Data Modelling & AI Data Structure & Algorithm

Sum of element whose prime factors are present in array

23 June 2025

0

Given an array arr[] of non-negative integers where 2 ? arr[i] ? 10⁶. The task is to find the sum of all those elements from the array whose prime factors are present in the same array. Examples:

Input: arr[] = {2, 3, 10} Output: 5 Factor of 2 is 2 which is present in the array Factor of 3 is 3, also present in the array Factors of 10 are 2 and 5, out of which only 2 is present in the array. So, sum = 2 + 3 = 5 Input: arr[] = {5, 11, 55, 25, 100} Output: 96

Approach: The idea is to first calculate the least prime factor till maximum element of the array with Prime Factorization using Sieve.

Now, Iterate the array and for an element arr[i]
If arr[i] != 1:
- If leastPrimeFactor(arr[i]) is present in the array then update arr[i] / leastPrimeFactor(arr[i]) and go to step 2.
- Else go to step 1.
Else update sum = sum + originalVal(arr[i]).
Print the sum in the end.

Below is the implementation of the above approach:

C++

// C++ program to find the sum of the elements of an array
// whose prime factors are present in the same array
#include <bits/stdc++.h>
using namespace std;
 
#define MAXN 1000001
 
// Stores smallest prime factor for every number
int spf[MAXN];
 
// Function to calculate SPF (Smallest Prime Factor)
// for every number till MAXN
void sieve()
{
    spf[1] = 1;
    for (int i = 2; i < MAXN; i++)
 
        // Marking smallest prime factor for every
        // number to be itself.
        spf[i] = i;
 
    // Separately marking spf for every even
    // number as 2
    for (int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
 
    for (int i = 3; i * i < MAXN; i++) {
 
        // If i is prime
        if (spf[i] == i) {
 
            // Marking SPF for all numbers divisible by i
            for (int j = i * i; j < MAXN; j += i)
 
                // Marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
        }
    }
}
 
// Function to return the sum of the elements of an array
// whose prime factors are present in the same array
int sumFactors(int arr[], int n)
{
 
    // Function call to calculate smallest prime factors of
    // all the numbers upto MAXN
    sieve();
 
    // Create map for each element
    std::map<int, int> map;
 
    for (int i = 0; i < n; ++i)
        map[arr[i]] = 1;
 
    int sum = 0;
 
    for (int i = 0; i < n; ++i) {
        int num = arr[i];
 
        // If smallest prime factor of num is present in array
        while (num != 1 && map[spf[num]] == 1) {
            num /= spf[num];
        }
 
        // Each factor of arr[i] is present in the array
        if (num == 1)
            sum += arr[i];
    }
 
    return sum;
}
 
// Driver program
int main()
{
    int arr[] = { 5, 11, 55, 25, 100 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call to print required answer
    cout << sumFactors(arr, n);
    return 0;
}

Java

// Java program to find the sum of the elements of an array 
// whose prime factors are present in the same array 
 
import java.util.*;  
 
public class GFG{
 
final static int MAXN = 1000001 ;
 
// Stores smallest prime factor for every number 
static int spf[] = new int [MAXN]; 
 
    // Function to calculate SPF (Smallest Prime Factor) 
    // for every number till MAXN 
    static void sieve() 
    { 
        spf[1] = 1; 
        for (int i = 2; i < MAXN; i++) 
     
            // Marking smallest prime factor for every 
            // number to be itself. 
            spf[i] = i; 
     
        // Separately marking spf for every even 
        // number as 2 
        for (int i = 4; i < MAXN; i += 2) 
            spf[i] = 2; 
     
        for (int i = 3; i * i < MAXN; i++) { 
     
            // If i is prime 
            if (spf[i] == i) { 
     
                // Marking SPF for all numbers divisible by i 
                for (int j = i * i; j < MAXN; j += i) 
     
                    // Marking spf[j] if it is not 
                    // previously marked 
                    if (spf[j] == j) 
                        spf[j] = i; 
            } 
        } 
    } 
     
    // Function to return the sum of the elements of an array 
    // whose prime factors are present in the same array 
    static int sumFactors(int arr[], int n) 
    { 
     
        // Function call to calculate smallest prime factors of 
        // all the numbers upto MAXN 
        sieve(); 
     
        // Create map for each element 
        Map map=new HashMap();
         
        for(int i = 0 ; i < MAXN ; ++i)
            map.put(i,0) ;
             
        for (int i = 0; i < n; ++i) 
            map.put(arr[i],1); 
         
     
        int sum = 0; 
     
        for (int i = 0; i < n; ++i) { 
            int num = arr[i]; 
     
            // If smallest prime factor of num is present in array 
            while (num != 1 && (int)(map.get(spf[num])) == 1) { 
                num /= spf[num]; 
            } 
     
            // Each factor of arr[i] is present in the array 
            if (num == 1) 
                sum += arr[i]; 
        } 
     
        return sum; 
    } 
     
    // Driver program 
     public static void main(String []args)
    { 
        int arr[] = { 5, 11, 55, 25, 100 }; 
        int n = arr.length ;
     
        // Function call to print required answer 
         System.out.println(sumFactors(arr, n)) ;
    } 
    // This code is contributed by Ryuga
}

Python3

# Python program to find the sum of the 
# elements of an array whose prime factors
# are present in the same array 
from collections import defaultdict
 
MAXN = 1000001
MAXN_sqrt = int(MAXN ** (0.5))
 
# Stores smallest prime factor
# for every number 
spf = [None] * (MAXN) 
 
# Function to calculate SPF (Smallest 
# Prime Factor) for every number till MAXN 
def sieve():
 
    spf[1] = 1
    for i in range(2, MAXN): 
 
        # Marking smallest prime factor 
        # for every number to be itself. 
        spf[i] = i 
 
    # Separately marking spf for every 
    # even number as 2 
    for i in range(4, MAXN, 2): 
        spf[i] = 2
 
    for i in range(3, MAXN_sqrt): 
 
        # If i is prime 
        if spf[i] == i:
 
            # Marking SPF for all numbers 
            # divisible by i 
            for j in range(i * i, MAXN, i): 
 
                # Marking spf[j] if it is  
                # not previously marked 
                if spf[j] == j:
                    spf[j] = i 
         
# Function to return the sum of the elements 
# of an array whose prime factors are present 
# in the same array 
def sumFactors(arr, n): 
 
    # Function call to calculate smallest 
    # prime factors of all the numbers upto MAXN 
    sieve() 
 
    # Create map for each element 
    Map = defaultdict(lambda:0) 
 
    for i in range(0, n): 
        Map[arr[i]] = 1
 
    Sum = 0
 
    for i in range(0, n): 
        num = arr[i] 
         
        # If smallest prime factor of num
        # is present in array 
        while num != 1 and Map[spf[num]] == 1: 
            num = num // spf[num] 
         
        # Each factor of arr[i] is present 
        # in the array 
        if num == 1: 
            Sum += arr[i] 
     
    return Sum
 
# Driver Code
if __name__ == "__main__":
 
    arr = [5, 11, 55, 25, 100] 
    n = len(arr) 
 
    # Function call to print 
    # required answer 
    print(sumFactors(arr, n)) 
 
# This code is contributed by Rituraj Jain

C#

// C# program to find the sum of the elements 
// of an array whose prime factors are present
// in the same array 
using System;
using System.Collections.Generic;
 
class GFG
{
    static int MAXN = 1000001;
     
    // Stores smallest prime factor for every number 
    static int []spf = new int [MAXN]; 
 
    // Function to calculate SPF (Smallest Prime Factor) 
    // for every number till MAXN 
    static void sieve() 
    { 
        spf[1] = 1; 
        for (int i = 2; i < MAXN; i++) 
     
            // Marking smallest prime factor for 
            // every number to be itself. 
            spf[i] = i; 
     
        // Separately marking spf for every even 
        // number as 2 
        for (int i = 4; i < MAXN; i += 2) 
            spf[i] = 2; 
     
        for (int i = 3; i * i < MAXN; i++) 
        { 
     
            // If i is prime 
            if (spf[i] == i)
            { 
     
                // Marking SPF for all numbers divisible by i 
                for (int j = i * i; j < MAXN; j += i) 
     
                    // Marking spf[j] if it is not 
                    // previously marked 
                    if (spf[j] == j) 
                        spf[j] = i; 
            } 
        } 
    } 
     
    // Function to return the sum of the elements 
    // of an array whose prime factors are present 
    // in the same array 
    static int sumFactors(int []arr, int n) 
    { 
     
        // Function call to calculate smallest 
        // prime factors of all the numbers upto MAXN 
        sieve(); 
     
        // Create map for each element 
        Dictionary<int, int> map = new Dictionary<int, int>();
         
        for(int i = 0 ; i < MAXN ; ++i)
            map.Add(i, 0);
             
        for (int i = 0; i < n; ++i)
        {
            if(map.ContainsKey(arr[i]))
            {
                map[arr[i]] = 1; 
            }
            else
            {
                map.Add(arr[i], 1);
            }
        }
         
        int sum = 0; 
     
        for (int i = 0; i < n; ++i) 
        { 
            int num = arr[i]; 
     
            // If smallest prime factor of num 
            // is present in array 
            while (num != 1 && 
             (int)(map[spf[num]]) == 1) 
            { 
                num /= spf[num]; 
            } 
     
            // Each factor of arr[i] is present
            // in the array 
            if (num == 1) 
                sum += arr[i]; 
        } 
        return sum; 
    } 
     
    // Driver Code
    public static void Main(String []args)
    { 
        int []arr = { 5, 11, 55, 25, 100 }; 
        int n = arr.Length;
     
        // Function call to print required answer 
        Console.WriteLine(sumFactors(arr, n));
    } 
}
 
// This code is contributed by PrinciRaj1992

Javascript

// JavaScript program to find the sum of the 
// elements of an array whose prime factors
// are present in the same array 
let MAXN = 1000001
let MAXN_sqrt = Math.floor(MAXN ** (0.5))
 
// Stores smallest prime factor
// for every number 
let spf = new Array(MAXN) 
 
// Function to calculate SPF (Smallest 
// Prime Factor) for every number till MAXN 
function sieve()
{
    spf[1] = 1
    for (var i = 2; i < MAXN; i++) 
    {
        // Marking smallest prime factor 
        // for every number to be itself. 
        spf[i] = i 
    }
     
    // Separately marking spf for every 
    // even number as 2 
    for (var i = 4; i <  MAXN; i += 2) 
        spf[i] = 2
 
    for (var i = 3; i < MAXN_sqrt; i++) 
    {
        // If i is prime 
        if (spf[i] == i)
        {
            // Marking SPF for all numbers 
            // divisible by i 
            for (var j = i * i; j < MAXN; j += i) 
            {
                // Marking spf[j] if it is  
                // not previously marked 
                if (spf[j] == j)
                    spf[j] = i 
            }
        }
    }
}
         
// Function to return the sum of the elements 
// of an array whose prime factors are present 
// in the same array 
function sumFactors(arr, n)
{
    // Function call to calculate smallest 
    // prime factors of all the numbers upto MAXN 
    sieve() 
 
    // Create map for each element 
    let Map = {}
 
    for (var i = 0; i < n; i++) 
        Map[arr[i]] = 1
 
    let Sum = 0
 
    for (var i = 0; i < n; i++)
    {
        let num = arr[i] 
         
        // If smallest prime factor of num
        // is present in array 
        while (num != 1 && Map.hasOwnProperty(spf[num]))
            num = Math.floor(num / spf[num])
         
        // Each factor of arr[i] is present 
        // in the array 
        if (num == 1) 
            Sum += arr[i] 
    }
     
    return Sum
}
 
 
// Driver Code
let arr = [5, 11, 55, 25, 100] 
let n = arr.length 
 
// Function call to print 
// required answer 
console.log(sumFactors(arr, n)) 
 
// This code is contributed by phasing17

Output:

Time Complexity: O(MAXN*log(log(MAXN)))

Auxiliary Space: O(MAXN)

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Sum of element whose prime factors are present in array

C++

Java

Python3

C#

Javascript

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