Prime factors of LCM of array elements (original) (raw)

Last Updated : 23 Jul, 2025

Given an array arr[] such that 1 <= arr[i] <= 10^12, the task is to find prime factors of LCM of array elements.

Examples:

Input : arr[] = {1, 2, 3, 4, 5, 6, 7, 8} Output : 2 3 5 7 // LCM of n elements is 840 and 840 = 22235*7 // so prime factors would be 2, 3, 5, 7

Input : arr[] = {20, 10, 15, 60} Output : 2 3 5 // LCM of n elements is 60 and 60 = 223*5, // so prime factors would be 2,3,5

A simple solution for this problem is to find LCM of n elements in array. First initialize lcm = 1, then iterate for each element in array and find the lcm of previous result with new element using formula LCM(a, b) = (a * b) / gcd(a, b) i.e., lcm = (lcm * arr[i]) / gcd(lcm, arr[i]). After finding LCM of all n elements we can calculate all prime factors of LCM.
Since here constraints are large, we can not implement above method to solve this problem because while calculating LCM(a, b) we need to calculate a*b and if a,b both are of value 10^12 so it will exceed the limit of integer size. We proceed for this problem in another way using sieve of sundaram and prime factorization of a number. As we know if LCM(a,b) = k so any prime factor of a or b will also be the prime factor of 'k'.

• Take an array factor[] of size 10^6 and initialize it with 0 because prime factor of any number are always less than and equal to its square root and in our constraint arr[i] <= 10^12.
• Generate all primes less than and equal to 10^6 and store them in another array.
• Now one by one calculate all prime factors of each number in array and mark them as 1 in factor[] array.
• Now traverse factor[] array and print all indexes which are marked as 1 because these will be prime factors of lcm of n numbers in given array.

Below is the implementation of above idea.

Try It Yourself

C++ `

// C++ program to find prime factors of LCM of array elements #include <bits/stdc++.h> using namespace std;

const int MAX = 1000000; typedef long long int ll;

// array to store all prime less than and equal to 10^6 vector primes;

// utility function for sieve of sundaram void sieve() { int n = MAX;

// In general Sieve of Sundaram, produces primes smaller
// than (2*x + 2) for a number given number x. Since
// we want primes smaller than n, we reduce n to half
int nNew = (n)/2;

// This array is used to separate numbers of the form
// i+j+2ij from others where 1 <= i <= j
bool marked[nNew + 100];

// Initialize all elements as not marked
memset(marked, false, sizeof(marked));

// Main logic of Sundaram. Mark all numbers which do not
// generate prime number by doing 2*i+1
int tmp=sqrt(n);
for (int i=1; i<=(tmp-1)/2; i++)
    for (int j=(i*(i+1))<<1; j<=nNew; j=j+2*i+1)
        marked[j] = true;

// Since 2 is a prime number
primes.push_back(2);

// Print other primes. Remaining primes are of the form
// 2*i + 1 such that marked[i] is false.
for (int i=1; i<=nNew; i++)
    if (marked[i] == false)
        primes.push_back(2*i + 1);

}

// Function to find prime factors of n elements of // given array void primeLcm(ll arr[], int n ) { // factors[] --> array to mark all prime factors of // lcm of array elements int factors[MAX] = {0};

// One by one calculate prime factors of number
// and mark them in factors[] array
for (int i=0; i<n; i++)
{
    // copy --> duplicate of original element to
    //          perform operation
    ll copy = arr[i];

    // sqr --> square root of current number 'copy'
    // because all prime factors are always less
    // than and equal to square root of given number
    int sqr = sqrt(copy);

    // check divisibility with prime factor
    for (int j=0; primes[j]<=sqr; j++)
    {
        // if current prime number is factor of 'copy'
        if (copy%primes[j] == 0)
        {
            // divide with current prime factor until
            // it can divide the number
            while (copy%primes[j] == 0)
                copy = copy/primes[j];

            // mark current prime factor as 1 in
            // factors[] array
            factors[primes[j]] = 1;
        }
    }

    // After calculating exponents of all prime factors
    // either value of 'copy' will be 1 because of
    // complete divisibility or remaining value of
    // 'copy' will be surely a prime , so we will
    // also mark this prime as a factor
    if (copy > 1)
        factors[copy] = 1;
}

// if 2 is prime factor of lcm of all elements
// in given array
if (factors[2] == 1)
    cout << 2 << " ";

// traverse to print all prime factors of lcm of
// all elements in given array
for (int i=3; i<=MAX; i=i+2)
    if (factors[i] == 1)
        cout << i << " ";

}

// Driver program to run the case int main() { sieve(); ll arr[] = {20, 10, 15, 60}; int n = sizeof(arr)/sizeof(arr[0]); primeLcm(arr, n); return 0; }

Java

// Java program to find prime // factors of LCM of array elements import java.util.*;

class GFG {

static int MAX = 1000000;

// array to store all prime less // than and equal to 10^6 static ArrayList primes = new ArrayList();

// utility function for sieve of sundaram static void sieve() { int n = MAX;

// In general Sieve of Sundaram, 
// produces primes smaller than 
// (2*x + 2) for a number given 
// number x. Since we want primes
// smaller than n, we reduce n to half
int nNew = (n) / 2;

// This array is used to separate
// numbers of the form i+j+2ij 
// from others where 1 <= i <= j
boolean[] marked = new boolean[nNew + 100];


// Main logic of Sundaram. Mark all
// numbers which do not generate 
// prime number by doing 2*i+1
int tmp = (int)Math.sqrt(n);
for (int i = 1; i <= (tmp - 1) / 2; i++)
    for (int j = (i * (i + 1)) << 1; 
            j <= nNew; j = j + 2 * i + 1)
        marked[j] = true;

// Since 2 is a prime number
primes.add(2);

// Print other primes. Remaining 
// primes are of the form 2*i + 1 
// such that marked[i] is false.
for (int i = 1; i <= nNew; i++)
    if (marked[i] == false)
        primes.add(2 * i + 1);

}

// Function to find prime factors // of n elements of given array static void primeLcm(int[] arr, int n ) { // factors[] --> array to mark all prime // factors of lcm of array elements int[] factors = new int[MAX];

// One by one calculate prime factors of number
// and mark them in factors[] array
for (int i = 0; i < n; i++)
{
    // copy --> duplicate of original element to
    //  perform operation
    int copy = arr[i];

    // sqr --> square root of current number 'copy'
    // because all prime factors are always less
    // than and equal to square root of given number
    int sqr = (int)Math.sqrt(copy);

    // check divisibility with prime factor
    for (int j = 0; primes.get(j) <= sqr; j++)
    {
        // if current prime number is factor of 'copy'
        if (copy % primes.get(j) == 0)
        {
            // divide with current prime factor until
            // it can divide the number
            while (copy % primes.get(j) == 0)
                copy = copy / primes.get(j);

            // mark current prime factor as 1 in
            // factors[] array
            factors[primes.get(j)] = 1;
        }
    }

    // After calculating exponents of all prime factors
    // either value of 'copy' will be 1 because of
    // complete divisibility or remaining value of
    // 'copy' will be surely a prime , so we will
    // also mark this prime as a factor
    if (copy > 1)
        factors[copy] = 1;
}

// if 2 is prime factor of lcm of all elements
// in given array
if (factors[2] == 1)
    System.out.print("2 ");

// traverse to print all prime factors of lcm of
// all elements in given array
for (int i = 3; i <= MAX; i = i + 2)
    if (factors[i] == 1)
        System.out.print(i+" ");

}

// Driver code public static void main (String[] args) { sieve(); int[] arr = {20, 10, 15, 60}; int n = arr.length; primeLcm(arr, n); } }

// This code is contributed by chandan_jnu

Python3

Python3 program to find prime factors

of LCM of array elements

import math; MAX = 10000;

array to store all prime less than

and equal to 10^6

primes = [];

utility function for sieve of sundaram

def sieve():

n = MAX; 

# In general Sieve of Sundaram, produces 
# primes smaller than (2*x + 2) for a 
# number given number x. Since we want 
# primes smaller than n, we reduce n to half 
nNew = int(n / 2); 

# This array is used to separate numbers of 
# the form i+j+2ij from others where 1 <= i <= j 
marked = [False] * (nNew + 100); 


# Main logic of Sundaram. Mark all numbers 
# which do not generate prime number by
# doing 2*i+1 
tmp = int(math.sqrt(n)); 
for i in range(1, int((tmp - 1) / 2) + 1): 
    for j in range((i * (i + 1)) << 1, 
                    nNew + 1, 2 * i + 1): 
        marked[j] = True; 

# Since 2 is a prime number 
primes.append(2); 

# Print other primes. Remaining primes 
# are of the form 2*i + 1 such that 
# marked[i] is false. 
for i in range(1, nNew + 1): 
    if (marked[i] == False): 
        primes.append(2 * i + 1); 

Function to find prime factors of

n elements of given array

def primeLcm(arr, n ):

# factors[] --> array to mark all prime 
# factors of lcm of array elements 
factors = [0] * (MAX); 

# One by one calculate prime factors of 
# number and mark them in factors[] array 
for i in range(n):
    
    # copy --> duplicate of original
    # element to perform operation 
    copy = arr[i]; 

    # sqr --> square root of current number 
    # 'copy' because all prime factors are 
    # always less than and equal to square 
    # root of given number 
    sqr = int(math.sqrt(copy)); 

    # check divisibility with prime factor
    j = 0;
    while (primes[j] <= sqr): 
    
        # if current prime number is 
        # factor of 'copy' 
        if (copy % primes[j] == 0): 
            
            # divide with current prime factor 
            # until it can divide the number 
            while (copy % primes[j] == 0): 
                copy = int(copy / primes[j]); 

            # mark current prime factor as 1 
            # in factors[] array 
            factors[primes[j]] = 1; 
        j += 1; 

    # After calculating exponents of all prime 
    # factors either value of 'copy' will be 1 
    # because of complete divisibility or 
    # remaining value of 'copy' will be surely 
    # a prime, so we will also mark this prime
    # as a factor 
    if (copy > 1): 
        factors[copy] = 1; 

# if 2 is prime factor of lcm of 
# all elements in given array 
if (factors[2] == 1): 
    print("2 ", end = ""); 

# traverse to print all prime factors of 
# lcm of all elements in given array 
for i in range(3, MAX + 1, 2): 
    if (factors[i] == 1): 
        print(i, end = " "); 

Driver Code

sieve(); arr = [20, 10, 15, 60]; n = len(arr); primeLcm(arr, n);

This code is contributed by chandan_jnu

C#

// C# program to find prime // factors of LCM of array elements using System; using System.Collections;

class GFG {

static int MAX = 1000000;

// array to store all prime less // than and equal to 10^6 static ArrayList primes = new ArrayList();

// utility function for sieve of sundaram static void sieve() { int n = MAX;

// In general Sieve of Sundaram, 
// produces primes smaller than 
// (2*x + 2) for a number given 
// number x. Since we want primes
// smaller than n, we reduce n to half
int nNew = (n) / 2;

// This array is used to separate
// numbers of the form i+j+2ij 
// from others where 1 <= i <= j
bool[] marked = new bool[nNew + 100];


// Main logic of Sundaram. Mark all
// numbers which do not generate 
// prime number by doing 2*i+1
int tmp = (int)Math.Sqrt(n);
for (int i = 1; i <= (tmp - 1) / 2; i++)
    for (int j = (i * (i + 1)) << 1; 
            j <= nNew; j = j + 2 * i + 1)
        marked[j] = true;

// Since 2 is a prime number
primes.Add(2);

// Print other primes. Remaining 
// primes are of the form 2*i + 1 
// such that marked[i] is false.
for (int i = 1; i <= nNew; i++)
    if (marked[i] == false)
        primes.Add(2 * i + 1);

}

// Function to find prime factors // of n elements of given array static void primeLcm(int[] arr, int n ) { // factors[] --> array to // mark all prime factors // of lcm of array elements int[] factors = new int[MAX];

// One by one calculate prime
// factors of number and mark
// them in factors[] array
for (int i = 0; i < n; i++)
{
    // copy --> duplicate of original 
    // element to perform operation
    int copy = arr[i];

    // sqr --> square root of current
    // number 'copy' because all prime
    // factors are always less than and 
    // equal to square root of given number
    int sqr = (int)Math.Sqrt(copy);

    // check divisibility with prime factor
    for (int j = 0; (int)primes[j] <= sqr; j++)
    {
        // if current prime number is factor of 'copy'
        if (copy % (int)primes[j] == 0)
        {
            // divide with current prime factor until
            // it can divide the number
            while (copy % (int)primes[j] == 0)
                copy = copy / (int)primes[j];

            // mark current prime factor as 1 in
            // factors[] array
            factors[(int)primes[j]] = 1;
        }
    }

    // After calculating exponents of all prime factors
    // either value of 'copy' will be 1 because of
    // complete divisibility or remaining value of
    // 'copy' will be surely a prime , so we will
    // also mark this prime as a factor
    if (copy > 1)
        factors[copy] = 1;
}

// if 2 is prime factor of lcm of all elements
// in given array
if (factors[2] == 1)
    Console.Write("2 ");

// traverse to print all prime factors of lcm of
// all elements in given array
for (int i = 3; i <= MAX; i = i + 2)
    if (factors[i] == 1)
        Console.Write(i+" ");

}

// Driver code static void Main() { sieve(); int[] arr = {20, 10, 15, 60}; int n = arr.Length; primeLcm(arr, n); } }

// This code is contributed by chandan_jnu

PHP

JavaScript

`

Output:

2 3 5