Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

C Program for Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph)

  • Last Updated : 22 Dec, 2021

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.

Union: Join two subsets into a single subset.

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.


// 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);
       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
        |  \
        |    \
        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" );
        printf( "graph doesn\'t contain cycle" );
    return 0;

Please refer complete article on Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph) for more details!

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!