V; i++) { visited[i] = false; recStack[i] = false; cyclePart[i] = false; } // Traverse each node for (int i = 0; i < V; i++) { // If current node is unvisited if (!visited[i]) { // Perform DFS Traversal if (printNodesNotInCycleUtil( i, visited, recStack, cyclePart)) { // Mark as cycle node // if it return true cyclePart[i] = 1; } } } // Traverse the cyclePart[] for (int i = 0; i < V; i++) { // If node i is not a part // of any cycle if (cyclePart[i] == 0) { cout << i << " "; } }}// Function that print the nodes for// the given directed graph that are// not present in any cyclevoid solve(int N, int E, int Edges[][2]){ // Initialize the graph g Graph g(N); // Create a directed Graph for (int i = 0; i < E; i++) { g.addEdge(Edges[i][0], Edges[i][1]); } // Function Call g.printNodesNotInCycle();}// Driver Codeint main(){ // Given Number of nodes int N = 6; // Given Edges int E = 7; int Edges[][2] = { { 0, 1 }, { 0, 2 }, { 1, 3 }, { 2, 1 }, { 2, 5 }, { 3, 0 }, { 4, 5 } }; // Function Call solve(N, E, Edges); return 0;} |
Java
// Java program for above approachimport java.util.*;import java.lang.*;class GFG{ static ArrayList<ArrayList<Integer>> adj; static int V; // Function to perform DFS Traversal // and return true if current node v // forms cycle static boolean printNodesNotInCycleUtil( int v, boolean visited[], boolean[] recStack, boolean[] cyclePart) { // If node v is unvisited if (visited[v] == false) { // Mark the current node as // visited and part of // recursion stack visited[v] = true; recStack[v] = true; // Traverse the Adjacency // List of current node v for (Integer child : adj.get(v)) { // If child node is unvisited if (!visited[child] && printNodesNotInCycleUtil( child, visited, recStack, cyclePart)) { // If child node is a part // of cycle node cyclePart[child] = true; return true; } // If child node is visited else if (recStack[child]) { cyclePart[child] = true; return true; } } } // Remove vertex from recursion stack recStack[v] = false; return false; } static void printNodesNotInCycle() { // Stores the visited node boolean[] visited = new boolean[V]; // Stores nodes in recursion stack boolean[] recStack = new boolean[V]; // Stores the nodes that are // part of any cycle boolean[] cyclePart = new boolean[V]; // Traverse each node for (int i = 0; i < V; i++) { // If current node is unvisited if (!visited[i]) { // Perform DFS Traversal if (printNodesNotInCycleUtil( i, visited, recStack, cyclePart)) { // Mark as cycle node // if it return true cyclePart[i] = true; } } } // Traverse the cyclePart[] for (int i = 0; i < V; i++) { // If node i is not a part // of any cycle if (!cyclePart[i]) { System.out.print(i+" "); } } } // Function that print the nodes for // the given directed graph that are // not present in any cycle static void solve(int N, int E, int Edges[][]) { adj = new ArrayList<>(); for(int i = 0; i < N; i++) adj.add(new ArrayList<>()); // Create a directed Graph for (int i = 0; i < E; i++) { adj.get(Edges[i][0]).add(Edges[i][1]); } // Function Call printNodesNotInCycle(); } // Driver function public static void main (String[] args) { // Given Number of nodes V = 6; // Given Edges int E = 7; int Edges[][] = { { 0, 1 }, { 0, 2 }, { 1, 3 }, { 2, 1 }, { 2, 5 }, { 3, 0 }, { 4, 5 } }; // Function Call solve(V, E, Edges); }}// This code is contributed by offbeat |
Python3
# Python3 program for the above approachclass Graph: # Function to initialize the graph def __init__(self, V): self.V = V self.adj = [[] for i in range(self.V)] # Function that adds directed edges # between node v with node w def addEdge(self, v, w): self.adj[v].append(w); # Function to perform DFS Traversal # and return True if current node v # forms cycle def printNodesNotInCycleUtil(self, v, visited,recStack, cyclePart): # If node v is unvisited if (visited[v] == False): # Mark the current node as # visited and part of # recursion stack visited[v] = True; recStack[v] = True; # Traverse the Adjacency # List of current node v for child in self.adj[v]: # If child node is unvisited if (not visited[child] and self.printNodesNotInCycleUtil(child, visited,recStack, cyclePart)): # If child node is a part # of cycle node cyclePart[child] = 1; return True; # If child node is visited elif (recStack[child]): cyclePart[child] = 1; return True; # Remove vertex from recursion stack recStack[v] = False; return False; # Function that print the nodes for # the given directed graph that are # not present in any cycle def printNodesNotInCycle(self): # Stores the visited node visited = [False for i in range(self.V)]; # Stores nodes in recursion stack recStack = [False for i in range(self.V)]; # Stores the nodes that are # part of any cycle cyclePart = [False for i in range(self.V)] # Traverse each node for i in range(self.V): # If current node is unvisited if (not visited[i]): # Perform DFS Traversal if(self.printNodesNotInCycleUtil( i, visited, recStack, cyclePart)): # Mark as cycle node # if it return True cyclePart[i] = 1; # Traverse the cyclePart[] for i in range(self.V): # If node i is not a part # of any cycle if (cyclePart[i] == 0) : print(i,end=' ') # Function that print the nodes for# the given directed graph that are# not present in any cycledef solve( N, E, Edges): # Initialize the graph g g = Graph(N); # Create a directed Graph for i in range(E): g.addEdge(Edges[i][0], Edges[i][1]); # Function Call g.printNodesNotInCycle(); # Driver Codeif __name__=='__main__': # Given Number of nodes N = 6; # Given Edges E = 7; Edges = [ [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 2, 1 ], [ 2, 5 ], [ 3, 0 ], [ 4, 5 ] ]; # Function Call solve(N, E, Edges); # This code is contributed by rutvik_56 |
C#
// C# program for above approachusing System;using System.Collections.Generic;public class GFG{ static List<List<int>> adj; static int V; // Function to perform DFS Traversal // and return true if current node v // forms cycle static bool printNodesNotInCycleUtil( int v, bool []visited, bool[] recStack, bool[] cyclePart) { // If node v is unvisited if (visited[v] == false) { // Mark the current node as // visited and part of // recursion stack visited[v] = true; recStack[v] = true; // Traverse the Adjacency // List of current node v foreach (int child in adj[v]) { // If child node is unvisited if (!visited[child] && printNodesNotInCycleUtil( child, visited, recStack, cyclePart)) { // If child node is a part // of cycle node cyclePart[child] = true; return true; } // If child node is visited else if (recStack[child]) { cyclePart[child] = true; return true; } } } // Remove vertex from recursion stack recStack[v] = false; return false; } static void printNodesNotInCycle() { // Stores the visited node bool[] visited = new bool[V]; // Stores nodes in recursion stack bool[] recStack = new bool[V]; // Stores the nodes that are // part of any cycle bool[] cyclePart = new bool[V]; // Traverse each node for (int i = 0; i < V; i++) { // If current node is unvisited if (!visited[i]) { // Perform DFS Traversal if (printNodesNotInCycleUtil( i, visited, recStack, cyclePart)) { // Mark as cycle node // if it return true cyclePart[i] = true; } } } // Traverse the cyclePart[] for (int i = 0; i < V; i++) { // If node i is not a part // of any cycle if (!cyclePart[i]) { Console.Write(i+" "); } } } // Function that print the nodes for // the given directed graph that are // not present in any cycle static void solve(int N, int E, int [,]Edges) { adj = new List<List<int>>(); for(int i = 0; i < N; i++) adj.Add(new List<int>()); // Create a directed Graph for (int i = 0; i < E; i++) { adj[Edges[i,0]].Add(Edges[i,1]); } // Function Call printNodesNotInCycle(); } // Driver function public static void Main(String[] args) { // Given Number of nodes V = 6; // Given Edges int E = 7; int [,]Edges = { { 0, 1 }, { 0, 2 }, { 1, 3 }, { 2, 1 }, { 2, 5 }, { 3, 0 }, { 4, 5 } }; // Function Call solve(V, E, Edges); }}// This code contributed by shikhasingrajput |
Javascript
<script>// JavaScript program for the above approachclass Graph{ // Function to initialize the graph constructor(V){ this.V = V this.adj = new Array(V).fill(0).map(()=>[]) } // Function that adds directed edges // between node v with node w addEdge(v, w){ this.adj[v].push(w); } // Function to perform DFS Traversal // and return true if current node v // forms cycle printNodesNotInCycleUtil(v, visited,recStack, cyclePart){ // If node v is unvisited if (visited[v] == false){ // Mark the current node as // visited and part of // recursion stack visited[v] = true; recStack[v] = true; // Traverse the Adjacency // List of current node v for(let child of this.adj[v]){ // If child node is unvisited if (visited[child] == false && this.printNodesNotInCycleUtil(child, visited,recStack, cyclePart)){ // If child node is a part // of cycle node cyclePart[child] = 1; return true; } // If child node is visited else if (recStack[child]){ cyclePart[child] = 1; return true; } } } // Remove vertex from recursion stack recStack[v] = false; return false; } // Function that print the nodes for // the given directed graph that are // not present in any cycle printNodesNotInCycle(){ // Stores the visited node let visited = new Array(this.V).fill(false); // Stores nodes in recursion stack let recStack = new Array(this.V).fill(false); // Stores the nodes that are // part of any cycle let cyclePart = new Array(this.V).fill(false) // Traverse each node for(let i=0;i<this.V;i++){ // If current node is unvisited if (visited[i] == false){ // Perform DFS Traversal if(this.printNodesNotInCycleUtil( i, visited, recStack, cyclePart)) // Mark as cycle node // if it return true cyclePart[i] = 1; } } // Traverse the cyclePart[] for(let i=0;i<this.V;i++){ // If node i is not a part // of any cycle if (cyclePart[i] == 0) document.write(i,' ') } }} // Function that print the nodes for// the given directed graph that are// not present in any cyclefunction solve(N, E, Edges){ // Initialize the graph g let g = new Graph(N); // Create a directed Graph for(let i=0;i<E;i++){ g.addEdge(Edges[i][0], Edges[i][1]); } // Function Call g.printNodesNotInCycle();} // Driver Code// Given Number of nodeslet N = 6;// Given Edgeslet E = 7;let Edges = [ [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 2, 1 ], [ 2, 5 ], [ 3, 0 ], [ 4, 5 ] ];// Function Callsolve(N, E, Edges)// This code is contributed by shinjanpatra</script> |
Output:
4 5
Time Complexity: O(V + E)
Space Complexity: O(V)
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