Given an array of integers. The task is to calculate the product of all the composite numbers in an array.
Note: 1 is neither prime nor composite.
Examples:
Input: arr[] = {2, 3, 4, 5, 6, 7} Output: 24 Composite numbers are 4 and 6. So, product = 24 Input: arr[] = {11, 13, 17, 20, 19} Output: 20
Naive Approach: A simple solution is to traverse the array and do a primality test on every element. If the element is not prime nor 1, multiply it to the running product.
Time Complexity: O(Nsqrt(N))
Efficient Approach: Using Sieve of Eratosthenes generate a boolean vector upto the size of the maximum element from the array which can be used to check whether a number is prime or not. Also add 0 and 1 as a prime so that they don’t get counted as composite numbers. Now traverse the array and find the product of those elements which are composite using the generated boolean vector.
Implementation:
C++
// C++ program to find the product // of all the composite numbers // in an array #include <bits/stdc++.h> using namespace std; // Function that returns the // the product of all composite numbers int compositeProduct( 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 ); // Set 0 and 1 as primes as // they don't need to be // counted as composite numbers prime[0] = true ; prime[1] = true ; 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 ; } } // Find the product of all // composite numbers in the arr[] int product = 1; for ( int i = 0; i < n; i++) if (!prime[arr[i]]) { product *= arr[i]; } return product; } // Driver code int main() { int arr[] = { 2, 3, 4, 5, 6, 7 }; int n = sizeof (arr) / sizeof (arr[0]); cout << compositeProduct(arr, n); return 0; } |
Java
// Java program to find the product // of all the composite numbers // in an array import java.util.*; class GFG { // Function that returns the // the product of all composite numbers static int compositeProduct( 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 ); // Set 0 and 1 as primes as // they don't need to be // counted as composite numbers prime[ 0 ] = true ; prime[ 1 ] = true ; 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 ; } } } // Find the product of all // composite numbers in the arr[] int product = 1 ; for ( int i = 0 ; i < n; i++) { if (!prime[arr[i]]) { product *= arr[i]; } } return product; } // Driver code public static void main(String[] args) { int arr[] = { 2 , 3 , 4 , 5 , 6 , 7 }; int n = arr.length; System.out.println(compositeProduct(arr, n)); } } // This code has been contributed by 29AjayKumar |
Python3
''' Python3 program to find product of all the composite numbers in given array''' import math as mt ''' function to find the product of all composite numbers in the given array ''' def compositeProduct(arr, n): # find the 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 = [ True for i in range (max_val + 1 )] ''' Set 0 and 1 as primes as they don't need to be counted as composite numbers ''' prime[ 0 ] = True prime[ 1 ] = True for p in range ( 2 , mt.ceil(mt.sqrt(max_val))): # Remaining part of SIEVE ''' if prime[p] is not changed, than it is prime ''' if prime[p]: # update all multiples of p for i in range (p * 2 , max_val + 1 , p): prime[i] = False # find the product of all composite numbers in the arr[] product = 1 for i in range (n): if prime[arr[i]] = = False : product * = arr[i] return product # Driver code arr = [ 2 , 3 , 4 , 5 , 6 , 7 ] n = len (arr) print (compositeProduct(arr, n)) # contributed by Mohit kumar 29 |
C#
// C# program to find the product // of all the composite numbers // in an array using System; using System.Linq; public class GFG { // Function that returns the // the product of all composite numbers static int compositeProduct( 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. bool [] prime = new bool [max_val + 1]; for ( int i = 0; i < max_val + 1; i++) prime[i] = true ; // Set 0 and 1 as primes as // they don't need to be // counted as composite numbers prime[0] = true ; prime[1] = true ; 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 ; } } } // Find the product of all // composite numbers in the arr[] int product = 1; for ( int i = 0; i < n; i++) { if (!prime[arr[i]]) { product *= arr[i]; } } return product; } // Driver code public static void Main() { int [] arr = { 2, 3, 4, 5, 6, 7 }; int n = arr.Length; Console.WriteLine(compositeProduct(arr, n)); } } /* This code contributed by PrinciRaj1992 */ |
PHP
<?php // PHP program to find the product // of all the composite numbers // in an array // Function that returns the // the product of all composite numbers function compositeProduct( $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 + 1, true); // Set 0 and 1 as primes as // they don't need to be // counted as composite numbers $prime [0] = true; $prime [1] = true; 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; } } // Find the product of all // composite numbers in the arr[] $product = 1; for ( $i = 0; $i < $n ; $i ++) if (! $prime [ $arr [ $i ]]) { $product *= $arr [ $i ]; } return $product ; } // Driver code $arr = array ( 2, 3, 4, 5, 6, 7 ); $n = count ( $arr ); echo compositeProduct( $arr , $n ); // This code is contributed by mits ?> |
Javascript
<script> // Javascript program to find the product // of all the composite numbers // in an array // Function that returns the // the product of all composite numbers function compositeProduct(arr, n) { // Find maximum value in the array let max_val = arr.sort((A, B) => B - A)[0]; // 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. let prime = new Array(max_val + 1).fill( true ); // Set 0 and 1 as primes as // they don't need to be // counted as composite numbers prime[0] = true ; prime[1] = true ; for (let 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 (let i = p * 2; i <= max_val; i += p) prime[i] = false ; } } // Find the product of all // composite numbers in the arr[] let product = 1; for (let i = 0; i < n; i++) if (!prime[arr[i]]) { product *= arr[i]; } return product; } // Driver code let arr = new Array( 2, 3, 4, 5, 6, 7 ); let n = arr.length; document.write(compositeProduct(arr, n)); // This code is contributed by gfgking </script> |
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Complexity Analysis:
- Time Complexity: O(n + max_val2)
- Auxiliary Space: O(max_val)
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