# Count and Sum of composite elements in an array

• Difficulty Level : Basic
• Last Updated : 12 Sep, 2022

Given an array ‘arr’ of positive integers, the task is to count the number of composite numbers in the array.

Note: 1 is neither Prime nor Composite.

Examples:

Input: arr[] = {1, 3, 4, 5, 7}
Output:
4 is the only composite number.

Input: arr[] = {1, 2, 3, 4, 5, 6, 7}
Output:

Naive Approach: A simple solution is to traverse the array and do a primality test on every element.

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 count of those elements which are composite using the generated boolean vector.

Below is the implementation of the above approach:

## C++

 // C++ program to count the// number of composite numbers// in the given array#include using namespace std; // Function that returns the// the count of composite numbersint compositeCount(int arr[], int n, int* sum){    // 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);     // 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;        }    }     // Count all composite    // numbers in the arr[]    int count = 0;    for (int i = 0; i < n; i++)        if (!prime[arr[i]]) {            count++;            *sum = *sum + arr[i];        }     return count;} // Driver codeint main(){     int arr[] = { 1, 2, 3, 4, 5, 6, 7 };    int n = sizeof(arr) / sizeof(arr[0]);    int sum = 0;     cout << "Count of Composite Numbers = "          << compositeCount(arr, n, &sum);     cout << "\nSum of Composite Numbers = " << sum;     return 0;}

## Java

 import java.util.*; // Java program to count the// number of composite numbers// in the given array class GFG{     static int sum = 0;         // Function that returns the    // the count of composite numbers    static int compositeCount(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.        Vector prime = new Vector(max_val + 1);        for (int i = 0; i < max_val + 1; i++)        {            prime.add(i, Boolean.TRUE);        }        // Set 0 and 1 as primes as        // they don't need to be        // counted as composite numbers        prime.add(0, Boolean.TRUE);        prime.add(1, Boolean.TRUE);        for (int p = 2; p * p <= max_val; p++)        {             // If prime[p] is not changed, then            // it is a prime            if (prime.get(p) == true)            {                 // Update all multiples of p                for (int i = p * 2; i <= max_val; i += p)                {                    prime.add(i, Boolean.FALSE);                }            }        }         // Count all composite        // numbers in the arr[]        int count = 0;        for (int i = 0; i < n; i++)        {            if (!prime.get(arr[i]))            {                count++;                sum = sum + arr[i];            }        }        return count;    }     // Driver code    public static void main(String[] args)    {        int arr[] = {1, 2, 3, 4, 5, 6, 7};        int n = arr.length;         System.out.print("Count of Composite Numbers = "                + compositeCount(arr, n));         System.out.print("\nSum of Composite Numbers = " + sum);    }} // This code has been contributed by 29AjayKumar

## Python3

 # Python3 program to count the# number of composite numbers# in the given array # Function that returns the# the count of composite numbersdef compositeCount(arr, n):    Sum = 0     # 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 = [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, max_val + 1):         if p * p > max_val:            break                     # 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             # Count all composite numbers    # in the arr[]    count = 0    for i in range(n):        if (prime[arr[i]] == False):            count += 1            Sum = Sum + arr[i]         return count, Sum # Driver codearr = [1, 2, 3, 4, 5, 6, 7 ]n = len(arr)count, Sum = compositeCount(arr, n) print("Count of Composite Numbers = ", count) print("Sum of Composite Numbers = ", Sum) // This code is contributed by Mohit Kumar

## C#

 // C# program to count the// number of composite numbers// in the given arrayusing System;using System.Linq;using System.Collections; class GFG{     static int sum1=0; // Function that returns the// the count of composite numbersstatic int compositeCount(int []arr, int n, int sum){    // 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];     // Set 0 and 1 as primes as    // they don't need to be    // counted as composite numbers    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] == false)        {             // Update all multiples of p            for (int i = p * 2; i <= max_val; i += p)                prime[i] = true;        }    }     // Count all composite    // numbers in the arr[]    int count = 0;    for (int i = 0; i < n; i++)        if (prime[arr[i]])        {            count++;            sum = sum + arr[i];        }    sum1 = sum;    return count;} // Driver codestatic void Main(){     int []arr = { 1, 2, 3, 4, 5, 6, 7 };    int n = arr.Length;    int sum = 0;     Console.Write("Count of Composite Numbers = "+                    compositeCount(arr, n, sum));     Console.Write("\nSum of Composite Numbers = "+sum1);}} // This code is contributed by mits



## Javascript



Output

Count of Composite Numbers = 2
Sum of Composite Numbers = 10

Complexity Analysis:

• Time complexity : O(n log(log n))
• Space complexity: O(n) since auxiliary space is being used

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