Union-Find Algorithm | (Union By Rank and Find by Optimized Path Compression)

28 Nov, 2021
Check whether a given graph contains a cycle or not.Example:  
Input: 
Output: Graph contains Cycle.

Input: 
Output: Graph does not contain Cycle.
Prerequisites: Disjoint Set (Or Union-Find), Union By Rank and Path CompressionWe have already discussed union-find to detect cycle. Here we discuss find by path compression, where it is slightly modified to work faster than the original method as we are skipping one level each time we are going up the graph. Implementation of find function is iterative, so there is no overhead involved.Time complexity of optimized find function is O(log*(n)), i.e iterated logarithm, which converges to O(1) for repeated calls.Refer this link for Proof of log*(n) complexity of Union-Find Explanation of find function: Take Example 1 to understand find function:(1)call find(8) for first time and mappings will be done like this:  
It took 3 mappings for find function to get the root of node 8. Mappings are illustrated below: From node 8, skipped node 7, Reached node 6. From node 6, skipped node 5, Reached node 4. From node 4, skipped node 2, Reached node 0.(2)call find(8) for second time and mappings will be done like this:  
It took 2 mappings for find function to get the root of node 8. Mappings are illustrated below: From node 8, skipped node 5, node 6 and node 7, Reached node 4. From node 4, skipped node 2, Reached node 0.(3)call find(8) for third time and mappings will be done like this:  
Finally, we see it took only 1 mapping for find function to get the root of node 8. Mappings are illustrated below: From node 8, skipped node 5, node 6, node 7, node 4, and node 2, Reached node 0. That is how it converges path from certain mappings to single mapping.Explanation of example 1:Initially array size and Arr look like: Arr[9] = {0, 1, 2, 3, 4, 5, 6, 7, 8} size[9] = {1, 1, 1, 1, 1, 1, 1, 1, 1}Consider the edges in the graph, and add them one by one to the disjoint-union set as follows: Edge 1: 0-1 find(0)=>0, find(1)=>1, both have different root parent Put these in single connected component as currently they doesn’t belong to different connected components. Arr[1]=0, size[0]=2;Edge 2: 0-2 find(0)=>0, find(2)=>2, both have different root parent Arr[2]=0, size[0]=3;Edge 3: 1-3 find(1)=>0, find(3)=>3, both have different root parent Arr[3]=0, size[0]=3;Edge 4: 3-4 find(3)=>1, find(4)=>4, both have different root parent Arr[4]=0, size[0]=4;Edge 5: 2-4 find(2)=>0, find(4)=>0, both have same root parent Hence, There is a cycle in graph. We stop further checking for cycle in graph.  
C++
Java
Python3
C#
Javascript
// CPP program to implement Union-Find with union// by rank and path compression.#include <bits/stdc++.h>using namespace std;  const int MAX_VERTEX = 101;  // Arr to represent parent of index iint Arr[MAX_VERTEX];  // Size to represent the number of nodes// in subgxrph rooted at index iint size[MAX_VERTEX];  // set parent of every node to itself and// size of node to onevoid initialize(int n){    for (int i = 0; i <= n; i++) {        Arr[i] = i;        size[i] = 1;    }}  // Each time we follow a path, find function// compresses it further until the path length// is greater than or equal to 1.int find(int i){    // while we reach a node whose parent is    // equal to itself    while (Arr[i] != i)    {        Arr[i] = Arr[Arr[i]]; // Skip one level        i = Arr[i]; // Move to the new level    }    return i;}  // A function that does union of two nodes x and y// where xr is root node  of x and yr is root node of yvoid _union(int xr, int yr){    if (size[xr] < size[yr]) // Make yr parent of xr    {        Arr[xr] = Arr[yr];        size[yr] += size[xr];    }    else // Make xr parent of yr    {        Arr[yr] = Arr[xr];        size[xr] += size[yr];    }}  // The main function to check whether a given// gxrph contains cycle or notint isCycle(vector<int> adj[], int V){    // Itexrte through all edges of gxrph, find    // nodes connecting them.    // If root nodes of both are same, then there is    // cycle in gxrph.    for (int i = 0; i < V; i++) {        for (int j = 0; j < adj[i].size(); j++) {            int x = find(i); // find root of i            int y = find(adj[i][j]); // find root of adj[i][j]              if (x == y)                return 1; // If same parent            _union(x, y); // Make them connect        }    }    return 0;}  // Driver progxrm to test above functionsint main(){    int V = 3;      // Initialize the values for arxry Arr and Size    initialize(V);      /* Let us create following gxrph         0        |  \        |    \        1-----2 */      vector<int> adj[V]; // Adjacency list for gxrph      adj[0].push_back(1);    adj[0].push_back(2);    adj[1].push_back(2);      // call is_cycle to check if it contains cycle    if (isCycle(adj, V))        cout << "Gxrph contains Cycle.\n";    else        cout << "Gxrph does not contain Cycle.\n";      return 0;}
// Java program to implement Union-Find with union// by rank and path compressionimport java.util.*;  class GFG {static int MAX_VERTEX = 101;  // Arr to represent parent of index istatic int []Arr = new int[MAX_VERTEX];  // Size to represent the number of nodes// in subgxrph rooted at index istatic int []size = new int[MAX_VERTEX];  // set parent of every node to itself and// size of node to onestatic void initialize(int n){    for (int i = 0; i <= n; i++)    {        Arr[i] = i;        size[i] = 1;    }}  // Each time we follow a path, find function// compresses it further until the path length// is greater than or equal to 1.static int find(int i){    // while we reach a node whose parent is    // equal to itself    while (Arr[i] != i)    {        Arr[i] = Arr[Arr[i]]; // Skip one level        i = Arr[i]; // Move to the new level    }    return i;}  // A function that does union of two nodes x and y// where xr is root node of x and yr is root node of ystatic void _union(int xr, int yr){    if (size[xr] < size[yr]) // Make yr parent of xr    {        Arr[xr] = Arr[yr];        size[yr] += size[xr];    }    else // Make xr parent of yr    {        Arr[yr] = Arr[xr];        size[xr] += size[yr];    }}  // The main function to check whether a given// gxrph contains cycle or notstatic int isCycle(Vector<Integer> adj[], int V){    // Itexrte through all edges of gxrph,     // find nodes connecting them.    // If root nodes of both are same,     // then there is cycle in gxrph.    for (int i = 0; i < V; i++)    {        for (int j = 0; j < adj[i].size(); j++)         {            int x = find(i); // find root of i                          // find root of adj[i][j]            int y = find(adj[i].get(j));               if (x == y)                return 1; // If same parent            _union(x, y); // Make them connect        }    }    return 0;}  // Driver Codepublic static void main(String[] args) {    int V = 3;      // Initialize the values for arxry Arr and Size    initialize(V);      /* Let us create following gxrph        0        | \        | \        1-----2 */      // Adjacency list for graph    Vector<Integer> []adj = new Vector[V];     for(int i = 0; i < V; i++)        adj[i] = new Vector<Integer>();      adj[0].add(1);    adj[0].add(2);    adj[1].add(2);      // call is_cycle to check if it contains cycle    if (isCycle(adj, V) == 1)        System.out.print("Graph contains Cycle.\n");    else        System.out.print("Graph does not contain Cycle.\n");    }}  // This code is contributed by PrinciRaj1992
# Python3 program to implement Union-Find # with union by rank and path compression.  # set parent of every node to itself # and size of node to one def initialize(n):    global Arr, size    for i in range(n + 1):        Arr[i] = i         size[i] = 1  # Each time we follow a path, find # function compresses it further # until the path length is greater # than or equal to 1. def find(i):    global Arr, size          # while we reach a node whose     # parent is equal to itself     while (Arr[i] != i):        Arr[i] = Arr[Arr[i]] # Skip one level         i = Arr[i] # Move to the new level    return i  # A function that does union of two # nodes x and y where xr is root node # of x and yr is root node of y def _union(xr, yr):    global Arr, size    if (size[xr] < size[yr]): # Make yr parent of xr         Arr[xr] = Arr[yr]         size[yr] += size[xr]    else: # Make xr parent of yr        Arr[yr] = Arr[xr]         size[xr] += size[yr]  # The main function to check whether # a given graph contains cycle or not def isCycle(adj, V):    global Arr, size          # Itexrte through all edges of gxrph,     # find nodes connecting them.     # If root nodes of both are same,     # then there is cycle in gxrph.    for i in range(V):        for j in range(len(adj[i])):            x = find(i) # find root of i             y = find(adj[i][j]) # find root of adj[i][j]               if (x == y):                return 1 # If same parent             _union(x, y) # Make them connect    return 0  # Driver CodeMAX_VERTEX = 101  # Arr to represent parent of index i Arr = [None] * MAX_VERTEX   # Size to represent the number of nodes # in subgxrph rooted at index i size = [None] * MAX_VERTEX   V = 3  # Initialize the values for arxry # Arr and Size initialize(V)   # Let us create following gxrph #     0 # | \ # | \ # 1-----2   # Adjacency list for graph adj = [[] for i in range(V)]   adj[0].append(1) adj[0].append(2) adj[1].append(2)   # call is_cycle to check if it # contains cycle if (isCycle(adj, V)):     print("Graph contains Cycle.") else:    print("Graph does not contain Cycle.")  # This code is contributed by PranchalK
// C# program to implement Union-Find // with union by rank and path compressionusing System;using System.Collections.Generic;      class GFG {static int MAX_VERTEX = 101;  // Arr to represent parent of index istatic int []Arr = new int[MAX_VERTEX];  // Size to represent the number of nodes// in subgxrph rooted at index istatic int []size = new int[MAX_VERTEX];  // set parent of every node to itself // and size of node to onestatic void initialize(int n){    for (int i = 0; i <= n; i++)    {        Arr[i] = i;        size[i] = 1;    }}  // Each time we follow a path, // find function compresses it further // until the path length is greater than // or equal to 1.static int find(int i){    // while we reach a node whose     // parent is equal to itself    while (Arr[i] != i)    {        Arr[i] = Arr[Arr[i]]; // Skip one level        i = Arr[i]; // Move to the new level    }    return i;}  // A function that does union of // two nodes x and y where xr is // root node of x and yr is root node of ystatic void _union(int xr, int yr){    if (size[xr] < size[yr]) // Make yr parent of xr    {        Arr[xr] = Arr[yr];        size[yr] += size[xr];    }    else // Make xr parent of yr    {        Arr[yr] = Arr[xr];        size[xr] += size[yr];    }}  // The main function to check whether // a given graph contains cycle or notstatic int isCycle(List<int> []adj, int V){    // Itexrte through all edges of graph,     // find nodes connecting them.    // If root nodes of both are same,     // then there is cycle in graph.    for (int i = 0; i < V; i++)    {        for (int j = 0; j < adj[i].Count; j++)         {            int x = find(i); // find root of i                          // find root of adj[i][j]            int y = find(adj[i][j]);               if (x == y)                return 1; // If same parent            _union(x, y); // Make them connect        }    }    return 0;}  // Driver Codepublic static void Main(String[] args) {    int V = 3;      // Initialize the values for     // array Arr and Size    initialize(V);      /* Let us create following graph        0        | \        | \        1-----2 */      // Adjacency list for graph    List<int> []adj = new List<int>[V];     for(int i = 0; i < V; i++)        adj[i] = new List<int>();      adj[0].Add(1);    adj[0].Add(2);    adj[1].Add(2);      // call is_cycle to check if it contains cycle    if (isCycle(adj, V) == 1)        Console.Write("Graph contains Cycle.\n");    else        Console.Write("Graph does not contain Cycle.\n");    }}  // This code is contributed by Rajput-Ji
<script>  // Javascript program to implement Union-Find with union// by rank and path compression.  var MAX_VERTEX = 101;  // Arr to represent parent of index ivar Arr = Array(MAX_VERTEX).fill(0);  // Size to represent the number of nodes// in subgxrph rooted at index ivar size = Array(MAX_VERTEX).fill(0);  // set parent of every node to itself and// size of node to onefunction initialize(n){    for (var i = 0; i <= n; i++) {        Arr[i] = i;        size[i] = 1;    }}  // Each time we follow a path, find function// compresses it further until the path length// is greater than or equal to 1.function find(i){    // while we reach a node whose parent is    // equal to itself    while (Arr[i] != i)    {        Arr[i] = Arr[Arr[i]]; // Skip one level        i = Arr[i]; // Move to the new level    }    return i;}  // A function that does union of two nodes x and y// where xr is root node  of x and yr is root node of yfunction _union(xr, yr){    if (size[xr] < size[yr]) // Make yr parent of xr    {        Arr[xr] = Arr[yr];        size[yr] += size[xr];    }    else // Make xr parent of yr    {        Arr[yr] = Arr[xr];        size[xr] += size[yr];    }}  // The main function to check whether a given// gxrph contains cycle or notfunction isCycle(adj, V){    // Itexrte through all edges of gxrph, find    // nodes connecting them.    // If root nodes of both are same, then there is    // cycle in gxrph.    for (var i = 0; i < V; i++) {        for (var j = 0; j < adj[i].length; j++) {            var x = find(i); // find root of i            var y = find(adj[i][j]); // find root of adj[i][j]              if (x == y)                return 1; // If same parent            _union(x, y); // Make them connect        }    }    return 0;}  // Driver progxrm to test above functionsvar V = 3;  // Initialize the values for arxry Arr and Sizeinitialize(V);/* Let us create following gxrph     0    |  \    |    \    1-----2 */var adj = Array.from(Array(V), ()=>Array()); // Adjacency list for gxrphadj[0].push(1);adj[0].push(2);adj[1].push(2);  // call is_cycle to check if it contains cycleif (isCycle(adj, V))    document.write("Graph contains Cycle.<br>");else    document.write("Graph does not contain Cycle.<br>");  // This code is contributed by rutvik_56.</script>
Output:  
Graph contains Cycle.
Time Complexity(Find) : O(log*(n)) Time Complexity(union) : O(1) 
PranchalKatiyar
rahul_15
princiraj1992
Rajput-Ji
rutvik_56
union-find
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