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 nint 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 sumint 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 codeint 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 approachclass 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 ndef 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 sumdef 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 codearr = [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 approachusing 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> |
Output:
12
Time Complexity: O(n * max(arr)), where max(arr) is the largest element of the array arr.
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!
