# Sum of array elements which are prime factors of a given number

Given an array arr[] of size N and a positive integer K, the task is to find the sum of all array elements which are prime factors of K.

Examples:

Input: arr[] = {1, 2, 3, 5, 6, 7, 15}, K = 35
Output: 12
Explanation: From the given array, 5 and 7 are prime factors of 35. Therefore, required sum = 5 + 7 = 12.

Input: arr[] = {1, 3, 5, 7}, K = 42
Output: 10
Explanation: From the given array, 3 and 7 are prime factors of 42. Therefore, required sum = 3 + 7 = 10.

Approach: The idea is to traverse the array and for each array element, check if it is a prime factor of K or not. Add those elements to the sum, for which the condition satisfies. Follow the steps below to solve the problem:

• Initialize a variable, say sum, to store the required sum.
• Traverse the given array, and for each array element, perform the following operations:
• After complete traversal of the array, print the value of the sum as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to check if a` `// number is prime or not` `bool` `isPrime(``int` `n)` `{` `    ``// Corner cases` `    ``if` `(n <= 1)` `        ``return` `false``;` `    ``if` `(n <= 3)` `        ``return` `true``;`   `    ``// Check if n is a` `    ``// multiple of 2 or 3` `    ``if` `(n % 2 == 0 || n % 3 == 0)` `        ``return` `false``;`   `    ``// Above condition allows to check only` `    ``// for every 6th number, starting from 5` `    ``for` `(``int` `i = 5; i * i <= n; i = i + 6)`   `        ``// If n is a multiple of i and i + 2` `        ``if` `(n % i == 0 || n % (i + 2) == 0)` `            ``return` `false``;`   `    ``return` `true``;` `}`   `// Function to find the sum of array` `// elements which are prime factors of K` `void` `primeFactorSum(``int` `arr[], ``int` `n, ``int` `k)` `{`   `    ``// Stores the required sum` `    ``int` `sum = 0;`   `    ``// Traverse the given array` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// If current element is a prime` `        ``// factor of k, add it to the sum` `        ``if` `(k % arr[i] == 0 && isPrime(arr[i])) {` `            ``sum = sum + arr[i];` `        ``}` `    ``}`   `    ``// Print the result` `    ``cout << sum;` `}`   `// Driver Code` `int` `main()` `{`   `    ``// Given arr[]` `    ``int` `arr[] = { 1, 2, 3, 5, 6, 7, 15 };`   `    ``// Store the size of the array` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``int` `K = 35;`   `    ``primeFactorSum(arr, N, K);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;`   `class` `GFG ` `{`   `    ``// Function to check if a` `    ``// number is prime or not` `    ``static` `boolean` `isPrime(``int` `n)` `    ``{` `      `  `        ``// Corner cases` `        ``if` `(n <= ``1``)` `            ``return` `false``;` `        ``if` `(n <= ``3``)` `            ``return` `true``;`   `        ``// Check if n is a` `        ``// multiple of 2 or 3` `        ``if` `(n % ``2` `== ``0` `|| n % ``3` `== ``0``)` `            ``return` `false``;`   `        ``// Above condition allows to check only` `        ``// for every 6th number, starting from 5` `        ``for` `(``int` `i = ``5``; i * i <= n; i = i + ``6``)`   `            ``// If n is a multiple of i and i + 2` `            ``if` `(n % i == ``0` `|| n % (i + ``2``) == ``0``)` `                ``return` `false``;`   `        ``return` `true``;` `    ``}`   `    ``// Function to find the sum of array` `    ``// elements which are prime factors of K` `    ``static` `void` `primeFactorSum(``int` `arr[], ``int` `n, ``int` `k)` `    ``{`   `        ``// Stores the required sum` `        ``int` `sum = ``0``;`   `        ``// Traverse the given array` `        ``for` `(``int` `i = ``0``; i < n; i++) {`   `            ``// If current element is a prime` `            ``// factor of k, add it to the sum` `            ``if` `(k % arr[i] == ``0` `&& isPrime(arr[i])) ` `            ``{` `                ``sum = sum + arr[i];` `            ``}` `        ``}`   `        ``// Print the result` `        ``System.out.println(sum);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``// Given arr[]` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``5``, ``6``, ``7``, ``15` `};`   `        ``// Store the size of the array` `        ``int` `N = arr.length;` `        ``int` `K = ``35``;` `        ``primeFactorSum(arr, N, K);` `    ``}` `}`   `// This code is contributed by Kingash.`

## Python3

 `# Python3 program for the above approach`   `# Function to check if a` `# number is prime or not` `def` `isPrime(n):` `    `  `    ``# Corner cases` `    ``if` `(n <``=` `1``):` `        ``return` `False` `    ``if` `(n <``=` `3``):` `        ``return` `True`   `    ``# Check if n is a` `    ``# multiple of 2 or 3` `    ``if` `(n ``%` `2` `=``=` `0` `or` `n ``%` `3` `=``=` `0``):` `        ``return` `False`   `    ``# Above condition allows to check only` `    ``# for every 6th number, starting from 5` `    ``i ``=` `5` `    `  `    ``while` `(i ``*` `i <``=` `n ):`   `        ``# If n is a multiple of i and i + 2` `        ``if` `(n ``%` `i ``=``=` `0` `or` `n ``%` `(i ``+` `2``) ``=``=` `0``):` `            ``return` `False` `        `  `        ``i ``=` `i ``+` `6` `        `  `    ``return` `True`   `# Function to find the sum of array` `# elements which are prime factors of K` `def` `primeFactorSum(arr, n, k):`   `    ``# Stores the required sum` `    ``sum` `=` `0`   `    ``# Traverse the given array` `    ``for` `i ``in` `range``(n):`   `        ``# If current element is a prime` `        ``# factor of k, add it to the sum` `        ``if` `(k ``%` `arr[i] ``=``=` `0` `and` `isPrime(arr[i])):` `            ``sum` `=` `sum` `+` `arr[i]` `        `  `    ``# Print the result` `    ``print``(``sum``)`   `# Driver Code`   `# Given arr[]` `arr ``=` `[ ``1``, ``2``, ``3``, ``5``, ``6``, ``7``, ``15` `]`   `# Store the size of the array` `N ``=` `len``(arr)`   `K ``=` `35`   `primeFactorSum(arr, N, K)`   `# This code is contributed by code_hunt`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG ` `{`   `    ``// Function to check if a` `    ``// number is prime or not` `    ``static` `bool` `isPrime(``int` `n)` `    ``{` `      `  `        ``// Corner cases` `        ``if` `(n <= 1)` `            ``return` `false``;` `        ``if` `(n <= 3)` `            ``return` `true``;`   `        ``// Check if n is a` `        ``// multiple of 2 or 3` `        ``if` `(n % 2 == 0 || n % 3 == 0)` `            ``return` `false``;`   `        ``// Above condition allows to check only` `        ``// for every 6th number, starting from 5` `        ``for` `(``int` `i = 5; i * i <= n; i = i + 6)`   `            ``// If n is a multiple of i and i + 2` `            ``if` `(n % i == 0 || n % (i + 2) == 0)` `                ``return` `false``;`   `        ``return` `true``;` `    ``}`   `    ``// Function to find the sum of array` `    ``// elements which are prime factors of K` `    ``static` `void` `primeFactorSum(``int` `[]arr, ``int` `n, ``int` `k)` `    ``{`   `        ``// Stores the required sum` `        ``int` `sum = 0;`   `        ``// Traverse the given array` `        ``for` `(``int` `i = 0; i < n; i++) {`   `            ``// If current element is a prime` `            ``// factor of k, add it to the sum` `            ``if` `(k % arr[i] == 0 && isPrime(arr[i])) ` `            ``{` `                ``sum = sum + arr[i];` `            ``}` `        ``}`   `        ``// Print the result` `        ``Console.Write(sum);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{`   `        ``// Given arr[]` `        ``int` `[]arr = { 1, 2, 3, 5, 6, 7, 15 };`   `        ``// Store the size of the array` `        ``int` `N = arr.Length;` `        ``int` `K = 35;` `        ``primeFactorSum(arr, N, K);` `    ``}` `}`   `// This code is contributed by ukasp.`

## Javascript

 ``

Output:

`12`

Time Complexity: O(N*?X), where X is the largest element in the array
Auxiliary Space: O(1)

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