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 numbersint 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 codeint 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 arrayimport 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 ofall the composite numbers in given array'''import math as mt'''function to find the product of all compositenumbers 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 codearr =[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 arrayusing 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 numbersfunction 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 numbersfunction 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 codelet 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> |
Output
24
Complexity Analysis:
- Time Complexity: O(n + max_val2)
- Auxiliary Space: O(max_val)
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