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:
- Check if the array element is a prime factor of K or not.
- If found to be true, add the current element to the sum.
- 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 <bits/stdc++.h>using namespace std;// Function to check if a// number is prime or notbool 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 Kvoid 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 Codeint 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 approachimport 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 notdef 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 Kdef 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 arrayN = len(arr)K = 35primeFactorSum(arr, N, K)# This code is contributed by code_hunt |
C#
// C# program for the above approachusing 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
<script>// Javascript program for the above approach// Function to check if a// number is prime or notfunction 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 || n % 3 == 0) return false; var i; // Above condition allows to check only // for every 6th number, starting from 5 for (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 Kfunction primeFactorSum(arr, n, k){ // Stores the required sum var sum = 0; var i; // Traverse the given array for (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 document.write(sum);}// Driver Code // Given arr[] var arr = [1, 2, 3, 5, 6, 7, 15] // Store the size of the array var N = arr.length; var K = 35; primeFactorSum(arr, N, K);</script> |
Output:
12
Time Complexity: O(N*?X), where X is the largest element in the array
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
