# 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.

- Increment count of visited nodes by 1.
- Decrease in-degree by 1 for all its neighboring nodes.
- 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 indegree[dest]++

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 indegree[dest]++;

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; ` ` ` `public` `: ` ` ` `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) ` `{ ` ` ` `adj[u].push_back(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]) ` ` ` `in_degree[v]++; ` ` ` `} ` ` ` ` ` `// 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) ` ` ` `q.push(i); ` ` ` ` ` `// 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(); ` ` ` `q.pop(); ` ` ` `top_order.push_back(u); ` ` ` ` ` `// 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) ` ` ` `q.push(*itr); ` ` ` ` ` `cnt++; ` ` ` `} ` ` ` ` ` `// Check if there was a cycle ` ` ` `if` `(cnt != V) ` ` ` `return` `true` `; ` ` ` `else` ` ` `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"` `; ` ` ` `else` ` ` `cout << ` `"No"` `; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

Yes

Time Complexity : O(V+E)

## Recommended Posts:

- Detect Cycle in a Directed Graph
- Detect Cycle in a directed graph using colors
- Detect cycle in an undirected graph using BFS
- Detect cycle in an undirected graph
- Detect a negative cycle in a Graph | (Bellman Ford)
- Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph)
- Hierholzer's Algorithm for directed graph
- Euler Circuit in a Directed Graph
- Clone a Directed Acyclic Graph
- Longest Path in a Directed Acyclic Graph | Set 2
- Find if there is a path between two vertices in a directed graph
- Shortest Path in Directed Acyclic Graph
- All Topological Sorts of a Directed Acyclic Graph
- Longest Path in a Directed Acyclic Graph
- Determine whether a universal sink exists in a directed graph

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