# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ program to find the Absolute Difference  ` `// between the Product of Non-Prime numbers  ` `// and Prime numbers of an Array  ` `   `  `#include   ` `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 P1 and  ` `    ``// the product of non primes in P2  ` `    ``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;  ` `}  `

 `// 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 P1 and  ` `        ``// the product of non primes in P2  ` `        ``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   `

 `# 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 P1 and  ` `    ``# the product of non primes in P2  ` `    ``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  `

 `// 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 P1 and  ` `        ``// the product of non primes in P2  ` `        ``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  `

 ` `

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, gp6

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