Prerequisites:
Disjoint Set data structure is used to keep track of a set of elements partitioned into a number of disjoint (non-overlapping) subsets.
In this article, we will learn about constructing the same Data Structure dynamically. This data structure basically helps in situations where we cannot simply use arrays for creating disjoint sets because of large inputs of order 109.
To illustrate this, consider the following problem. We need to find the total number of connected components in the Graph when the total Number of Vertices can be up to 10^9.
Examples:
Input : Edges : { { 1, 2 },
{ 2, 3 },
{ 4, 5 } }
Output : 2
Explanation: {1, 2, 3} forms a component and
{4, 5} forms another component.
The idea to solve this problem is, we will maintain two hash tables (implemented using unordered_maps in C++). One for parent and other for degree. Parent[V] will give the parent of the component which the Vertex V is part of and Degree will give the number of vertices in that component.
Initially, both Parent and Degree will be empty. We will keep inserting vertices to the maps as sequentially.
See the code and the explanation simultaneously for a better understanding. Below are the methods used in the code to solve the above problem:
- getParent(V): This method will give the parent of the vertex V. Here we recursively find the parent of the vertex V( see code), meanwhile we assign all the vertex in that component to have the same parent.( In a disjoint set data structure all the vertex in the same component have the same parent.)
- Union(): When we add an edge and the two vertexes are of different components we call the Union() method to join both components. Here the parent of the component formed after joining both components will be the parent of the component among the two which had more vertexes before the union. The degree of the new component is updated accordingly.
- getTotalComponent(): Vertex in the same component will have the same parent.
We use unordered_set (STL) to count the total number of components. As we have maintained the Data Structure as Dynamic, there can be any vertex that has not been added to any of the components hence they are different components alone. So the total number of components will be given by,
Total no of Component = Total Vertices - Number of Vertices
in parent (Map) + Number of Component
formed from the Vertexes inserted
in the graph.
Below is the implementation of the above idea:
C++
// Dynamic Disjoint Set Data Structure// Union-Find#include <bits/stdc++.h>using namespace std;int N;int Edges[3][2];// Dynamic Disjoint Set Data Structurestruct DynamicDisjointSetDS { // We will add the vertex to the edge // only when it is asked to i.e. maintain // a dynamic DS. unordered_map<int, int> parent, degree; // Total number of Vertex in the Graph int N; // Constructor DynamicDisjointSetDS(int n) { N = n; } // Get Parent of vertex V int getParent(int vertex) { // If the vertex is already inserted // in the graph if (parent.find(vertex) != parent.end()) { if (parent[vertex] != vertex) { parent[vertex] = getParent(parent[vertex]); return parent[vertex]; } } // if the vertex is operated for the first // time else { // insert the vertex and assign its // parent to itself parent.insert(make_pair(vertex, vertex)); // Degree of the vertex degree.insert(make_pair(vertex, 1)); } return vertex; } // Union by Rank void Union(int vertexA, int vertexB) { // Parent of Vertex A int x = getParent(vertexA); // Parent of Vertex B int y = getParent(vertexB); // if both have same parent // Do Nothing if (x == y) return; // Merging the component // Assigning the parent of smaller Component // as the parent of the bigger Component. if (degree[x] > degree[y]) { parent[y] = x; degree[x] = degree[x] + degree[y]; } else { parent[x] = y; degree[y] = degree[y] + degree[x]; } } // Count total Component in the Graph int GetTotalComponent() { // To count the total Component formed // from the inserted vertex in the Graph unordered_set<int> total; // Iterate through the parent for (auto itr = parent.begin(); itr != parent.end(); itr++) { // Add the parent of each Vertex // to the set total.insert(getParent(itr->first)); } // Total Component = Total Vertexes - // Number of Vertex in the parent + // Number of Component formed from // the Vertexes inserted in the Graph return N - parent.size() + total.size(); }};// Solvevoid Solve(){ // Declaring the Dynamic Disjoint Set DS DynamicDisjointSetDS dsu(N); // Traversing through the Edges for (int i = 0; i < 3; i++) { // If the Vertexes in the Edges // have same parent do nothing if (dsu.getParent(Edges[i][0]) == dsu.getParent(Edges[i][1])) { continue; } // else Do Union of both the Components. else { dsu.Union(Edges[i][0], Edges[i][1]); } } // Get total Components cout << dsu.GetTotalComponent();}// Driver Codeint main(){ // Total Number of Vertexes N = 5; /* Edges * 1 <--> 2 * 2 <--> 3 * 4 <--> 5 */ Edges[0][0] = 1; Edges[0][1] = 2; Edges[1][0] = 2; Edges[1][1] = 3; Edges[2][0] = 4; Edges[2][1] = 3; // Solve Solve(); return 0;} |
Java
// Dynamic Disjoint Set Data Structure// Union-Findimport java.util.*;// Dynamic Disjoint Set Data Structureclass DynamicDisjointSetDS { // We will add the vertex to the edge // only when it is asked to i.e. maintain // a dynamic DS. Map<Integer, Integer> parent, degree; // Total number of Vertex in the Graph int N; // Constructor DynamicDisjointSetDS(int n) { N = n; parent = new HashMap<>(); degree = new HashMap<>(); } // Get Parent of vertex V int getParent(int vertex) { // If the vertex is already inserted // in the graph if (parent.containsKey(vertex)) { if (parent.get(vertex) != vertex) { parent.put(vertex, getParent(parent.get(vertex))); return parent.get(vertex); } } // if the vertex is operated for the first // time else { // insert the vertex and assign its // parent to itself parent.put(vertex, vertex); // Degree of the vertex degree.put(vertex, 1); } return vertex; } // Union by Rank void union(int vertexA, int vertexB) { // Parent of Vertex A int x = getParent(vertexA); // Parent of Vertex B int y = getParent(vertexB); // if both have same parent // Do Nothing if (x == y) { return; } // Merging the component // Assigning the parent of smaller Component // as the parent of the bigger Component. if (degree.get(x) > degree.get(y)) { parent.put(y, x); degree.put(x, degree.get(x) + degree.get(y)); } else { parent.put(x, y); degree.put(y, degree.get(y) + degree.get(x)); } } // Count total Component in the Graph int getTotalComponent() { // To count the total Component formed // from the inserted vertex in the Graph Set<Integer> total = new HashSet<>(); // Iterate through the parent for (Map.Entry<Integer, Integer> entry : parent.entrySet()) { // Add the parent of each Vertex // to the set total.add(getParent(entry.getKey())); } // Total Component = Total Vertexes - // Number of Vertex in the parent + // Number of Component formed from // the Vertexes inserted in the Graph return N - parent.size() + total.size(); }}// Driver Codepublic class Main { public static void main(String[] args) { // Total Number of Vertexes int N = 5; /* Edges * 1 <--> 2 * 2 <--> 3 * 4 <--> 5 */ int[][] Edges = { { 1, 2 }, { 2, 3 }, { 4, 3 } }; DynamicDisjointSetDS dsu = new DynamicDisjointSetDS(N); for (int i = 0; i < 3; i++) { if (dsu.getParent(Edges[i][0]) == dsu.getParent(Edges[i][1])) { continue; } else { dsu.union(Edges[i][0], Edges[i][1]); } } System.out.println(dsu.getTotalComponent()); }} |
Python3
# Dynamic Disjoint Set Data Structure # Union-Find # Dynamic Disjoint Set Data Structure class DynamicDisjointSetDS: # Constructor def __init__(self, n): # Total number of Vertex in the Graph self.N = n # We will add the vertex to the edge # only when it is asked to i.e. maintain # a dynamic DS. self.parent = {} self.degree = {} # Get Parent of vertex V def getParent(self, vertex): # If the vertex is already inserted # in the graph if vertex in self.parent: if self.parent[vertex] != vertex: self.parent[vertex] = \ self.getParent(self.parent[vertex]) return self.parent[vertex] # if the vertex is operated # for the first time else: # insert the vertex and assign # its parent to itself self.parent[vertex] = vertex # Degree of the vertex self.degree[vertex] = 1 return vertex # Union by Rank def Union(self, vertexA, vertexB): # Parent of Vertex A x = self.getParent(vertexA) # Parent of Vertex B y = self.getParent(vertexB) # if both have same parent # Do Nothing if x == y: return # Merging the component # Assigning the parent of smaller Component # as the parent of the bigger Component. if self.degree[x] > self.degree[y]: self.parent[y] = x self.degree[x] = (self.degree[x] + self.degree[y]) else: self.parent[x] = y self.degree[y] = (self.degree[y] + self.degree[x]) # Count total Component in the Graph def GetTotalComponent(self): # To count the total Component formed # from the inserted vertex in the Graph total = set() # Iterate through the parent for itr in self.parent: # Add the parent of each Vertex # to the set total.add(self.getParent(itr)) # Total Component = Total Vertexes - # Number of Vertex in the parent + # Number of Component formed from # the Vertexes inserted in the Graph return self.N - len(self.parent) + len(total) # Solve def Solve(N): # Declaring the Dynamic Disjoint Set DS dsu = DynamicDisjointSetDS(N) # Traversing through the Edges for i in range(0, 3): # If the Vertexes in the Edges # have same parent do nothing if (dsu.getParent(Edges[i][0]) == dsu.getParent(Edges[i][1])): continue # else Do Union of both the Components. else: dsu.Union(Edges[i][0], Edges[i][1]) # Get total Components print(dsu.GetTotalComponent())# Driver Code if __name__ == "__main__": # Total Number of Vertexes N = 5 Edges = [[1, 2], [2, 3], [4, 3]] # Solve Solve(N) # This code is contributed by # Rituraj Jain |
C#
// Dynamic Disjoint Set Data Structure// Union-Findusing System;using System.Collections.Generic;public class GFG { static int N; static int[, ] Edges = new int[3, 2]; // Dynamic Disjoint Set Data Structure public class DynamicDisjointSetDS { // We will add the vertex to the edge // only when it is asked to i.e. maintain // a dynamic DS. Dictionary<int, int> parent = new Dictionary<int, int>(); Dictionary<int, int> degree = new Dictionary<int, int>(); // Total number of Vertex in the Graph int N; // Constructor public DynamicDisjointSetDS(int n) { N = n; } // Get Parent of vertex V public int GetParent(int vertex) { // If the vertex is already inserted // in the graph if (parent.ContainsKey(vertex)) { if (parent[vertex] != vertex) { parent[vertex] = GetParent(parent[vertex]); return parent[vertex]; } } // if the vertex is operated for the first // time else { // insert the vertex and assign its // parent to itself parent[vertex] = vertex; // Degree of the vertex degree[vertex] = 1; } return vertex; } // Union by Rank public void Union(int vertexA, int vertexB) { // Parent of Vertex A int x = GetParent(vertexA); // Parent of Vertex B int y = GetParent(vertexB); // if both have same parent // Do Nothing if (x == y) return; // Merging the component // Assigning the parent of smaller Component // as the parent of the bigger Component. if (degree[x] > degree[y]) { parent[y] = x; degree[x] += degree[y]; } else { parent[x] = y; degree[y] += degree[x]; } } // Count total Component in the Graph public int GetTotalComponent() { // To count the total Component formed // from the inserted vertex in the Graph HashSet<int> total = new HashSet<int>(); // Iterate through the parent foreach(int key in new List<int>(parent.Keys)) { // Add the parent of each Vertex // to the set total.Add(GetParent(key)); } // Total Component = Total Vertexes - // Number of Vertex in the parent + // Number of Component formed from // the Vertexes inserted in the Graph return N - parent.Count + total.Count; } } // Solve public static void Solve() { // Declaring the Dynamic Disjoint Set DS DynamicDisjointSetDS dsu = new DynamicDisjointSetDS(N); // Traversing through the Edges for (int i = 0; i < 3; i++) { // If the Vertexes in the Edges // have same parent do nothing if (dsu.GetParent(Edges[i, 0]) == dsu.GetParent(Edges[i, 1])) { continue; } // else Do Union of both the Components. else { dsu.Union(Edges[i, 0], Edges[i, 1]); } } // Get total Components Console.WriteLine(dsu.GetTotalComponent()); } // Driver Code public static void Main() { // Total Number of Vertexes N = 5; /* Edges * 1 <--> 2 * 2 <--> 3 * 4 <--> 5 */ Edges[0, 0] = 1; Edges[0, 1] = 2; Edges[1, 0] = 2; Edges[1, 1] = 3; Edges[2, 0] = 4; Edges[2, 1] = 3; // Solve Solve(); }}// This code is contributed by prasad264 |
Javascript
// Dynamic Disjoint Set Data Structure// Union-Findconst N = 5;const Edges = [[1, 2], [2, 3], [4, 5]];// Dynamic Disjoint Set Data Structureclass DynamicDisjointSetDS { // Constructor constructor(n) { this.N = n; this.parent = new Map(); this.degree = new Map(); } // Get Parent of vertex V getParent(vertex) { // If the vertex is already inserted // in the graph if (this.parent.has(vertex)) { if (this.parent.get(vertex) !== vertex) { this.parent.set(vertex, this.getParent(this.parent.get(vertex))); return this.parent.get(vertex); } } // if the vertex is operated for the first // time else { // insert the vertex and assign its // parent to itself this.parent.set(vertex, vertex); // Degree of the vertex this.degree.set(vertex, 1); } return vertex; } // Union by Rank Union(vertexA, vertexB) { // Parent of Vertex A const x = this.getParent(vertexA); // Parent of Vertex B const y = this.getParent(vertexB); // if both have same parent // Do Nothing if (x === y) { return; } // Merging the component // Assigning the parent of smaller Component // as the parent of the bigger Component. if (this.degree.get(x) > this.degree.get(y)) { this.parent.set(y, x); this.degree.set(x, this.degree.get(x) + this.degree.get(y)); } else { this.parent.set(x, y); this.degree.set(y, this.degree.get(y) + this.degree.get(x)); } } // Count total Component in the Graph GetTotalComponent() { // To count the total Component formed // from the inserted vertex in the Graph const total = new Set(); // Iterate through the parent for (const [key, value] of this.parent) { // Add the parent of each Vertex // to the set total.add(this.getParent(key)); } // Total Component = Total Vertexes - // Number of Vertex in the parent + // Number of Component formed from // the Vertexes inserted in the Graph return this.N - this.parent.size + total.size; }}// Solvefunction solve() { // Declaring the Dynamic Disjoint Set DS const dsu = new DynamicDisjointSetDS(N); // Traversing through the Edges for (let i = 0; i < Edges.length; i++) { // If the Vertexes in the Edges // have same parent do nothing if (dsu.getParent(Edges[i][0]) === dsu.getParent(Edges[i][1])) { continue; } // else Do Union of both the Components. else { dsu.Union(Edges[i][0], Edges[i][1]); } } // Get total Components console.log(dsu.GetTotalComponent());}// Driver Codesolve(); |
Output:
2
Time Complexity: O(E * alpha(N)), where E is the number of edges in the graph, N is the number of vertices in the graph, and alpha is the inverse Ackermann function, which has a value less than 5 for all practical values of N.
Space Complexity: O(N), where N is the total number of vertices in the graph.
Note: If the number of vertices is even larger, we can implement the same code just by changing the data type from int to long.
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