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Find the maximum component size after addition of each edge to the graph

  • Difficulty Level : Medium
  • Last Updated : 24 Sep, 2021

Given an array arr[][] which contains the edges of a graph to be used to construct an undirected graph G with N nodes, the task is to find the maximum component size in the graph after each edge is added while constructing the graph.

Examples: 

Input: N = 4, arr[][] = {{1, 2}, {3, 4}, {2, 3}} 
Output: 2 2 4 
Explanation: 
Initially, the graph has 4 individual nodes 1, 2, 3 and 4. 
After the first edge is added : 1 – 2, 3, 4 -> maximum component size = 2 
After the second edge is added : 1 – 2, 3 – 4 -> maximum component size = 2 
After the third edge is added : 1 – 2 – 3 – 4 -> maximum component size = 4

Input: N = 4, arr[][] = {{2, 3}, {1, 2}, {1, 5}, {2, 4}} 
Output: 2 3 4 5 
 

Naive Approach: The naive approach for this problem is to add the edges sequentially and at each step apply depth-first search algorithm to find the size of the largest component. 



Below is the implementation of the above approach:

C++




// C++ program to find the
// maximum comake_paironent size
// after addition of each
// edge to the graph
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to perform
// Depth First Search
// on the given graph
int dfs(int u, int visited[],
        vector<int>* adj)
{
    // Mark visited
    visited[u] = 1;
    int size = 1;
 
    // Add each child's
    // comake_paironent size
    for (auto child : adj[u]) {
        if (!visited[child])
            size += dfs(child,
                        visited, adj);
    }
    return size;
}
 
// Function to find the maximum
// comake_paironent size
// after addition of each
// edge to the graph
void maxSize(vector<pair<int, int> > e,
             int n)
{
    // Graph in the adjacency
    // list format
    vector<int> adj[n];
 
    // Visited array
    int visited[n];
 
    vector<int> answer;
 
    // At each step, add a new
    // edge and apply dfs on all
    // the nodes to find the maximum
    // comake_paironent size
    for (auto edge : e) {
 
        // Add this edge to undirected graph
        adj[edge.first - 1].push_back(
            edge.second - 1);
        adj[edge.second - 1].push_back(
            edge.first - 1);
 
        // Mark all the nodes
        // as unvisited
        memset(visited, 0,
               sizeof(visited));
 
        int maxAns = 0;
 
        // Loop to perform DFS
        // and find the size
        // of the maximum comake_paironent
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                maxAns = max(maxAns,
                             dfs(i, visited, adj));
            }
        }
        answer.push_back(maxAns);
    }
 
    // Print the answer
    for (auto i : answer) {
        cout << i << " ";
    }
}
 
// Driver code
int main()
{
    int N = 4;
    vector<pair<int, int> > E;
    E.push_back(make_pair(1, 2));
    E.push_back(make_pair(3, 4));
    E.push_back(make_pair(2, 3));
 
    maxSize(E, N);
 
    return 0;
}

Java




// Java program to find the maximum
// comake_paironent size after
// addition of each edge to the graph
import java.util.*;
 
@SuppressWarnings("unchecked")
class GFG{
     
static class pair
{
    int Key, Value;
     
    pair(int Key, int Value)
    {
        this.Key = Key;
        this.Value = Value;
    }
}
       
// Function to perform Depth First
// Search on the given graph
static int dfs(int u, int []visited,
               ArrayList []adj)
{
     
    // Mark visited
    visited[u] = 1;
    int size = 1;
   
    // Add each child's
    // comake_paironent size
    for(int child : (ArrayList<Integer>)adj[u])
    {
        if (visited[child] == 0)
            size += dfs(child,
                        visited, adj);
    }
    return size;
}
   
// Function to find the maximum
// comake_paironent size after
// addition of each edge to the graph
static void maxSize(ArrayList e,
                    int n)
{
     
    // Graph in the adjacency
    // list format
    ArrayList []adj = new ArrayList[n];
      
    for(int i = 0; i < n; i++)
    {
        adj[i] = new ArrayList();
    }
      
    // Visited array
    int []visited = new int[n];
   
    ArrayList answer = new ArrayList();
   
    // At each step, add a new
    // edge and apply dfs on all
    // the nodes to find the maximum
    // comake_paironent size
    for(pair edge : (ArrayList<pair>)e)
    {
         
        // Add this edge to undirected graph
        adj[edge.Key - 1].add(
           edge.Value - 1);
        adj[edge.Value - 1].add(
              edge.Key - 1);
               
        // Mark all the nodes
        // as unvisited
        Arrays.fill(visited,0);
   
        int maxAns = 0;
   
        // Loop to perform DFS and find the
        // size of the maximum comake_paironent
        for(int i = 0; i < n; i++)
        {
            if (visited[i] == 0)
            {
                maxAns = Math.max(maxAns,
                              dfs(i, visited, adj));
            }
        }
        answer.add(maxAns);
    }
     
    // Print the answer
    for(int i : (ArrayList<Integer>) answer)
    {
        System.out.print(i + " ");
    }
}
   
// Driver code
public static void main(String[] args)
{
    int N = 4;
     
    ArrayList E = new ArrayList();
    E.add(new pair(1, 2));
    E.add(new pair(3, 4));
    E.add(new pair(2, 3));
   
    maxSize(E, N);
}
}
 
// This code is contributed by pratham76

Python3




# Python3 program to find the
# maximum comake_paironent size
# after addition of each
# edge to the graph
 
# Function to perform
# Depth First Search
# on the given graph
def dfs(u, visited, adj):
   
    # Mark visited
    visited[u] = 1
    size = 1
   
    # Add each child's
    # comake_paironent size
    for child in adj[u]:
        if (visited[child] == 0):
            size += dfs(child, visited, adj)
    return size
   
# Function to find the maximum
# comake_paironent size
# after addition of each
# edge to the graph
def maxSize(e, n):
   
    # Graph in the adjacency
    # list format
    adj = []
      
    for i in range(n):
        adj.append([])
      
    # Visited array
    visited = [0]*(n)
   
    answer = []
   
    # At each step, add a new
    # edge and apply dfs on all
    # the nodes to find the maximum
    # comake_paironent size
    for edge in e:
        # Add this edge to undirected graph
        adj[edge[0] - 1].append(edge[1] - 1)
        adj[edge[1] - 1].append(edge[0] - 1)
   
        # Mark all the nodes
        # as unvisited
        visited = [0]*(n)
   
        maxAns = 0
   
        # Loop to perform DFS
        # and find the size
        # of the maximum comake_paironent
        for i in range(n):
            if (visited[i] == 0):
                maxAns = max(maxAns, dfs(i, visited, adj))
        answer.append(maxAns)
   
    # Print the answer
    for i in answer:
        print(i, "", end = "")
 
N = 4
E = []
E.append([1, 2])
E.append([3, 4])
E.append([2, 3])
 
maxSize(E, N)
 
# This code is contributed by divyesh072019.

C#




// C# program to find the
// maximum comake_paironent size
// after addition of each
// edge to the graph
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
      
// Function to perform
// Depth First Search
// on the given graph
static int dfs(int u, int []visited,
               ArrayList []adj)
{
     
    // Mark visited
    visited[u] = 1;
    int size = 1;
  
    // Add each child's
    // comake_paironent size
    foreach (int child in adj[u])
    {
        if (visited[child] == 0)
            size += dfs(child,
                        visited, adj);
    }
    return size;
}
  
// Function to find the maximum
// comake_paironent size
// after addition of each
// edge to the graph
static void maxSize(ArrayList e,
                    int n)
{
     
    // Graph in the adjacency
    // list format
    ArrayList []adj = new ArrayList[n];
     
    for(int i = 0; i < n; i++)
    {
        adj[i] = new ArrayList();
    }
     
    // Visited array
    int []visited = new int[n];
  
    ArrayList answer = new ArrayList();
  
    // At each step, add a new
    // edge and apply dfs on all
    // the nodes to find the maximum
    // comake_paironent size
    foreach(KeyValuePair<int, int> edge in e)
    {
         
        // Add this edge to undirected graph
        adj[edge.Key - 1].Add(
           edge.Value - 1);
        adj[edge.Value - 1].Add(
              edge.Key - 1);
  
        // Mark all the nodes
        // as unvisited
        Array.Fill(visited,0);
  
        int maxAns = 0;
  
        // Loop to perform DFS
        // and find the size
        // of the maximum comake_paironent
        for(int i = 0; i < n; i++)
        {
            if (visited[i] == 0)
            {
                maxAns = Math.Max(maxAns,
                              dfs(i, visited, adj));
            }
        }
        answer.Add(maxAns);
    }
  
    // Print the answer
    foreach(int i in answer)
    {
        Console.Write(i + " ");
    }
}
  
// Driver code
public static void Main(string[] args)
{
    int N = 4;
    ArrayList E = new ArrayList();
    E.Add(new KeyValuePair<int, int>(1, 2));
    E.Add(new KeyValuePair<int, int>(3, 4));
    E.Add(new KeyValuePair<int, int>(2, 3));
  
    maxSize(E, N);
}
}
 
// This code is contributed by rutvik_56
Output: 
2 2 4

 

Time Complexity: O(|E| * N)

Efficient Approach: The idea is to use the concept of Disjoint Set (Union by rank and Path compression) to solve the problem more efficiently. 

  • Each node is initially a disjoint set within itself. As and when the edges are added, the disjoint sets are merged together forming larger components. In the disjoint set implementation, we will make the ranking system based on component sizes i.e when merging of two components is performed the larger component’s root will be considered the final root after the merge operation.
  • One way to find the largest size component after each edge addition is to traverse the size array (size[i] represents the size of the component in which node ‘i’ belongs), but this is inefficient when the number of nodes in the graph is high.
  • A more efficient way is to store the component sizes of all the root in some ordered data structure like sets.
  • When two components are merged, all we need to do is remove the previous component sizes from the set and add the combined component size. So at each step, we would be able to find the largest component size in logarithmic complexity.

Below is the implementation of the above approach:

C++




// C++ implementation to find the maximum
// component size after the addition of
// each edge to the graph
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Variables for implementing DSU
int par[100005];
int size[100005];
 
// Root of the component of node i
int root(int i)
{
    if (par[i] == i)
        return i;
 
    // Finding the root and applying
    // path compression
    else
        return par[i] = root(par[i]);
}
 
// Function to merge two components
void merge(int a, int b)
{
 
    // Find the roots of both
    // the components
    int p = root(a);
    int q = root(b);
 
    // If both the nodes already belong
    // to the same compenent
    if (p == q)
        return;
 
    // Union by rank, the rank in
    // this case is the size of
    // the component.
    // Smaller size will be
    // merged into larger,
    // so the larger's root
    // will be the final root
    if (size[p] > size[q])
        swap(p, q);
 
    par[p] = q;
    size[q] += size[p];
}
 
// Function to find the
// maximum component size
// after the addition of
// each edge to the graph
void maxSize(vector<pair<int, int> > e, int n)
{
 
    // Initialising the disjoint set
    for (int i = 1; i < n + 1; i++) {
 
        // Each node is the root and
        // each component size is 1
        par[i] = i;
        size[i] = 1;
    }
 
    vector<int> answer;
 
    // A multiset is being used to store
    // the size of the components
    // because multiple components
    // can have same sizes
    multiset<int> compSizes;
    for (int i = 1; i <= n; i++)
        compSizes.insert(size[i]);
 
    // At each step; add a new edge,
    // merge the components
    // and find the max
    // sized component
    for (auto edge : e) {
 
        // Merge operation is required only when
        // both the nodes don't belong to the
        // same component
        if (root(edge.first) != root(edge.second)) {
 
            // Sizes of the compenents
            int size1 = size[root(edge.first)];
            int size2 = size[root(edge.second)];
 
            // Remove the previous component sizes
            compSizes.erase(compSizes.find(size1));
            compSizes.erase(compSizes.find(size2));
 
            // Perform the merge operation
            merge(edge.first, edge.second);
 
            // Insert the combined size
            compSizes.insert(size1 + size2);
        }
 
        // Maximum value in the multiset is
        // the max component size
        answer.push_back(*compSizes.rbegin());
    }
 
    // Printing the answer
    for (int i = 0; i < answer.size(); i++) {
        cout << answer[i] << " ";
    }
}
 
// Driver code
int main()
{
    int N = 4;
    vector<pair<int, int> > E;
    E.push_back(make_pair(1, 2));
    E.push_back(make_pair(3, 4));
    E.push_back(make_pair(2, 3));
 
    maxSize(E, N);
 
    return 0;
}
Output: 
2 2 4

 

Time Complexity: O(|E| * log(N)) 
Auxiliary Space: O(N) 
 




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