A *disjoint-set data structure* is a data structure that keeps track of a set of elements partitioned into a number of disjoint (non-overlapping) subsets. A *union-find algorithm* is an algorithm that performs two useful operations on such a data structure:* Find:* Determine which subset a particular element is in. This can be used for determining if two elements are in the same subset.

*Join two subsets into a single subset.*

**Union:**In this post, we will discuss the application of Disjoint Set Data Structure. The application is to check whether a given graph contains a cycle or not.

*Union-Find Algorithm*can be used to check whether an undirected graph contains cycle or not. Note that we have discussed an algorithm to detect cycle. This is another method based on

*Union-Find*. This method assumes that the graph doesn’t contain any self-loops.

We can keep track of the subsets in a 1D array, let’s call it parent[].

Let us consider the following graph:

For each edge, make subsets using both the vertices of the edge. If both the vertices are in the same subset, a cycle is found.

Initially, all slots of parent array are initialized to -1 (means there is only one item in every subset).

0 1 2 -1 -1 -1

Now process all edges one by one.*Edge 0-1:* Find the subsets in which vertices 0 and 1 are. Since they are in different subsets, we take the union of them. For taking the union, either make node 0 as parent of node 1 or vice-versa.

0 1 2 <----- 1 is made parent of 0 (1 is now representative of subset {0, 1}) 1 -1 -1

*Edge 1-2:* 1 is in subset 1 and 2 is in subset 2. So, take union.

0 1 2 <----- 2 is made parent of 1 (2 is now representative of subset {0, 1, 2}) 1 2 -1

*Edge 0-2:* 0 is in subset 2 and 2 is also in subset 2. Hence, including this edge forms a cycle.

How subset of 0 is same as 2?

0->1->2 // 1 is parent of 0 and 2 is parent of 1

Based on the above explanation, below are implementations:

## C++

`// A union-find algorithm to detect cycle in a graph` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// a structure to represent an edge in graph` `class` `Edge ` `{` `public` `:` ` ` `int` `src, dest;` `};` `// a structure to represent a graph` `class` `Graph ` `{` `public` `:` ` ` `// V-> Number of vertices, E-> Number of edges` ` ` `int` `V, E;` ` ` `// graph is represented as an array of edges` ` ` `Edge* edge;` `};` `// Creates a graph with V vertices and E edges` `Graph* createGraph(` `int` `V, ` `int` `E)` `{` ` ` `Graph* graph = ` `new` `Graph();` ` ` `graph->V = V;` ` ` `graph->E = E;` ` ` `graph->edge = ` `new` `Edge[graph->E * ` `sizeof` `(Edge)];` ` ` `return` `graph;` `}` `// A utility function to find the subset of an element i` `int` `find(` `int` `parent[], ` `int` `i)` `{` ` ` `if` `(parent[i] == -1)` ` ` `return` `i;` ` ` `return` `find(parent, parent[i]);` `}` `// A utility function to do union of two subsets` `void` `Union(` `int` `parent[], ` `int` `x, ` `int` `y)` `{` ` ` `parent[x] = y;` `}` `// The main function to check whether a given graph contains` `// cycle or not` `int` `isCycle(Graph* graph)` `{` ` ` `// Allocate memory for creating V subsets` ` ` `int` `* parent = ` `new` `int` `[graph->V * ` `sizeof` `(` `int` `)];` ` ` `// Initialize all subsets as single element sets` ` ` `memset` `(parent, -1, ` `sizeof` `(` `int` `) * graph->V);` ` ` `// Iterate through all edges of graph, find subset of` ` ` `// both vertices of every edge, if both subsets are` ` ` `// same, then there is cycle in graph.` ` ` `for` `(` `int` `i = 0; i < graph->E; ++i) {` ` ` `int` `x = find(parent, graph->edge[i].src);` ` ` `int` `y = find(parent, graph->edge[i].dest);` ` ` `if` `(x == y)` ` ` `return` `1;` ` ` `Union(parent, x, y);` ` ` `}` ` ` `return` `0;` `}` `// Driver code` `int` `main()` `{` ` ` `/* Let us create the following graph` ` ` `0` ` ` `| \` ` ` `| \` ` ` `1---2 */` ` ` `int` `V = 3, E = 3;` ` ` `Graph* graph = createGraph(V, E);` ` ` `// add edge 0-1` ` ` `graph->edge[0].src = 0;` ` ` `graph->edge[0].dest = 1;` ` ` `// add edge 1-2` ` ` `graph->edge[1].src = 1;` ` ` `graph->edge[1].dest = 2;` ` ` `// add edge 0-2` ` ` `graph->edge[2].src = 0;` ` ` `graph->edge[2].dest = 2;` ` ` `if` `(isCycle(graph))` ` ` `cout << ` `"graph contains cycle"` `;` ` ` `else` ` ` `cout << ` `"graph doesn't contain cycle"` `;` ` ` `return` `0;` `}` `// This code is contributed by rathbhupendra` |

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## C

`// A union-find algorithm to detect cycle in a graph` `#include <stdio.h>` `#include <stdlib.h>` `#include <string.h>` `// a structure to represent an edge in graph` `struct` `Edge` `{` ` ` `int` `src, dest;` `};` `// a structure to represent a graph` `struct` `Graph` `{` ` ` `// V-> Number of vertices, E-> Number of edges` ` ` `int` `V, E;` ` ` `// graph is represented as an array of edges` ` ` `struct` `Edge* edge;` `};` `// Creates a graph with V vertices and E edges` `struct` `Graph* createGraph(` `int` `V, ` `int` `E)` `{` ` ` `struct` `Graph* graph = ` ` ` `(` `struct` `Graph*) ` `malloc` `( ` `sizeof` `(` `struct` `Graph) );` ` ` `graph->V = V;` ` ` `graph->E = E;` ` ` `graph->edge = ` ` ` `(` `struct` `Edge*) ` `malloc` `( graph->E * ` `sizeof` `( ` `struct` `Edge ) );` ` ` `return` `graph;` `}` `// A utility function to find the subset of an element i` `int` `find(` `int` `parent[], ` `int` `i)` `{` ` ` `if` `(parent[i] == -1)` ` ` `return` `i;` ` ` `return` `find(parent, parent[i]);` `}` `// A utility function to do union of two subsets ` `void` `Union(` `int` `parent[], ` `int` `x, ` `int` `y)` `{` ` ` `int` `xset = find(parent, x);` ` ` `int` `yset = find(parent, y);` ` ` `if` `(xset!=yset){` ` ` `parent[xset] = yset;` ` ` `}` `}` `// The main function to check whether a given graph contains ` `// cycle or not` `int` `isCycle( ` `struct` `Graph* graph )` `{` ` ` `// Allocate memory for creating V subsets` ` ` `int` `*parent = (` `int` `*) ` `malloc` `( graph->V * ` `sizeof` `(` `int` `) );` ` ` `// Initialize all subsets as single element sets` ` ` `memset` `(parent, -1, ` `sizeof` `(` `int` `) * graph->V);` ` ` `// Iterate through all edges of graph, find subset of both` ` ` `// vertices of every edge, if both subsets are same, then ` ` ` `// there is cycle in graph.` ` ` `for` `(` `int` `i = 0; i < graph->E; ++i)` ` ` `{` ` ` `int` `x = find(parent, graph->edge[i].src);` ` ` `int` `y = find(parent, graph->edge[i].dest);` ` ` `if` `(x == y)` ` ` `return` `1;` ` ` `Union(parent, x, y);` ` ` `}` ` ` `return` `0;` `}` `// Driver program to test above functions` `int` `main()` `{` ` ` `/* Let us create the following graph ` ` ` `0 ` ` ` `| \ ` ` ` `| \ ` ` ` `1---2 */` ` ` `int` `V = 3, E = 3;` ` ` `struct` `Graph* graph = createGraph(V, E);` ` ` `// add edge 0-1` ` ` `graph->edge[0].src = 0;` ` ` `graph->edge[0].dest = 1;` ` ` `// add edge 1-2` ` ` `graph->edge[1].src = 1;` ` ` `graph->edge[1].dest = 2;` ` ` `// add edge 0-2` ` ` `graph->edge[2].src = 0;` ` ` `graph->edge[2].dest = 2;` ` ` `if` `(isCycle(graph))` ` ` `printf` `( ` `"graph contains cycle"` `);` ` ` `else` ` ` `printf` `( ` `"graph doesn't contain cycle"` `);` ` ` `return` `0;` `}` |

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## Java

`// Java Program for union-find algorithm to detect cycle in a graph` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;` `class` `Graph` `{` ` ` `int` `V, E; ` `// V-> no. of vertices & E->no.of edges` ` ` `Edge edge[]; ` `// /collection of all edges` ` ` `class` `Edge` ` ` `{` ` ` `int` `src, dest;` ` ` `};` ` ` `// Creates a graph with V vertices and E edges` ` ` `Graph(` `int` `v,` `int` `e)` ` ` `{` ` ` `V = v;` ` ` `E = e;` ` ` `edge = ` `new` `Edge[E];` ` ` `for` `(` `int` `i=` `0` `; i<e; ++i)` ` ` `edge[i] = ` `new` `Edge();` ` ` `}` ` ` `// A utility function to find the subset of an element i` ` ` `int` `find(` `int` `parent[], ` `int` `i)` ` ` `{` ` ` `if` `(parent[i] == -` `1` `)` ` ` `return` `i;` ` ` `return` `find(parent, parent[i]);` ` ` `}` ` ` `// A utility function to do union of two subsets` ` ` `void` `Union(` `int` `parent[], ` `int` `x, ` `int` `y)` ` ` `{` ` ` `int` `xset = find(parent, x);` ` ` `int` `yset = find(parent, y);` ` ` `parent[xset] = yset;` ` ` `}` ` ` `// The main function to check whether a given graph` ` ` `// contains cycle or not` ` ` `int` `isCycle( Graph graph)` ` ` `{` ` ` `// Allocate memory for creating V subsets` ` ` `int` `parent[] = ` `new` `int` `[graph.V];` ` ` `// Initialize all subsets as single element sets` ` ` `for` `(` `int` `i=` `0` `; i<graph.V; ++i)` ` ` `parent[i]=-` `1` `;` ` ` `// Iterate through all edges of graph, find subset of both` ` ` `// vertices of every edge, if both subsets are same, then` ` ` `// there is cycle in graph.` ` ` `for` `(` `int` `i = ` `0` `; i < graph.E; ++i)` ` ` `{` ` ` `int` `x = graph.find(parent, graph.edge[i].src);` ` ` `int` `y = graph.find(parent, graph.edge[i].dest);` ` ` `if` `(x == y)` ` ` `return` `1` `;` ` ` `graph.Union(parent, x, y);` ` ` `}` ` ` `return` `0` `;` ` ` `}` ` ` `// Driver Method` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `/* Let us create the following graph ` ` ` `0 ` ` ` `| \ ` ` ` `| \ ` ` ` `1---2 */` ` ` `int` `V = ` `3` `, E = ` `3` `;` ` ` `Graph graph = ` `new` `Graph(V, E);` ` ` `// add edge 0-1` ` ` `graph.edge[` `0` `].src = ` `0` `;` ` ` `graph.edge[` `0` `].dest = ` `1` `;` ` ` `// add edge 1-2` ` ` `graph.edge[` `1` `].src = ` `1` `;` ` ` `graph.edge[` `1` `].dest = ` `2` `;` ` ` `// add edge 0-2` ` ` `graph.edge[` `2` `].src = ` `0` `;` ` ` `graph.edge[` `2` `].dest = ` `2` `;` ` ` `if` `(graph.isCycle(graph)==` `1` `)` ` ` `System.out.println( ` `"graph contains cycle"` `);` ` ` `else` ` ` `System.out.println( ` `"graph doesn't contain cycle"` `);` ` ` `}` `}` |

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## Python

`# Python Program for union-find algorithm to detect cycle in a undirected graph` `# we have one egde for any two vertex i.e 1-2 is either 1-2 or 2-1 but not both` ` ` `from` `collections ` `import` `defaultdict` ` ` `#This class represents a undirected graph using adjacency list representation` `class` `Graph:` ` ` ` ` `def` `__init__(` `self` `,vertices):` ` ` `self` `.V` `=` `vertices ` `#No. of vertices` ` ` `self` `.graph ` `=` `defaultdict(` `list` `) ` `# default dictionary to store graph` ` ` ` ` `# function to add an edge to graph` ` ` `def` `addEdge(` `self` `,u,v):` ` ` `self` `.graph[u].append(v)` ` ` ` ` `# A utility function to find the subset of an element i` ` ` `def` `find_parent(` `self` `, parent,i):` ` ` `if` `parent[i] ` `=` `=` `-` `1` `:` ` ` `return` `i` ` ` `if` `parent[i]!` `=` `-` `1` `:` ` ` `return` `self` `.find_parent(parent,parent[i])` ` ` `# A utility function to do union of two subsets` ` ` `def` `union(` `self` `,parent,x,y):` ` ` `x_set ` `=` `self` `.find_parent(parent, x)` ` ` `y_set ` `=` `self` `.find_parent(parent, y)` ` ` `parent[x_set] ` `=` `y_set` ` ` ` ` ` ` `# The main function to check whether a given graph` ` ` `# contains cycle or not` ` ` `def` `isCyclic(` `self` `):` ` ` ` ` `# Allocate memory for creating V subsets and` ` ` `# Initialize all subsets as single element sets` ` ` `parent ` `=` `[` `-` `1` `]` `*` `(` `self` `.V)` ` ` `# Iterate through all edges of graph, find subset of both` ` ` `# vertices of every edge, if both subsets are same, then` ` ` `# there is cycle in graph.` ` ` `for` `i ` `in` `self` `.graph:` ` ` `for` `j ` `in` `self` `.graph[i]:` ` ` `x ` `=` `self` `.find_parent(parent, i) ` ` ` `y ` `=` `self` `.find_parent(parent, j)` ` ` `if` `x ` `=` `=` `y:` ` ` `return` `True` ` ` `self` `.union(parent,x,y)` `# Create a graph given in the above diagram` `g ` `=` `Graph(` `3` `)` `g.addEdge(` `0` `, ` `1` `)` `g.addEdge(` `1` `, ` `2` `)` `g.addEdge(` `2` `, ` `0` `)` `if` `g.isCyclic():` ` ` `print` `"Graph contains cycle"` `else` `:` ` ` `print` `"Graph does not contain cycle "` ` ` `#This code is contributed by Neelam Yadav` |

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## C#

`// C# Program for union-find ` `// algorithm to detect cycle ` `// in a graph` `using` `System;` `class` `Graph{` `// V-> no. of vertices & ` `// E->no.of edges ` `public` `int` `V, E; ` `// collection of all edges` `public` `Edge []edge; ` `class` `Edge` `{` ` ` `public` `int` `src, dest;` `};` `// Creates a graph with V ` `// vertices and E edges` `public` `Graph(` `int` `v,` `int` `e)` `{` ` ` `V = v;` ` ` `E = e;` ` ` `edge = ` `new` `Edge[E];` ` ` ` ` `for` `(` `int` `i = 0; i < e; ++i)` ` ` `edge[i] = ` `new` `Edge();` `}` `// A utility function to find ` `// the subset of an element i` `int` `find(` `int` `[]parent, ` `int` `i)` `{` ` ` `if` `(parent[i] == -1)` ` ` `return` `i;` ` ` `return` `find(parent, ` ` ` `parent[i]);` `}` `// A utility function to do ` `// union of two subsets` `void` `Union(` `int` `[]parent, ` ` ` `int` `x, ` `int` `y)` `{` ` ` `int` `xset = find(parent, x);` ` ` `int` `yset = find(parent, y);` ` ` `parent[xset] = yset;` `}` `// The main function to check ` `// whether a given graph` `// contains cycle or not` `int` `isCycle(Graph graph)` `{` ` ` `// Allocate memory for ` ` ` `// creating V subsets` ` ` `int` `[]parent = ` ` ` `new` `int` `[graph.V];` ` ` `// Initialize all subsets as ` ` ` `// single element sets` ` ` `for` `(` `int` `i = 0; i < graph.V; ++i)` ` ` `parent[i] =- 1;` ` ` `// Iterate through all edges of graph, ` ` ` `// find subset of both vertices of every ` ` ` `// edge, if both subsets are same, then ` ` ` `// there is cycle in graph.` ` ` `for` `(` `int` `i = 0; i < graph.E; ++i)` ` ` `{` ` ` `int` `x = graph.find(parent, ` ` ` `graph.edge[i].src);` ` ` `int` `y = graph.find(parent, ` ` ` `graph.edge[i].dest);` ` ` `if` `(x == y)` ` ` `return` `1;` ` ` `graph.Union(parent, x, y);` ` ` `}` ` ` `return` `0;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `/* Let us create the following graph ` ` ` `0 ` ` ` `| \ ` ` ` `| \ ` ` ` `1---2 */` ` ` `int` `V = 3, E = 3;` ` ` `Graph graph = ` `new` `Graph(V, E);` ` ` `// add edge 0-1` ` ` `graph.edge[0].src = 0;` ` ` `graph.edge[0].dest = 1;` ` ` `// add edge 1-2` ` ` `graph.edge[1].src = 1;` ` ` `graph.edge[1].dest = 2;` ` ` `// add edge 0-2` ` ` `graph.edge[2].src = 0;` ` ` `graph.edge[2].dest = 2;` ` ` `if` `(graph.isCycle(graph) == 1)` ` ` `Console.WriteLine(` `"graph contains cycle"` `);` ` ` `else` ` ` `Console.WriteLine(` `"graph doesn't contain cycle"` `);` `}` `}` `// This code is contributed by Princi Singh` |

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**Output: **

graph contains cycle

Note that the implementation of *union()* and *find()* is naive and takes O(n) time in the worst case. These methods can be improved to O(Logn) using *Union by Rank or Height*. We will soon be discussing *Union by Rank* in a separate post.

https://youtu.be/mHz-mx-8lJ8?list=PLqM7alHXFySEaZgcg7uRYJFBnYMLti-nh **Related Articles : **

Union-Find Algorithm | Set 2 (Union By Rank and Path Compression)

Disjoint Set Data Structures (Java Implementation)

Greedy Algorithms | Set 2 (Kruskal’s Minimum Spanning Tree Algorithm)

Job Sequencing Problem | Set 2 (Using Disjoint Set)

This article is compiled by Aashish Barnwal and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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