Sum of all perfect numbers present in an array
Given an array arr[] containing N positive integer. The task is to find the sum of all the perfect numbers from the array.
A number is perfect if it is equal to the sum of its proper divisors i.e. the sum of its positive divisors excluding the number itself.
Examples:
Input: arr[] = {3, 6, 9}
Output: 6
Proper divisor sum of 3 = 1
Proper divisor sum of 6 = 1 + 2 + 3 = 6
Proper divisor sum of 9 = 1 + 3 = 4
Input: arr[] = {17, 6, 10, 6, 4}
Output: 12
Approach: Initialize sum = 0 and for every element of the array, find the sum of its proper divisors say sumFactors. If arr[i] = sumFactors then update the resultant sum as sum = sum + arr[i]. Print the sum in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <iostream> using namespace std; // Function to return the sum of // all the proper factors of n int sumOfFactors( int n) { int sum = 0; for ( int f = 1; f <= n / 2; f++) { // f is the factor of n if (n % f == 0) { sum += f; } } return sum; } // Function to return the required sum int getSum( int arr[], int n) { // To store the sum int sum = 0; for ( int i = 0; i < n; i++) { // If current element is non-zero and equal // to the sum of proper factors of itself if (arr[i] > 0 && arr[i] == sumOfFactors(arr[i])) { sum += arr[i]; } } return sum; } // Driver code int main() { int arr[10] = { 17, 6, 10, 6, 4 }; int n = sizeof (arr) / sizeof (arr[0]); cout << (getSum(arr, n)); return 0; } |
Java
// Java implementation of the above approach class GFG { // Function to return the sum of // all the proper factors of n static int sumOfFactors( int n) { int sum = 0 ; for ( int f = 1 ; f <= n / 2 ; f++) { // f is the factor of n if (n % f == 0 ) { sum += f; } } return sum; } // Function to return the required sum static int getSum( int [] arr, int n) { // To store the sum int sum = 0 ; for ( int i = 0 ; i < n; i++) { // If current element is non-zero and equal // to the sum of proper factors of itself if (arr[i] > 0 && arr[i] == sumOfFactors(arr[i])) { sum += arr[i]; } } return sum; } // Driver code public static void main(String[] args) { int [] arr = { 17 , 6 , 10 , 6 , 4 }; int n = arr.length; System.out.print(getSum(arr, n)); } } |
Python3
# Python3 implementation of the above approach # Function to return the sum of # all the proper factors of n def sumOfFactors(n): sum = 0 for f in range ( 1 , n / / 2 + 1 ): # f is the factor of n if (n % f = = 0 ): sum + = f return sum # Function to return the required sum def getSum(arr, n): # To store the sum sum = 0 for i in range (n): # If current element is non-zero and equal # to the sum of proper factors of itself if (arr[i] > 0 and arr[i] = = sumOfFactors(arr[i])) : sum + = arr[i] return sum # Driver code arr = [ 17 , 6 , 10 , 6 , 4 ] n = len (arr) print (getSum(arr, n)) # This code is contributed by Mohit Kumar |
C#
// C# implementation of the above approach using System; class GFG { // Function to return the sum of // all the proper factors of n static int sumOfFactors( int n) { int sum = 0; for ( int f = 1; f <= n / 2; f++) { // f is the factor of n if (n % f == 0) { sum += f; } } return sum; } // Function to return the required sum static int getSum( int [] arr, int n) { // To store the sum int sum = 0; for ( int i = 0; i < n; i++) { // If current element is non-zero and equal // to the sum of proper factors of itself if (arr[i] > 0 && arr[i] == sumOfFactors(arr[i])) { sum += arr[i]; } } return sum; } // Driver code static public void Main () { int [] arr = { 17, 6, 10, 6, 4 }; int n = arr.Length; Console.WriteLine(getSum(arr, n)); } } // This code is contributed by @ajit_0023 |
Javascript
<script> // Java script implementation of the above approach // Function to return the sum of // all the proper factors of n function sumOfFactors( n) { let sum = 0; for (let f = 1; f <= n / 2; f++) { // f is the factor of n if (n % f == 0) { sum += f; } } return sum; } // Function to return the required sum function getSum( arr, n) { // To store the sum let sum = 0; for (let i = 0; i < n; i++) { // If current element is non-zero and equal // to the sum of proper factors of itself if (arr[i] > 0 && arr[i] == sumOfFactors(arr[i])) { sum += arr[i]; } } return sum; } // Driver code let arr = [ 17, 6, 10, 6, 4 ]; let n = arr.length; document.write(getSum(arr, n)); //contributed by bobby </script> |
12
Time Complexity: O(n * max(arr)), where max(arr) is the largest element of the array arr.
Auxiliary Space: O(1)
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