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Largest component size in a graph formed by connecting non-co-prime nodes

  • Difficulty Level : Expert
  • Last Updated : 03 Jun, 2021
Geek Week

Given a graph with N nodes and their values defined in array A, the task is to find the largest component size in a graph by connecting non-co-prime nodes. An edge is between two nodes U and V if they are non-co-prime, which means that the greatest common divisor of A[U] and A[V] should be greater than 1.

Examples:  

Input : A = [4, 6, 15, 35]
Output : 4
Graph will be :
         4
         |
         6
         |
         15
         |
         35

Input : A = [2, 3, 6, 7, 4, 12, 21, 39]
Output : 8

Naive Approach: 
We can iterate over all the pairs of nodes and check whether they are co-prime or not. If they are not co-prime, we will connect them. Once the graph is created, we will apply Depth First Search to find the maximum component size.

Below is the implementation of the above approach:  

C++




#include <bits/stdc++.h>
using namespace std;
  
int dfs(int u, vector<int>* adj, int vis[])
{
    // mark this node as visited
    vis[u] = 1;
    int componentSize = 1;
  
    // apply dfs and add nodes belonging to this component
    for (auto it : adj[u]) {
        if (!vis[it]) {
            componentSize += dfs(it, adj, vis);
        }
    }
    return componentSize;
}
  
int maximumComponentSize(int a[], int n)
{
    // create graph and store in adjacency list form
    vector<int> adj[n];
  
    // iterate over all pair of nodes
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            // if not co-prime
            if (__gcd(a[i], a[j]) > 1)
                // build undirected graph
                adj[i].push_back(j);
            adj[j].push_back(i);
        }
    }
    int answer = 0;
  
    // visited array for dfs
    int vis[n];
    for(int k=0;k<n;k++){
      vis[k]=0;
    }
    for (int i = 0; i < n; i++) {
        if (!vis[i]) {
            answer = max(answer, dfs(i, adj, vis));
        }
    }
    return answer;
}
  
// Driver Code
int main()
{
    int n = 8;
    int A[] = { 2, 3, 6, 7, 4, 12, 21, 39 };
    cout << maximumComponentSize(A, n);
    return 0;
}

Java




import java.util.*;
 
class GFG{
 
static int dfs(int u, Vector<Integer> []adj,
               int vis[])
{
     
    // Mark this node as visited
    vis[u] = 1;
    int componentSize = 1;
 
    // Apply dfs and add nodes belonging
    // to this component
    for(int it : adj[u])
    {
        if (vis[it] == 0)
        {
            componentSize += dfs(it, adj, vis);
        }
    }
    return componentSize;
}
 
static int maximumComponentSize(int a[], int n)
{
     
    // Create graph and store in adjacency
    // list form
    @SuppressWarnings("unchecked")
    Vector<Integer> []adj = new Vector[n];
    for(int i = 0; i < adj.length; i++)
        adj[i] = new Vector<Integer>();
         
    // Iterate over all pair of nodes
    for(int i = 0; i < n; i++)
    {
        for(int j = i + 1; j < n; j++)
        {
             
            // If not co-prime
            if (__gcd(a[i], a[j]) > 1)
             
                // Build undirected graph
                adj[i].add(j);
                 
            adj[j].add(i);
        }
    }
    int answer = 0;
 
    // Visited array for dfs
    int []vis = new int[n];
    for(int k = 0; k < n; k++)
    {
        vis[k] = 0;
    }
     
    for(int i = 0; i < n; i++)
    {
        if (vis[i] == 0)
        {
            answer = Math.max(answer,
                              dfs(i, adj, vis));
        }
    }
    return answer;
}
 
static int __gcd(int a, int b)
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 8;
    int A[] = { 2, 3, 6, 7, 4, 12, 21, 39 };
     
    System.out.print(maximumComponentSize(A, n));
}
}
 
// This code is contributed by Amit Katiyar

Python3




from math import gcd
def dfs(u, adj, vis):
 
    # mark this node as visited
    vis[u] = 1
    componentSize = 1
 
    # apply dfs and add nodes belonging to this component
    for x in adj[u]:
        if (vis[x] == 0):
            componentSize += dfs(x, adj, vis)
    return componentSize
 
def maximumComponentSize(a,n):
 
    # create graph and store in adjacency list form
    adj = [[] for i in range(n)]
 
    # iterate over all pair of nodes
    for i in range(n):
        for j in range(i + 1, n):
            # if not co-prime
            if (gcd(a[i], a[j]) > 1):
                # build undirected graph
                adj[i].append(j)
            adj[j].append(i)
    answer = 0
 
    # visited array for dfs
    vis = [0 for i in range(n)]
    for i in range(n):
        if (vis[i]==False):
            answer = max(answer, dfs(i, adj, vis))
    return answer
 
# Driver Code
if __name__ == '__main__':
    n = 8
    A = [2, 3, 6, 7, 4, 12, 21, 39]
    print(maximumComponentSize(A, n))
 
# This code is contributed by Bhupendra_Singh

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
static int dfs(int u,
               List<int> []adj,
               int []vis)
{   
  // Mark this node as visited
  vis[u] = 1;
  int componentSize = 1;
 
  // Apply dfs and add nodes belonging
  // to this component
  foreach(int it in adj[u])
  {
    if (vis[it] == 0)
    {
      componentSize += dfs(it,
                           adj, vis);
    }
  }
  return componentSize;
}
 
  static int maximumComponentSize(int []a,
                                  int n)
  {   
    // Create graph and store in adjacency
    // list form
 
    List<int> []adj = new List<int>[n];
    for(int i = 0; i < adj.Length; i++)
      adj[i] = new List<int>();
 
    // Iterate over all pair of nodes
    for(int i = 0; i < n; i++)
    {
      for(int j = i + 1; j < n; j++)
      {           
        // If not co-prime
        if (__gcd(a[i], a[j]) > 1)
 
          // Build undirected graph
          adj[i].Add(j);
 
        adj[j].Add(i);
      }
    }
    int answer = 0;
 
    // Visited array for dfs
    int []vis = new int[n];
    for(int k = 0; k < n; k++)
    {
      vis[k] = 0;
    }
 
    for(int i = 0; i < n; i++)
    {
      if (vis[i] == 0)
      {
        answer = Math.Max(answer,
                          dfs(i,
                              adj, vis));
      }
    }
    return answer;
}
 
static int __gcd(int a, int b)
  return b == 0 ? a :
         __gcd(b, a % b);    
}
 
// Driver Code
public static void Main(String[] args)
{
  int n = 8;
  int []A = {2, 3, 6, 7,
             4, 12, 21, 39};
  Console.Write(maximumComponentSize(A, n));
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// Javascript program to implement
// the above approach
 
function dfs(u, adj, vis)
{  
      // Mark this node as visited
    vis[u] = 1;
    let componentSize = 1;
  
      // Apply dfs and add nodes belonging
      // to this component
    for(let it in adj[u])
    {
          if (vis[it] == 0)
        {
              componentSize += dfs(it,
                           adj, vis);
        }
      }
      return componentSize;
}
  
function maximumComponentSize(a, n)
{  
    // Create graph and store in adjacency
    // list form
  
    let adj = new Array(n);
    for (var i = 0; i < adj.length; i++) {
        adj[i] = new Array(2);
    }
    for (var i = 0; i < adj.length; i++) {
        for (var j = 0; j < adj.length; j++) {
            adj[i][j] = 0;
        }
    }
  
    // Iterate over all pair of nodes
    for(let i = 0; i < n; i++)
    {
        for(let j = i + 1; j < n; j++)
          {          
            // If not co-prime
            if (__gcd(a[i], a[j]) > 1)
  
              // Build undirected graph
              adj[i].push(j);
  
            adj[j].push(i);
          }
    }
    let answer = 0;
  
    // Visited array for dfs
    let vis = Array.from({length: n}, (_, i) => 0);
    for(let k = 0; k < n; k++)
    {
          vis[k] = 0;
    }
  
    for(let i = 0; i < n; i++)
    {
          if (vis[i] == 0)
          {
            answer = Math.max(answer,
                          dfs(i,
                              adj, vis));
          }
       }
    return answer;
}
  
function __gcd(a, b)
{
      return b == 0 ? a :
         __gcd(b, a % b);   
}
 
// Driver Code
     
    let n = 8;
    let A = [2, 3, 6, 7, 4, 12, 21, 39];
      
    document.write(maximumComponentSize(A, n));
                
</script>

Output:



8

Time complexity: O(N2)

Efficient Approach

  • For any two numbers to be non-co-prime, they must have at least one common factor. So, instead of traversing through all the pairs, it’s better to prime factorize each node value. The idea is to then club together numbers with common factors as a single group.
  • Prime factorization can be done efficiently using the Sieve of Eratosthenes. For clubbing together of nodes we will use a Disjoint set data structure (Union by Rank and Path Compression).
  • The following information will be stored :

par[i] -> represents the parent of node i 
size[i] -> represents the size of the component node i belongs to 
id[p] -> represents which node prime number p was first seen as a factor of A[i] 

  • For each node value, we will factorize and store the prime factors in set S. Iterate each element of S. If the prime number is seen for the first time as a factor of some number (id[p] is zero), then mark this prime id with the current index. If this prime has been marked, then simply merge this node with that of id[p].
    This way, all nodes will belong to some component finally, and size[i] will be the size of component node i belongs to.

Below is the implementation of the above approach:

C++




#include <bits/stdc++.h>
  
using namespace std;
  
// smallest prime factor
int spf[100005];
  
// Sieve of Eratosthenes
void sieve()
{
    for (int i = 2; i < 100005; i++) {
        // is spf[i] = 0, then it's prime
        if (spf[i] == 0) {
            spf[i] = i;
 
            for (int j = 2 * i; j < 100005; j += i) {
 
                // smallest prime factor of all multiples is i
                if (spf[j] == 0)
                    spf[j] = i;
            }
        }
    }
}
  
// Prime factorise n,
// and store prime factors in set s
void factorize(int n, set<int>& s)
{
  
    while (n > 1) {
        int z = spf[n];
        s.insert(z);
        while (n % z == 0)
            n /= z;
    }
}
  
// for implementing DSU
int id[100005];
int par[100005];
int sizeContainer[100005];
  
// root of component of node i
int root(int i)
{
    if (par[i] == i)
        return i;
    // finding root as well as applying
    // path compression
    else
        return par[i] = root(par[i]);
}
  
// merging two components
void merge(int a, int b)
{
  
    // find roots of both components
    int p = root(a);
    int q = root(b);
  
    // if already belonging to the same compenent
    if (p == q)
        return;
  
    // Union by rank, the rank in this case is
    // sizeContainer of the component.
    // Smaller sizeContainer will be merged into
    // larger, so the larger's root will be
    // final root
    if (sizeContainer[p] > sizeContainer[q])
        swap(p, q);
  
    par[p] = q;
    sizeContainer[q] += sizeContainer[p];
}
  
// Function to find the maximum sized container
int maximumComponentsizeContainer(int a[], int n)
{
  
    // intitalise the parents,
    // and component sizeContainer
    for (int i = 0; i < 100005; i++) {
        // intitally all component sizeContainers are 1
        // ans each node it parent of itself
        par[i] = i;
        sizeContainer[i] = 1;
    }
  
    sieve();
  
    for (int i = 0; i < n; i++) {
        // store prime factors of a[i] in s
        set<int> s;
        factorize(a[i], s);
  
        for (auto it : s) {
            // if this prime is seen as a factor
            // for the first time
            if (id[it] == 0)
                id[it] = i + 1;
            // if not then merge with that component
            // in which this prime was previously seen
            else
                merge(i + 1, id[it]);
        }
    }
  
    int answer = 0;
 
    // maximum of sizeContainer of all components
    for (int i = 0; i < n; i++)
        answer = max(answer, sizeContainer[i]);
  
    return answer;
}
 
// Driver Code
int main()
{
    int n = 8;
    int A[] = { 2, 3, 6, 7, 4, 12, 21, 39 };
  
    cout << maximumComponentsizeContainer(A, n);
  
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
class GFG{
  
// smallest prime factor
static int []spf = new int[100005];
  
// Sieve of Eratosthenes
static void sieve()
{
    for (int i = 2; i < 100005; i++)
    {
        // is spf[i] = 0, then it's prime
        if (spf[i] == 0)
        {
            spf[i] = i;
            for (int j = 2 * i; j < 100005; j += i)
            {
                // smallest prime factor of all
                // multiples is i
                if (spf[j] == 0)
                    spf[j] = i;
            }
        }
    }
}
  
// Prime factorise n,
// and store prime factors in set s
static void factorize(int n, HashSet<Integer> s)
{
    while (n > 1)
    {
        int z = spf[n];
        s.add(z);
        while (n % z == 0)
            n /= z;
    }
}
  
// for implementing DSU
static int []id = new int[100005];
static int []par = new int[100005];
static int []sizeContainer = new int[100005];
  
// root of component of node i
static int root(int i)
{
    if (par[i] == i)
        return i;
    // finding root as well as applying
    // path compression
    else
        return par[i] = root(par[i]);
}
  
// merging two components
static void merge(int a, int b)
{
    // find roots of both components
    int p = root(a);
    int q = root(b);
  
    // if already belonging to the same compenent
    if (p == q)
        return;
  
    // Union by rank, the rank in this case is
    // sizeContainer of the component.
    // Smaller sizeContainer will be merged into
    // larger, so the larger's root will be
    // final root
    if (sizeContainer[p] > sizeContainer[q])
    {
        p = p + q;
        q = p - q;
        p = p - q;
    }
    par[p] = q;
    sizeContainer[q] += sizeContainer[p];
}
  
// Function to find the maximum sized container
static int maximumComponentsizeContainer(int a[],
                                         int n)
{
    // intitalise the parents,
    // and component sizeContainer
    for (int i = 0; i < 100005; i++)
    {
        // intitally all component sizeContainers are 1
        // ans each node it parent of itself
        par[i] = i;
        sizeContainer[i] = 1;
    }
    sieve();
  
    for (int i = 0; i < n; i++)
    {
        // store prime factors of a[i] in s
        HashSet<Integer> s = new HashSet<Integer>();
        factorize(a[i], s);
  
        for (int it : s)
        {
            // if this prime is seen as a factor
            // for the first time
            if (id[it] == 0)
                id[it] = i + 1;
            // if not then merge with that component
            // in which this prime was previously seen
            else
                merge(i + 1, id[it]);
        }
    }
    int answer = 0;
 
    // maximum of sizeContainer of all components
    for (int i = 0; i < n; i++)
        answer = Math.max(answer, sizeContainer[i]);
  
    return answer;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 8;
    int A[] = {2, 3, 6, 7,
               4, 12, 21, 39};
    System.out.print(
           maximumComponentsizeContainer(A, n));
}
}
 
// This code is contributed by shikhasingrajput

Python3




# Python3 program to implement
# the above approach
   
# smallest prime factor
spf = [0 for i in range(100005)]
   
# Sieve of Eratosthenes
def sieve():
 
    for i in range(2, 100005):
     
        # is spf[i] = 0, then it's prime
        if (spf[i] == 0):
         
            spf[i] = i;
            for j in range(2 * i, 100005, i):
             
                # smallest prime factor of all
                # multiples is i
                if (spf[j] == 0):
                    spf[j] = i;
              
# Prime factorise n,
# and store prime factors in set s
def factorize(n, s):
    while (n > 1):  
        z = spf[n];
        s.add(z);
        while (n % z == 0):
            n //= z;
      
# for implementing DSU
id = [0 for i in range(100005)]
par = [0 for i in range(100005)]
sizeContainer = [0 for i in range(100005)]
   
# root of component of node i
def root(i):
 
    if (par[i] == i):
        return i;
       
    # finding root as well as applying
    # path compression
    else:
        return root(par[i]);
        return par[i]
  
# merging two components
def merge(a, b):
 
    # find roots of both components
    p = root(a);
    q = root(b);
   
    # if already belonging to the same compenent
    if (p == q):
        return;
   
    # Union by rank, the rank in this case is
    # sizeContainer of the component.
    # Smaller sizeContainer will be merged into
    # larger, so the larger's root will be
    # final root
    if (sizeContainer[p] > sizeContainer[q]):   
        p = p + q;
        q = p - q;
        p = p - q;  
    par[p] = q;
    sizeContainer[q] += sizeContainer[p];
   
# Function to find the maximum sized container
def maximumComponentsizeContainer(a, n):
 
    # intitalise the parents,
    # and component sizeContainer
    for i in range(100005):
         
        # intitally all component sizeContainers are 1
        # ans each node it parent of itself
        par[i] = i;
        sizeContainer[i] = 1;
    sieve();
     
    for i in range(n):
 
        # store prime factors of a[i] in s
        s = set()       
        factorize(a[i], s); 
        for it in s:
 
            # if this prime is seen as a factor
            # for the first time
            if (id[it] == 0):
                id[it] = i + 1;
                 
            # if not then merge with that component
            # in which this prime was previously seen
            else:
                merge(i + 1, id[it]);       
    answer = 0;
  
    # maximum of sizeContainer of all components
    for i in range(n): 
        answer = max(answer, sizeContainer[i]);
    return answer;
  
# Driver Code
if __name__=='__main__':
     
    n = 8;
    A = [2, 3, 6, 7, 4, 12, 21, 39]
    print(maximumComponentsizeContainer(A, n));
 
# This code is contributed by Pratham76

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Smallest prime factor
static int []spf = new int[100005];
 
// Sieve of Eratosthenes
static void sieve()
{
    for(int i = 2; i < 100005; i++)
    {
         
        // Is spf[i] = 0, then it's prime
        if (spf[i] == 0)
        {
            spf[i] = i;
            for(int j = 2 * i; j < 100005; j += i)
            {
                 
                // Smallest prime factor of all
                // multiples is i
                if (spf[j] == 0)
                    spf[j] = i;
            }
        }
    }
}
 
// Prime factorise n, and store
// prime factors in set s
static void factorize(int n, HashSet<int> s)
{
    while (n > 1)
    {
        int z = spf[n];
        s.Add(z);
         
        while (n % z == 0)
            n /= z;
    }
}
 
// For implementing DSU
static int []id = new int[100005];
static int []par = new int[100005];
static int []sizeContainer = new int[100005];
 
// Root of component of node i
static int root(int i)
{
    if (par[i] == i)
        return i;
         
    // Finding root as well as applying
    // path compression
    else
        return par[i] = root(par[i]);
}
 
// Merging two components
static void merge(int a, int b)
{
     
    // Find roots of both components
    int p = root(a);
    int q = root(b);
 
    // If already belonging to
    // the same compenent
    if (p == q)
        return;
 
    // Union by rank, the rank in this case is
    // sizeContainer of the component.
    // Smaller sizeContainer will be merged into
    // larger, so the larger's root will be
    // readonly root
    if (sizeContainer[p] > sizeContainer[q])
    {
        p = p + q;
        q = p - q;
        p = p - q;
    }
    par[p] = q;
    sizeContainer[q] += sizeContainer[p];
}
 
// Function to find the maximum sized container
static int maximumComponentsizeContainer(int []a,
                                         int n)
{
     
    // Initialise the parents,
    // and component sizeContainer
    for(int i = 0; i < 100005; i++)
    {
         
        // Intitally all component sizeContainers
        // are 1 ans each node it parent of itself
        par[i] = i;
        sizeContainer[i] = 1;
    }
    sieve();
 
    for(int i = 0; i < n; i++)
    {
         
        // Store prime factors of a[i] in s
        HashSet<int> s = new HashSet<int>();
        factorize(a[i], s);
 
        foreach(int it in s)
        {
             
            // If this prime is seen as a factor
            // for the first time
            if (id[it] == 0)
                id[it] = i + 1;
                 
            // If not then merge with that component
            // in which this prime was previously seen
            else
                merge(i + 1, id[it]);
        }
    }
    int answer = 0;
 
    // Maximum of sizeContainer of all components
    for(int i = 0; i < n; i++)
        answer = Math.Max(answer, sizeContainer[i]);
 
    return answer;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 8;
    int []A = { 2, 3, 6, 7,
                4, 12, 21, 39 };
                 
    Console.Write(
        maximumComponentsizeContainer(A, n));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
  
// smallest prime factor
var spf = Array(100005).fill(0);
  
// Sieve of Eratosthenes
function sieve()
{
    for (var i = 2; i < 100005; i++) {
        // is spf[i] = 0, then it's prime
        if (spf[i] == 0) {
            spf[i] = i;
 
            for (var j = 2 * i; j < 100005; j += i) {
 
                // smallest prime factor of all multiples is i
                if (spf[j] == 0)
                    spf[j] = i;
            }
        }
    }
}
  
// Prime factorise n,
// and store prime factors in set s
function factorize(n, s)
{
  
    while (n > 1) {
        var z = spf[n];
        s.add(z);
        while (n % z == 0)
            n = parseInt(n/z);
    }
    return s;
}
  
// for implementing DSU
var id = Array(100005).fill(0);
var par =Array(100005).fill(0);
var sizeContainer = Array(100005).fill(0);
  
// root of component of node i
function root(i)
{
    if (par[i] == i)
        return i;
    // finding root as well as applying
    // path compression
    else
        return par[i] = root(par[i]);
}
  
// merging two components
function merge(a, b)
{
  
    // find roots of both components
    var p = root(a);
    var q = root(b);
  
    // if already belonging to the same compenent
    if (p == q)
        return;
  
    // Union by rank, the rank in this case is
    // sizeContainer of the component.
    // Smaller sizeContainer will be merged into
    // larger, so the larger's root will be
    // final root
    if (sizeContainer[p] > sizeContainer[q])
    {
        [p, q] = [q, p]
    }
  
    par[p] = q;
    sizeContainer[q] += sizeContainer[p];
}
  
// Function to find the maximum sized container
function maximumComponentsizeContainer(a, n)
{
  
    // intitalise the parents,
    // and component sizeContainer
    for (var i = 0; i < 100005; i++) {
        // intitally all component sizeContainers are 1
        // ans each node it parent of itself
        par[i] = i;
        sizeContainer[i] = 1;
    }
  
    sieve();
  
    for (var i = 0; i < n; i++) {
        // store prime factors of a[i] in s
        var s = new Set();
        s = factorize(a[i], s);
         
        s.forEach(it => {
             
            // if this prime is seen as a factor
            // for the first time
            if (id[it] == 0)
                id[it] = i + 1;
            // if not then merge with that component
            // in which this prime was previously seen
            else
                merge(i + 1, id[it]);
        });
    }
  
    var answer = 0;
 
    // maximum of sizeContainer of all components
    for (var i = 0; i < n; i++)
        answer = Math.max(answer, sizeContainer[i]);
  
    return answer;
}
 
// Driver Code
var n = 8;
var A = [2, 3, 6, 7, 4, 12, 21, 39];
 
document.write( maximumComponentsizeContainer(A, n));
 
 
</script>

Output:

 8

Time Complexity : O(N * log(max(A)))  

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