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# Count and Sum of composite elements in an array

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 numbers``int` `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<``bool``> prime(max_val + 1, ``true``);` `    ``// Set 0 and 1 as primes as``    ``// they don't need to be``    ``// counted as composite numbers``    ``prime = ``true``;``    ``prime = ``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 code``int` `main()``{` `    ``int` `arr[] = { 1, 2, 3, 4, 5, 6, 7 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``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 numbers``def` `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 code``arr ``=` `[``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 array``using` `System;``using` `System.Linq;``using` `System.Collections;` `class` `GFG``{``    ` `static` `int` `sum1=0;` `// Function that returns the``// the count of composite numbers``static` `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 = ``false``;``    ``prime = ``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 code``static` `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`

## PHP

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## Javascript

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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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