Given a graph, the task is to detect a cycle in the graph using degrees of the nodes in the graph and print all the nodes that are involved in any of the cycles. If there is no cycle in the graph then print -1.
Examples:
Input:
Output: 0 1 2
Approach: Recursively remove all vertices of degree 1. This can be done efficiently by storing a map of vertices to their degrees.
Initially, traverse the map and store all the vertices with degree = 1 in a queue. Traverse the queue as long as it is not empty. For each node in the queue, mark it as visited, and iterate through all the nodes that are connected to it (using the adjacency list), and decrement the degree of each of those nodes by one in the map. Add all nodes whose degree becomes equal to one to the queue. At the end of this algorithm, all the nodes that are unvisited are part of the cycle.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Graph class class Graph
{ public :
// No. of vertices of graph
int v;
// Adjacency List
vector< int > *l;
Graph( int v)
{
this ->v = v;
this ->l = new vector< int >[v];
}
void addedge( int i, int j)
{
l[i].push_back(j);
l[j].push_back(i);
}
}; // Function to find a cycle in the given graph if exists void findCycle( int n, int r, Graph g)
{ // HashMap to store the degree of each node
unordered_map< int , int > degree;
for ( int i = 0; i < g.v; i++)
degree[i] = g.l[i].size();
// Array to track visited nodes
int visited[g.v] = {0};
// Queue to store the nodes of degree 1
queue< int > q;
// Continuously adding those nodes whose
// degree is 1 to the queue
while ( true )
{
// Adding nodes to queue whose degree is 1
// and is not visited
for ( int i = 0; i < degree.size(); i++)
if (degree.at(i) == 1 and !visited[i])
q.push(i);
// If queue becomes empty then get out
// of the continuous loop
if (q.empty())
break ;
while (!q.empty())
{
// Remove the front element from the queue
int temp = q.front();
q.pop();
// Mark the removed element visited
visited[temp] = 1;
// Decrement the degree of all those nodes
// adjacent to removed node
for ( int i = 0; i < g.l[temp].size(); i++)
{
int value = degree[g.l[temp][i]];
degree[g.l[temp][i]] = --value;
}
}
}
int flag = 0;
// Checking all the nodes which are not visited
// i.e. they are part of the cycle
for ( int i = 0; i < g.v; i++)
if (visited[i] == 0)
flag = 1;
if (flag == 0)
cout << "-1" ;
else
{
for ( int i = 0; i < g.v; i++)
if (visited[i] == 0)
cout << i << " " ;
}
} // Driver Code int main()
{ // No of nodes
int n = 5;
// No of edges
int e = 5;
Graph g(n);
g.addedge(0, 1);
g.addedge(0, 2);
g.addedge(0, 3);
g.addedge(1, 2);
g.addedge(3, 4);
findCycle(n, e, g);
return 0;
} // This code is contributed by // sanjeev2552 |
// Java implementation of the approach import java.util.*;
// Graph class class Graph {
// No. of vertices of graph
int v;
// Adjacency List
@SuppressWarnings ( "unchecked" )
ArrayList<ArrayList<Integer>> l;
Graph( int v)
{
this .v = v;
this .l = new ArrayList<>();
for ( int i = 0 ; i < v; i++) {
l.add( new ArrayList<>());
}
}
void addedge( int i, int j)
{
l.get(i).add(j);
l.get(j).add(i);
}
} class GFG {
// Function to find a cycle in the given graph if exists
static void findCycle( int n, int e, Graph g)
{
// HashMap to store the degree of each node
HashMap<Integer, Integer> degree = new HashMap<>();
for ( int i = 0 ; i < n; i++)
degree.put(i, g.l.get(i).size());
// Array to track visited nodes
int visited[] = new int [g.v];
// Initially all nodes are not visited
for ( int i = 0 ; i < visited.length; i++)
visited[i] = 0 ;
// Queue to store the nodes of degree 1
Queue<Integer> q = new LinkedList<>();
// Continuously adding those nodes whose
// degree is 1 to the queue
while ( true ) {
// Adding nodes to queue whose degree is 1
// and is not visited
for ( int i = 0 ; i < degree.size(); i++){
if (( int )degree.get(i) == 1 && visited[i] == 0 )
q.add(i);
}
// If queue becomes empty then get out
// of the continuous loop
if (q.isEmpty())
break ;
while (!q.isEmpty()) {
// Remove the front element from the queue
int temp = ( int )q.poll();
// Mark the removed element visited
visited[temp] = 1 ;
// Decrement the degree of all those nodes
// adjacent to removed node
for ( int i = 0 ; i < g.l.get(temp).size(); i++) {
int value = ( int )degree.get(( int )g.l.get(temp).get(i));
degree.replace(g.l.get(temp).get(i), --value);
}
}
}
int flag = 0 ;
// Checking all the nodes which are not visited
// i.e. they are part of the cycle
for ( int i = 0 ; i < visited.length; i++)
if (visited[i] == 0 )
flag = 1 ;
if (flag == 0 )
System.out.print( "-1" );
else {
for ( int i = 0 ; i < visited.length; i++)
if (visited[i] == 0 )
System.out.print(i + " " );
}
}
// Driver code
public static void main(String[] args)
{
// No of nodes
int n = 5 ;
// No of edges
int e = 5 ;
Graph g = new Graph(n);
g.addedge( 0 , 1 );
g.addedge( 0 , 2 );
g.addedge( 0 , 3 );
g.addedge( 1 , 2 );
g.addedge( 3 , 4 );
findCycle(n, e, g);
}
} // This Code has been contributed by Mukul Sharma |
# Python3 implementation of the approach # Graph class class Graph:
def __init__( self , v):
# No. of vertices of graph
self .v = v
# Adjacency List
self .l = [ 0 ] * v
for i in range ( self .v):
self .l[i] = []
def addedge( self , i: int , j: int ):
self .l[i].append(j)
self .l[j].append(i)
# Function to find a cycle in the given graph if exists def findCycle(n: int , e: int , g: Graph) - > None :
# HashMap to store the degree of each node
degree = dict ()
for i in range ( len (g.l)):
degree[i] = len (g.l[i])
# Array to track visited nodes
visited = [ 0 ] * g.v
# Initially all nodes are not visited
for i in range ( len (visited)):
visited[i] = 0
# Queue to store the nodes of degree 1
q = list ()
# Continuously adding those nodes whose
# degree is 1 to the queue
while True :
# Adding nodes to queue whose degree is 1
# and is not visited
for i in range ( len (degree)):
if degree[i] = = 1 and visited[i] = = 0 :
q.append(i)
# If queue becomes empty then get out
# of the continuous loop
if len (q) = = 0 :
break
while q:
# Remove the front element from the queue
temp = q.pop()
# Mark the removed element visited
visited[temp] = 1
# Decrement the degree of all those nodes
# adjacent to removed node
for i in range ( len (g.l[temp])):
value = degree[g.l[temp][i]]
degree[g.l[temp][i]] = value - 1
flag = 0
# Checking all the nodes which are not visited
# i.e. they are part of the cycle
for i in range ( len (visited)):
if visited[i] = = 0 :
flag = 1
if flag = = 0 :
print ( "-1" )
else :
for i in range ( len (visited)):
if visited[i] = = 0 :
print (i, end = " " )
# Driver Code if __name__ = = "__main__" :
# No of nodes
n = 5
# No of edges
e = 5
g = Graph(n)
g.addedge( 0 , 1 )
g.addedge( 0 , 2 )
g.addedge( 0 , 3 )
g.addedge( 1 , 2 )
g.addedge( 3 , 4 )
findCycle(n, e, g)
# This code is contributed by # sanjeev2552 |
// C# implementation of the approach using System;
using System.Collections.Generic;
// Graph class public class Graph
{ // No. of vertices of graph
public int v;
// Adjacency List
public List< int > []l;
public Graph( int v)
{
this .v = v;
this .l = new List< int >[v];
for ( int i = 0; i < v; i++)
{
l[i] = new List< int >();
}
}
public void addedge( int i, int j)
{
l[i].Add(j);
l[j].Add(i);
}
} class GFG{
// Function to find a cycle in the // given graph if exists static void findCycle( int n, int e, Graph g)
{ // Dictionary to store the degree of each node
Dictionary< int ,
int > degree = new Dictionary< int ,
int >();
for ( int i = 0; i < g.l.Length; i++)
degree.Add(i, g.l[i].Count);
// Array to track visited nodes
int []visited = new int [g.v];
// Initially all nodes are not visited
for ( int i = 0; i < visited.Length; i++)
visited[i] = 0;
// Queue to store the nodes of degree 1
List< int > q = new List< int >();
// Continuously adding those nodes whose
// degree is 1 to the queue
while ( true )
{
// Adding nodes to queue whose degree is 1
// and is not visited
for ( int i = 0; i < degree.Count; i++)
if (( int )degree[i] == 1 && visited[i] == 0)
q.Add(i);
// If queue becomes empty then get out
// of the continuous loop
if (q.Count!=0)
break ;
while (q.Count != 0)
{
// Remove the front element from the queue
int temp = q[0];
q.RemoveAt(0);
// Mark the removed element visited
visited[temp] = 1;
// Decrement the degree of all those nodes
// adjacent to removed node
for ( int i = 0; i < g.l[temp].Count; i++)
{
int value = ( int )degree[( int )g.l[temp][i]];
degree[g.l[temp][i]] = value -= 1;
}
}
}
int flag = 0;
// Checking all the nodes which are not visited
// i.e. they are part of the cycle
for ( int i = 0; i < visited.Length; i++)
if (visited[i] == 0)
flag = 1;
if (flag == 0)
Console.Write( "-1" );
else
{
for ( int i = 0; i < visited.Length-2; i++)
if (visited[i] == 0)
Console.Write(i + " " );
}
} // Driver code public static void Main(String[] args)
{ // No of nodes
int n = 5;
// No of edges
int e = 5;
Graph g = new Graph(n);
g.addedge(0, 1);
g.addedge(0, 2);
g.addedge(0, 3);
g.addedge(1, 2);
g.addedge(3, 4);
findCycle(n, e, g);
} } // This code is contributed by Princi Singh |
<script> // Javascript implementation of the approach // Graph class class Graph { constructor(v)
{
// No. of vertices of graph
this .v = v;
// Adjacency List
this .l = Array.from(
Array( this .v), () => Array());
}
addedge(i, j)
{
this .l[i].push(j);
this .l[j].push(i);
}
} // Function to find a cycle in the // given graph if exists function findCycle(n, e, g)
{ // Dictionary to store the degree of each node
var degree = new Map();
for ( var i = 0; i < g.l.length; i++)
degree.set(i, g.l[i].length);
// Array to track visited nodes
var visited = Array(g.v).fill(0);
// Initially all nodes are not visited
for ( var i = 0; i < visited.length; i++)
visited[i] = 0;
// Queue to store the nodes of degree 1
var q = [];
// Continuously adding those nodes whose
// degree is 1 to the queue
while ( true )
{
// Adding nodes to queue whose degree is 1
// and is not visited
for ( var i = 0; i < degree.size; i++)
if (degree.has(i) && visited[i] == 0)
q.push(i);
// If queue becomes empty then get out
// of the continuous loop
if (q.length != 0)
break ;
while (q.length != 0)
{
// Remove the front element from the queue
var temp = q[0];
q.shift();
// Mark the removed element visited
visited[temp] = 1;
// Decrement the degree of all those nodes
// adjacent to removed node
for ( var i = 0; i < g.l[temp].length; i++)
{
var value = degree.get(g.l[temp][i]);
degree.set(g.l[temp][i], value - 1);
}
}
}
var flag = 0;
// Checking all the nodes which are not visited
// i.e. they are part of the cycle
for ( var i = 0; i < visited.length; i++)
if (visited[i] == 0)
flag = 1;
if (flag == 0)
document.write( "-1" );
else
{
for ( var i = 0; i < visited.length - 2; i++)
if (visited[i] == 0)
document.write(i + " " );
}
} // Driver code // No of nodes var n = 5;
// No of edges var e = 5;
var g = new Graph(n);
g.addedge(0, 1); g.addedge(0, 2); g.addedge(0, 3); g.addedge(1, 2); g.addedge(3, 4); findCycle(n, e, g); // This code is contributed by noob2000 </script> |
0 1 2
Time complexity : O(n + e) where n is the number of vertices and e is the number of edges in the graph
space complexity : O(n) as the most space-consuming data structure is the adjacency list