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 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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