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:
#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;
} |
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 |
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# 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 |
<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)
Auxiliary Space: O(N)
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:
#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 component
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++) {
// initially 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 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 component
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++)
{
// initially 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 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 component
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 ):
# initially 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# 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 component
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++)
{
// Initially 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 |
<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 component
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++) {
// initially 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)))
Auxiliary Space: O(105)