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

Difficulty Level : Medium
Last Updated : 31 Mar, 2021

In the previous post, we introduced union find algorithm and used it to detect cycle in a graph. We used following union() and find() operations for subsets.

C++

// Naive implementation of find
int find(int parent[], int i)
{
    if (parent[i] == -1)
        return i;
    return find(parent, parent[i]);
}
  
// Naive implementation of union()
void Union(int parent[], int x, int y)
{
    int xset = find(parent, x);
    int yset = find(parent, y);
    parent[xset] = yset;
}

Java

// Naive implementation of find
static int find(int parent[], int i)
{
    if (parent[i] == -1)
        return i;
    return find(parent, parent[i]);
}
   
// Naive implementation of union()
static void Union(int parent[], int x, int y)
{
    int xset = find(parent, x);
    int yset = find(parent, y);
    parent[xset] = yset;
}
 
// This code is contributed by divyesh072019

Python3

# Naive implementation of find
def find(parent, i):
     
    if (parent[i] == -1):
        return i
     
    return find(parent, parent[i])
 
# Naive implementation of union()
def Union(parent, x, y):
 
    xset = find(parent, x)
    yset = find(parent, y)
    parent[xset] = yset
 
# This code is contributed by rutvik_56

C#

// Naive implementation of find
static int find(int []parent, int i)
{
    if (parent[i] == -1)
        return i;
         
    return find(parent, parent[i]);
}
   
// Naive implementation of union()
static void Union(int []parent, int x, int y)
{
    int xset = find(parent, x);
    int yset = find(parent, y);
    parent[xset] = yset;
}
 
// This code is contributed by pratham76

Javascript

<script>
 
// Naive implementation of find
function find(parent, i)
{
    if (parent[i] == -1)
        return i;
         
    return find(parent, parent[i]);
}
   
// Naive implementation of union()
function Union(parent, x, y)
{
    let xset = find(parent, x);
    let yset = find(parent, y);
    parent[xset] = yset;
}
 
<script>

The above union() and find() are naive and the worst case time complexity is linear. The trees created to represent subsets can be skewed and can become like a linked list. Following is an example worst case scenario.

Let there be 4 elements 0, 1, 2, 3

Initially, all elements are single element subsets.
0 1 2 3 

Do Union(0, 1)
   1   2   3  
  /
 0

Do Union(1, 2)
     2   3   
    /
   1
 /
0

Do Union(2, 3)
         3    
        /
      2
     /
   1
 /
0

The above operations can be optimized to O(Log n) in worst case. The idea is to always attach smaller depth tree under the root of the deeper tree. This technique is called union by rank. The term rank is preferred instead of height because if path compression technique (we have discussed it below) is used, then rank is not always equal to height. Also, size (in place of height) of trees can also be used as rank. Using size as rank also yields worst case time complexity as O(Logn) (See this for proof)

Let us see the above example with union by rank
Initially, all elements are single element subsets.
0 1 2 3 

Do Union(0, 1)
   1   2   3  
  /
 0

Do Union(1, 2)
   1    3
 /  \
0    2

Do Union(2, 3)
    1    
 /  |  \
0   2   3

The second optimization to naive method is Path Compression. The idea is to flatten the tree when find() is called. When find() is called for an element x, root of the tree is returned. The find() operation traverses up from x to find root. The idea of path compression is to make the found root as parent of x so that we don’t have to traverse all intermediate nodes again. If x is root of a subtree, then path (to root) from all nodes under x also compresses.

Let the subset {0, 1, .. 9} be represented as below and find() is called
for element 3.
              9
         /    |    \  
        4     5      6
     /     \        /  \
    0        3     7    8
            /  \
           1    2  

When find() is called for 3, we traverse up and find 9 as representative
of this subset. With path compression, we also make 3 as the child of 9 so 
that when find() is called next time for 1, 2 or 3, the path to root is reduced.

               9
         /    /  \    \
        4    5    6     3 
     /           /  \   /  \
    0           7    8  1   2

The two techniques complement each other. The time complexity of each operation becomes even smaller than O(Logn). In fact, amortized time complexity effectively becomes small constant.

Following is union by rank and path compression based implementation to find a cycle in a graph.

C++

// A union by rank and path compression based program to
// detect cycle in a graph
#include <stdio.h>
#include <stdlib.h>
 
// a structure to represent an edge in the 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;
};
 
struct subset {
    int parent;
    int rank;
};
 
// 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 set of an element i
// (uses path compression technique)
int find(struct subset subsets[], int i)
{
    // find root and make root as parent of i (path
    // compression)
    if (subsets[i].parent != i)
        subsets[i].parent
            = find(subsets, subsets[i].parent);
 
    return subsets[i].parent;
}
 
// A function that does union of two sets of x and y
// (uses union by rank)
void Union(struct subset subsets[], int xroot, int yroot)
{
 
    // Attach smaller rank tree under root of high rank tree
    // (Union by Rank)
    if (subsets[xroot].rank < subsets[yroot].rank)
        subsets[xroot].parent = yroot;
    else if (subsets[xroot].rank > subsets[yroot].rank)
        subsets[yroot].parent = xroot;
 
    // If ranks are same, then make one as root and
    // increment its rank by one
    else {
        subsets[yroot].parent = xroot;
        subsets[xroot].rank++;
    }
}
 
// The main function to check whether a given graph contains
// cycle or not
int isCycle(struct Graph* graph)
{
    int V = graph->V;
    int E = graph->E;
 
    // Allocate memory for creating V sets
    struct subset* subsets
        = (struct subset*)malloc(V * sizeof(struct subset));
 
    for (int v = 0; v < V; ++v) {
        subsets[v].parent = v;
        subsets[v].rank = 0;
    }
 
    // Iterate through all edges of graph, find sets of both
    // vertices of every edge, if sets are same, then there
    // is cycle in graph.
    for (int e = 0; e < E; ++e) {
        int x = find(subsets, graph->edge[e].src);
        int y = find(subsets, graph->edge[e].dest);
 
        if (x == y)
            return 1;
 
        Union(subsets, x, y);
    }
    return 0;
}
 
// Driver code
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

// A union by rank and path compression
// based program to detect cycle in a graph
class Graph
{
    int V, E;
    Edge[] edge;
 
    Graph(int nV, int nE)
    {
        V = nV;
        E = nE;
        edge = new Edge[E];
        for (int i = 0; i < E; i++)
        {
            edge[i] = new Edge();
        }
    }
 
    // class to represent edge
    class Edge
    {
        int src, dest;
    }
 
    // class to represent Subset
    class subset
    {
        int parent;
        int rank;
    }
 
    // A utility function to find
    // set of an element i (uses
    // path compression technique)
    int find(subset[] subsets, int i)
    {
        if (subsets[i].parent != i)
            subsets[i].parent
                = find(subsets, subsets[i].parent);
        return subsets[i].parent;
    }
 
    // A function that does union
    // of two sets of x and y
    // (uses union by rank)
    void Union(subset[] subsets, int x, int y)
    {
        int xroot = find(subsets, x);
        int yroot = find(subsets, y);
 
        if (subsets[xroot].rank < subsets[yroot].rank)
            subsets[xroot].parent = yroot;
        else if (subsets[yroot].rank < subsets[xroot].rank)
            subsets[yroot].parent = xroot;
        else {
            subsets[xroot].parent = yroot;
            subsets[yroot].rank++;
        }
    }
 
    // The main function to check whether
    // a given graph contains cycle or not
    int isCycle(Graph graph)
    {
        int V = graph.V;
        int E = graph.E;
 
        subset[] subsets = new subset[V];
        for (int v = 0; v < V; v++) {
 
            subsets[v] = new subset();
            subsets[v].parent = v;
            subsets[v].rank = 0;
        }
 
        for (int e = 0; e < E; e++) {
            int x = find(subsets, graph.edge[e].src);
            int y = find(subsets, graph.edge[e].dest);
            if (x == y)
                return 1;
            Union(subsets, 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)
            System.out.println("Graph contains cycle");
        else
            System.out.println(
                "Graph doesn't contain cycle");
    }
}
 
// This code is contributed
// by ashwani khemani

Python

# A union by rank and path compression based
# program to detect cycle in a graph
from collections import defaultdict
 
# a structure to represent a graph
 
 
class Graph:
 
    def __init__(self, num_of_v):
        self.num_of_v = num_of_v
        self.edges = defaultdict(list)
 
    # graph is represented as an
    # array of edges
    def add_edge(self, u, v):
        self.edges[u].append(v)
 
 
class Subset:
    def __init__(self, parent, rank):
        self.parent = parent
        self.rank = rank
 
# A utility function to find set of an element
# node(uses path compression technique)
 
 
def find(subsets, node):
    if subsets[node].parent != node:
        subsets[node].parent = find(subsets, subsets[node].parent)
    return subsets[node].parent
 
# A function that does union of two sets
# of u and v(uses union by rank)
 
 
def union(subsets, u, v):
 
    # Attach smaller rank tree under root
    # of high rank tree(Union by Rank)
    if subsets[u].rank > subsets[v].rank:
        subsets[v].parent = u
    elif subsets[v].rank > subsets[u].rank:
        subsets[u].parent = v
 
    # If ranks are same, then make one as
    # root and increment its rank by one
    else:
        subsets[v].parent = u
        subsets[u].rank += 1
 
# The main function to check whether a given
# graph contains cycle or not
 
 
def isCycle(graph):
 
    # Allocate memory for creating sets
    subsets = []
 
    for u in range(graph.num_of_v):
        subsets.append(Subset(u, 0))
 
    # Iterate through all edges of graph,
    # find sets of both vertices of every
    # edge, if sets are same, then there
    # is cycle in graph.
    for u in graph.edges:
        u_rep = find(subsets, u)
 
        for v in graph.edges[u]:
            v_rep = find(subsets, v)
 
            if u_rep == v_rep:
                return True
            else:
                union(subsets, u_rep, v_rep)
 
 
# Driver Code
g = Graph(3)
 
# add edge 0-1
g.add_edge(0, 1)
 
# add edge 1-2
g.add_edge(1, 2)
 
# add edge 0-2
g.add_edge(0, 2)
 
if isCycle(g):
    print('Graph contains cycle')
else:
    print('Graph does not contain cycle')
 
# This code is contributed by
# Sampath Kumar Surine

C#

// A union by rank and path compression
// based program to detect cycle in a graph
using System;
 
class Graph {
    public int V, E;
    public Edge[] edge;
 
    public Graph(int nV, int nE)
    {
        V = nV;
        E = nE;
        edge = new Edge[E];
        for (int i = 0; i < E; i++)
        {
            edge[i] = new Edge();
        }
    }
 
    // class to represent edge
    public class Edge {
        public int src, dest;
    }
 
    // class to represent Subset
    class subset
    {
        public int parent;
        public int rank;
    }
 
    // A utility function to find
    // set of an element i (uses
    // path compression technique)
    int find(subset[] subsets, int i)
    {
        if (subsets[i].parent != i)
            subsets[i].parent
                = find(subsets, subsets[i].parent);
        return subsets[i].parent;
    }
 
    // A function that does union
    // of two sets of x and y
    // (uses union by rank)
    void Union(subset[] subsets, int x, int y)
    {
        int xroot = find(subsets, x);
        int yroot = find(subsets, y);
 
        if (subsets[xroot].rank < subsets[yroot].rank)
            subsets[xroot].parent = yroot;
        else if (subsets[yroot].rank < subsets[xroot].rank)
            subsets[yroot].parent = xroot;
        else {
            subsets[xroot].parent = yroot;
            subsets[yroot].rank++;
        }
    }
 
    // The main function to check whether
    // a given graph contains cycle or not
    int isCycle(Graph graph)
    {
        int V = graph.V;
        int E = graph.E;
 
        subset[] subsets = new subset[V];
        for (int v = 0; v < V; v++)
        {
            subsets[v] = new subset();
            subsets[v].parent = v;
            subsets[v].rank = 0;
        }
 
        for (int e = 0; e < E; e++)
        {
            int x = find(subsets, graph.edge[e].src);
            int y = find(subsets, graph.edge[e].dest);
            if (x == y)
                return 1;
            Union(subsets, x, y);
        }
        return 0;
    }
 
    // Driver Code
    static public 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 Arnab Kundu

Output

Graph contains cycle

References:
http://en.wikipedia.org/wiki/Disjoint-set_data_structure
IITD Video Lecture
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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