Absolute difference between the Product of Non-Prime numbers and Prime numbers of an Array

Given an array of positive numbers, the task is to calculate the absolute difference between product of non-prime numbers and prime numbers.

Note: 1 is neither prime nor non-prime.

Examples:

Input : arr[] = {1, 3, 5, 10, 15, 7}
Output : 45
Explanation : Product of non-primes = 150
              Product of primes = 105

Input : arr[] = {3, 4, 6, 7} 
Output : 3


Naive Approach: A simple solution is to traverse the array and keep checking for every element if it is prime or not. If number is prime, then multiply it to product P2 which represents the product of primes else check if its not 1 then multiply it to product of non-primes let’s say P1. After traversing the whole array, take the absolute difference between the two(P1-P2).
Time complexity: O(N*sqrt(N))

Efficient Approach: Generate all primes up to the maximum element of the array using the sieve of Eratosthenes and store them in a hash. Now, traverse the array and check if the number is present in the hash map. Then, multiply these numbers to product P2 else check if it’s not 1, then multiply it to product P1. After traversing the whole array, display the absolute difference between the two.

Below is the implementation of the above approach:

C++

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// C++ program to find the Absolute Difference 
// between the Product of Non-Prime numbers 
// and Prime numbers of an Array 
    
#include <bits/stdc++.h> 
using namespace std; 
    
// Function to find the difference between 
// the product of non-primes and the 
// product of primes of an array. 
int calculateDifference(int arr[], int n) 
    // Find maximum value in the array 
    int max_val = *max_element(arr, arr + n); 
    
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    // THAN OR EQUAL TO max_val 
    // Create a boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    vector<bool> prime(max_val + 1, true); 
    
    // Remaining part of SIEVE 
    prime[0] = false
    prime[1] = false
    for (int p = 2; p * p <= max_val; p++) { 
    
        // If prime[p] is not changed, then 
        // it is a prime 
        if (prime[p] == true) { 
    
            // Update all multiples of p 
            for (int i = p * 2; i <= max_val; i += p) 
                prime[i] = false
        
    
    
    // Store the product of primes in S1 and 
    // the product of non primes in S2 
    int P1 = 1, P2 = 1; 
    for (int i = 0; i < n; i++) { 
    
        if (prime[arr[i]]) { 
    
            // the number is prime 
            P1 *= arr[i]; 
        
        else if (arr[i] != 1) { 
    
            // the number is non-prime 
            P2 *= arr[i]; 
        
    
    
    // Return the absolute difference 
    return abs(P2 - P1); 
    
// Driver Code 
int main() 
    int arr[] = { 1, 3, 5, 10, 15, 7 }; 
    int n     = sizeof(arr) / sizeof(arr[0]); 
    
    // Find the absolute difference 
    cout << calculateDifference(arr, n); 
    
    return 0; 

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Java

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// Java program to find the Absolute Difference 
// between the Product of Non-Prime numbers 
// and Prime numbers of an Array 
import java.util.*;  
import java.util.Arrays; 
import java.util.Collections;
  
    
class GFG{
  
    // Function to find the difference between 
    // the product of non-primes and the 
    // product of primes of an array. 
    public static int calculateDifference(int []arr, int n) 
    
        // Find maximum value in the array 
  
        int max_val = Arrays.stream(arr).max().getAsInt();
        
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
        // THAN OR EQUAL TO max_val 
        // Create a boolean array "prime[0..n]". A 
        // value in prime[i] will finally be false 
        // if i is Not a prime, else true. 
        boolean[] prime = new boolean[max_val + 1]; 
        Arrays.fill(prime, true);
        
        // Remaining part of SIEVE 
        prime[0] = false
        prime[1] = false
        for (int p = 2; p * p <= max_val; p++) { 
        
            // If prime[p] is not changed, then 
            // it is a prime 
            if (prime[p] == true) { 
        
                // Update all multiples of p 
                for (int i = p * 2 ;i <= max_val ;i += p) 
                    prime[i] = false
            
        
        
        // Store the product of primes in S1 and 
        // the product of non primes in S2 
        int P1 = 1, P2 = 1
        for (int i = 0; i < n; i++) { 
        
            if (prime[arr[i]]) { 
        
                // the number is prime 
                P1 *= arr[i]; 
            
            else if (arr[i] != 1) { 
        
                // the number is non-prime 
                P2 *= arr[i]; 
            
        
        
        // Return the absolute difference 
        return Math.abs(P2 - P1); 
    
        
    // Driver Code 
    public static void main(String []args) 
    
        int[] arr = new int []{ 1, 3, 5, 10, 15, 7 }; 
        int n     = arr.length; 
        
        // Find the absolute difference 
        System.out.println(calculateDifference(arr, n)); 
        
        System.exit(0);
    
}
// This code is contributed 
// by Harshit Saini  

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Python3

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# Python3 program to find the Absolute Difference 
# between the Product of Non-Prime numbers 
# and Prime numbers of an Array 
  
    
# Function to find the difference between 
# the product of non-primes and the 
# product of primes of an array. 
def calculateDifference(arr, n): 
    # Find maximum value in the array 
    max_val = max(arr)
    
    # USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    # THAN OR EQUAL TO max_val 
    # Create a boolean array "prime[0..n]". A 
    # value in prime[i] will finally be false 
    # if i is Not a prime, else true. 
  
    prime    = (max_val + 1) * [True]
    
    # Remaining part of SIEVE 
    prime[0] = False
    prime[1] = False
    p = 2
  
    while p * p <= max_val: 
  
        # If prime[p] is not changed, then 
        # it is a prime 
        if prime[p] == True
    
            # Update all multiples of p 
            for i in range(p * 2, max_val+1, p): 
                prime[i] = False
        p += 1
    
    # Store the product of primes in S1 and 
    # the product of non primes in S2 
    P1 = 1 ; P2 = 1
    for i in range(n):
  
        if prime[arr[i]]:
            # the number is prime 
            P1 *= arr[i]
  
        elif arr[i] != 1
            # the number is non-prime 
            P2 *= arr[i]
    
    # Return the absolute difference 
    return abs(P2 - P1)
    
# Driver Code 
if __name__ == '__main__':
    arr   = [ 1, 3, 5, 10, 15, 7
    n     = len(arr)
    
    # Find the absolute difference 
    print(calculateDifference(arr, n))
# This code is contributed 
# by Harshit Saini 

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C#

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// C# program to find the Absolute Difference 
// between the Product of Non-Prime numbers 
// and Prime numbers of an Array 
using System;
using System.Linq;
    
class GFG{
  
    // Function to find the difference between 
    // the product of non-primes and the 
    // product of primes of an array. 
    static int calculateDifference(int []arr, int n) 
    
        // Find maximum value in the array 
        int max_val = arr.Max();
        
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
        // THAN OR EQUAL TO max_val 
        // Create a boolean array "prime[0..n]". A 
        // value in prime[i] will finally be false 
        // if i is Not a prime, else true. 
        var prime   = Enumerable.Repeat(true,
                                    max_val+1).ToArray();
        
        // Remaining part of SIEVE 
        prime[0] = false
        prime[1] = false
        for (int p = 2; p * p <= max_val; p++) { 
        
            // If prime[p] is not changed, then 
            // it is a prime 
            if (prime[p] == true) { 
        
                // Update all multiples of p 
                for (int i = p * 2; i <= max_val; i += p) 
                    prime[i] = false
            
        
        
        // Store the product of primes in S1 and 
        // the product of non primes in S2 
        int P1 = 1, P2 = 1; 
        for (int i = 0; i < n; i++) { 
        
            if (prime[arr[i]]) { 
        
                // the number is prime 
                P1 *= arr[i]; 
            
            else if (arr[i] != 1) { 
        
                // the number is non-prime 
                P2 *= arr[i]; 
            
        
        
        // Return the absolute difference 
        return Math.Abs(P2 - P1); 
    
        
    // Driver Code 
    public static void Main() 
    
        int[] arr = new int []{ 1, 3, 5, 10, 15, 7 }; 
        int n     = arr.Length; 
        
        // Find the absolute difference 
        Console.WriteLine(calculateDifference(arr, n)); 
    
}
// This code is contributed 
// by Harshit Saini 

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PHP

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<?php
// PHP program to find the Absolute Difference 
// between the Product of Non-Prime numbers 
// and Prime numbers of an Array 
    
  
// Function to find the difference between 
// the product of non-primes and the 
// product of primes of an array. 
function calculateDifference($arr, $n){ 
  
    // Find maximum value in the array 
    $max_val = max($arr); 
    
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    // THAN OR EQUAL TO max_val 
    // Create a boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    $prime   = array_fill(0 ,$max_val ,true); 
    
    // Remaining part of SIEVE 
    $prime[0] = false; 
    $prime[1] = false; 
    for ($p = 2; $p * $p <= $max_val; $p++) { 
    
        // If prime[p] is not changed, then 
        // it is a prime 
        if ($prime[$p] == true) { 
    
            // Update all multiples of p 
            for ($i = $p * 2; $i <= $max_val; $i += $p
                $prime[$i] = false; 
        
    
    
    // Store the product of primes in S1 and 
    // the product of non primes in S2 
    $P1 = 1;
    $P2 = 1; 
    for ($i = 0; $i < $n; $i++) { 
    
        if ($prime[$arr[$i]]) { 
    
            // the number is prime 
            $P1 *= $arr[$i]; 
        
        else if ($arr[$i] != 1) { 
    
            // the number is non-prime 
            $P2 *= $arr[$i]; 
        
    
    
    // Return the absolute difference 
    return abs($P2 - $P1); 
    
// Driver Code 
$arr = array( 1, 3, 5, 10, 15, 7 ); 
$n   = count($arr, COUNT_NORMAL);
  
// Find the absolute difference 
echo CalculateDifference($arr, $n);
  
// This code is contributed 
// by Harshit Saini 
?>

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Output:

45

Time Complexity: O(N * log(log(N))
Space Complexity: O(MAX(N, max_val)), where max_val is the maximum value of an element in the given array.



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Improved By : Harshit Saini