Introduction to Disjoint Set Data Structure or Union-Find Algorithm

A disjoint-set data structure is defined as 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. Here first we have to check if the two subsets belong to same set. If no, then we cannot perform union. 

UNION and FIND Operations for Disjoint Sets

A relation over a set of elements a1, a2,…a n  can be divided into equivalent classes. The equivalent class of an element a is the subset of S that contains all the elements of S that are related to a.

Divide a set of elements into equivalent classes through the two operations

1. UNION

2. FIND

A set is divided into subsets. Each subset contains related elements. If we come to know that the two element ai and aj are related, then we can do the followings:

1. Find the subset : Si containing ai

2.Find the subset : Sj containing aj

3. If Si and Sj are two independent subsets

then we create a new subset by taking union of Si and Sj

New subset = Si ∪ Sj

This algorithm is dynamic as during the course of the algorithm, the sets can change via the union operation.

Example:

Let us check an example to understand how the data structure is applied. For this consider the following problem statement

Problem: Given an undirected graph, the task is to check if the graph contains a cycle or not.

Examples:

Input: The following is the graph

Output: Yes
Explanation: There is a cycle of vertices {0, 1, 2}.

We already have discussed an algorithm to detect cycle in directed graph. Here Union-Find Algorithm can be used to check whether an undirected graph contains cycle or not. The idea is that, 

Initially create subsets containing only a single node which are the parent of itself. Now while traversing through the edges, if the two end nodes of the edge belongs to the same set then they form a cycle. Otherwise, perform union to merge the subsets together.

Note: This method assumes that the graph doesn’t contain any self-loops.

Illustration:

Follow the below illustration for a better understanding

Let us consider the following graph: 

Use an array to keep track of the subsets and which nodes belong to that subset. Let the array be parent[].

Initially, all slots of parent array are initialized to hold the same values as the node.

parent[] = {0, 1, 2}. Also when the value of the node and its parent are same, that is the root of that subset of nodes.

Now process all edges one by one.
Edge 0-1:
        => Find the subsets in which vertices 0 and 1 are. 
        => 0 and 1 belongs to subset 0 and 1.
        => Since they are in different subsets, take the union of them. 
        => For taking the union, either make node 0 as parent of node 1 or vice-versa. 
        => 1 is made parent of 0 (1 is now representative of subset {0, 1})
        => parent[] = {1, 1, 2}

Edge 1-2:
        => 1 is in subset 1 and 2 is in subset 2.
        => Since they are in different subsets, take union.
        => Make 2 as parent of 1. (2 is now representative of subset {0, 1, 2})
        => parent[] = {1, 2, 2}

Edge 0-2:
        => 0 is in subset 2 and 2 is also in subset 2. 
        => Because 1 is parent of 0 and 2 is parent of 1. So 0 also belongs to subset 2
        => Hence, including this edge forms a cycle. 

Therefore, the above graph contains a cycle.

Follow the below steps to implement the idea:

• Initially create a parent[] array to keep track of the subsets.
• Traverse through all the edges:
  • Check to which subset each of the nodes belong to by finding the parent[] array till the node and the parent are the same.
  • If the two nodes belong to the same subset then they belong to a cycle.
  • Otherwise, perform union operation on those two subsets.
• If no cycle is found, return false.

Below is the implementation of the above approach.

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] == i)
        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];

    // Initialize all subsets as single element sets
    for(int i = 0; i < graph->V; i++) {
        parent[i] = i;
    }

    // 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

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)
{
    parent[y] = x;
}

// 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);

    // Initialize all subsets as single element sets
    memset(parent, -1, sizeof(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 && (x!=-1 && y!=-1))
            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;
}

Java

// Java Program for union-find algorithm to detect cycle in
// a graph
import java.io.*;
import java.lang.*;
import java.util.*;

public 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] == i)
            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];

        // Initialize all subsets as single element sets
        for (int i = 0; i < graph.V; ++i)
            parent[i] = i;

        // 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");
    }
}

Python3

# Python Program for union-find algorithm
# to detect cycle in a undirected graph
# we have one edge 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] == i:
            return i
        if parent[i] != i:
            return self.find_parent(parent, parent[i])

    # A utility function to do union of two subsets
    def union(self, parent, x, y):
        parent[x] = y

    # 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 = [0]*(self.V)
        for i in range(self.V):
            parent[i] = i

        # 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

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;

    public 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] == i)
            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];

        // Initialize all subsets as
        // single element sets
        for (int i = 0; i < graph.V; ++i)
            parent[i] = i;

        // 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

Javascript

<script>

// Javascript program for union-find
// algorithm to detect cycle
// in a graph

// V-> no. of vertices &
// E->no.of edges
var V, E;

// Collection of all edges
var edge;

class Edge
{
    constructor()
    {
        this.src = 0;
        this.dest = 0;
    }
};

// Creates a graph with V
// vertices and E edges
function initialize(v,e)
{
    V = v;
    E = e;
    edge = Array.from(Array(E), () => Array());
}

// A utility function to find
// the subset of an element i
function find(parent, i)
{
    if (parent[i] == i)
        return i;

    return find(parent, parent[i]);
}

// A utility function to do
// union of two subsets
function Union(parent, x, y)
{
    parent[x] = y;
}

// The main function to check
// whether a given graph
// contains cycle or not
function isCycle()
{

    // Allocate memory for
    // creating V subsets
    var parent = Array(V).fill(0);

    // Initialize all subsets as
    // single element sets
    for(var i = 0; i < V; ++i)
        parent[i] = i;

    // 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 (var i = 0; i < E; ++i)
    {
        var x = find(parent,
                     edge[i].src);
        var y = find(parent,
                     edge[i].dest);

        if (x == y)
            return 1;

        Union(parent, x, y);
    }
    return 0;
}

// Driver code
/* Let us create the following graph
      0
      | \
      |  \
      1---2 */
var V = 3, E = 3;
initialize(V, E);

// Add edge 0-1
edge[0].src = 0;
edge[0].dest = 1;

// Add edge 1-2
edge[1].src = 1;
edge[1].dest = 2;

// Add edge 0-2
edge[2].src = 0;
edge[2].dest = 2;

if (isCycle() == 1)
    document.write("Graph contains cycle");
else
    document.write("Graph doesn't contain cycle");

// This code is contributed by rutvik_56

</script>

Output

Graph contains cycle

The time and space complexity of the given code is as follows:

Time Complexity:

Creating the graph takes O(V + E) time, where V is the number of vertices and E is the number of edges.
Finding the subset of an element takes O(log V) time in the worst case, where V is the number of vertices. The worst case occurs when the tree is skewed, and the depth of the tree is V.
Union of two subsets takes O(1) time.
The loop iterating through all edges takes O(E) time.
Therefore, the overall time complexity of the algorithm is O(E log V). However, in practice, it can be much faster than O(E log V) because the worst-case scenario of finding the subset of an element does not happen often.

Space Complexity:

The space complexity of creating the graph is O(E).
The space complexity of creating the parent array is O(V).
The space complexity of the algorithm is O(max(V,E)) because at any point in time, there can be at most max(V,E) subsets.
Therefore, the overall space complexity of the algorithm is O(max(V,E)).