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++
// 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 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 Codeint 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
// Java program to find the Absolute Difference// between the Product of Non-Prime numbers// and Prime numbers of an Arrayimport 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
# 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 Codeif __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#
// C# program to find the Absolute Difference// between the Product of Non-Prime numbers// and Prime numbers of an Arrayusing 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 |
PHP
<?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 P1 and // the product of non primes in P2 $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 differenceecho CalculateDifference($arr, $n);// This code is contributed// by Harshit Saini?> |
Javascript
<script>// Javascript 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 var max_val = Math.max.apply(Math,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. var prime = Array(max_val + 1).fill(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 P1 and // the product of non primes in P2 var 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 Math.abs(P2 - P1); } // Driver Code var arr = [ 1, 3, 5, 10, 15, 7 ]; var n = arr.length; // Find the absolute difference document.write(calculateDifference(arr, n));// This code contributed by gauravrajput1</script> |
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.
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
