Detect Cycle in a Directed Graph using BFS

Given a directed graph, check whether the graph contains a cycle or not. Your function should return true if the given graph contains at least one cycle, else return false. For example, the following graph contains two cycles 0->1->2->3->0 and 2->4->2, so your function must return true.

We have discussed a DFS based solution to detect cycle in a directed graph. In this post, BFS based solution is discussed.

The idea is to simply use Kahn’s algorithm for Topological Sorting

Steps involved in detecting cycle in a directed graph using BFS.

Step-1: Compute in-degree (number of incoming edges) for each of the vertex present in the graph and initialize the count of visited nodes as 0.

Step-2: Pick all the vertices with in-degree as 0 and add them into a queue (Enqueue operation)

Step-3: Remove a vertex from the queue (Dequeue operation) and then.

  1. Increment count of visited nodes by 1.
  2. Decrease in-degree by 1 for all its neighboring nodes.
  3. If in-degree of a neighboring nodes is reduced to zero, then add it to the queue.

Step 4: Repeat Step 3 until the queue is empty.

Step 5: If count of visited nodes is not equal to the number of nodes in the graph has cycle, otherwise not.

How to find in-degree of each node?
There are 2 ways to calculate in-degree of every vertex:
Take an in-degree array which will keep track of
1) Traverse the array of edges and simply increase the counter of the destination node by 1.

for each node in Nodes
    indegree[node] = 0;
for each edge(src,dest) in Edges

Time Complexity: O(V+E)

2) Traverse the list for every node and then increment the in-degree of all the nodes connected to it by 1.

    for each node in Nodes
        If (list[node].size()!=0) then
        for each dest in list

Time Complexity: The outer for loop will be executed V number of times and the inner for loop will be executed E number of times, Thus overall time complexity is O(V+E).

The overall time complexity of the algorithm is O(V+E)





// A C++ program to check if there is a cycle in 
// directed graph using BFS.
#include <bits/stdc++.h>
using namespace std;
// Class to represent a graph
class Graph {
    int V; // No. of vertices'
    // Pointer to an array containing adjacency lisr
    list<int>* adj;
    Graph(int V); // Constructor
    // function to add an edge to graph
    void addEdge(int u, int v);
    // Returns true if there is a cycle in the graph
    // else false.
    bool isCycle();
Graph::Graph(int V)
    this->V = V;
    adj = new list<int>[V];
void Graph::addEdge(int u, int v)
// This function returns true if there is a cycle
// in directed graph, else returns false.
bool Graph::isCycle()
    // Create a vector to store indegrees of all
    // vertices. Initialize all indegrees as 0.
    vector<int> in_degree(V, 0);
    // Traverse adjacency lists to fill indegrees of
    // vertices. This step takes O(V+E) time
    for (int u = 0; u < V; u++) {
        for (auto v : adj[u])
    // Create an queue and enqueue all vertices with
    // indegree 0
    queue<int> q;
    for (int i = 0; i < V; i++)
        if (in_degree[i] == 0)
    // Initialize count of visited vertices
    int cnt = 0;
    // Create a vector to store result (A topological
    // ordering of the vertices)
    vector<int> top_order;
    // One by one dequeue vertices from queue and enqueue
    // adjacents if indegree of adjacent becomes 0
    while (!q.empty()) {
        // Extract front of queue (or perform dequeue)
        // and add it to topological order
        int u = q.front();
        // Iterate through all its neighbouring nodes
        // of dequeued node u and decrease their in-degree
        // by 1
        list<int>::iterator itr;
        for (itr = adj[u].begin(); itr != adj[u].end(); itr++)
            // If in-degree becomes zero, add it to queue
            if (--in_degree[*itr] == 0)
    // Check if there was a cycle
    if (cnt != V) 
        return true;
        return false;
// Driver program to test above functions
int main()
    // Create a graph given in the above diagram
    Graph g(6);
    g.addEdge(0, 1);
    g.addEdge(1, 2);
    g.addEdge(2, 0);
    g.addEdge(3, 4);
    g.addEdge(4, 5);
    if (g.isCycle())
        cout << "Yes";
        cout << "No";
    return 0;




Time Complexity : O(V+E)

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