# Sum of all subsets whose sum is a Perfect Number from a given array

• Difficulty Level : Expert
• Last Updated : 22 Nov, 2021

Given an array arr[] consisting of N integers, the task is to find the sum of all subsets from an array, whose sum is a Perfect Number.

Examples:

Input: arr[] = {5, 4, 6}
Output: 6
Explanation:
All possible subsets from the array arr[] are:
{5} → Sum = 5
{4} → Sum = 4.
{6} → Sum = 6.
{5, 4} → Sum = 9.
{5, 6} → Sum = 11.
{4, 6} → Sum = 10.
{5, 4, 6} → Sum = 15.
Out of all subset sums, only 6 sums are found to be2` a perfect number.

Input: arr[] = {28, 6, 23, 3, 3}
Output: 28 6 6

Recursive Approach: The idea is to generate all possible subsets from the given array and print the sum of those subsets whose sum is a perfect number.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to check is a given number``// is a  perfect number or not``int` `isPerfect(``int` `x)``{``    ``// Stores the sum of its divisors``    ``int` `sum_div = 1;` `    ``// Add all divisors of x to sum_div``    ``for` `(``int` `i = 2; i <= x / 2; ++i) {``        ``if` `(x % i == 0) {``            ``sum_div += i;``        ``}``    ``}` `    ``// If the sum of divisors is equal``    ``// to the given number, return true``    ``if` `(sum_div == x) {``        ``return` `1;``    ``}` `    ``// Otherwise, return false``    ``else``        ``return` `0;``}` `// Function to find sum of all``// subsets from an array whose``// sum is a perfect number``void` `subsetSum(``int` `arr[], ``int` `l,``               ``int` `r, ``int` `sum = 0)``{``    ``// Print the current subset sum``    ``// if it is a perfect number``    ``if` `(l > r) {` `        ``// Check if sum is a``        ``// perfect number or not``        ``if` `(isPerfect(sum)) {``            ``cout << sum << ``" "``;``        ``}``        ``return``;``    ``}` `    ``// Calculate sum of the subset``    ``// including arr[l]``    ``subsetSum(arr, l + 1, r, sum + arr[l]);` `    ``// Calculate sum of the subset``    ``// excluding arr[l]``    ``subsetSum(arr, l + 1, r, sum);``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 5, 4, 6 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``subsetSum(arr, 0, N - 1);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG{``    ` `// Function to check is a given number``// is a  perfect number or not``static` `int` `isPerfect(``int` `x)``{``    ` `    ``// Stores the sum of its divisors``    ``int` `sum_div = ``1``;``  ` `    ``// Add all divisors of x to sum_div``    ``for``(``int` `i = ``2``; i <= x / ``2``; ++i)``    ``{``        ``if` `(x % i == ``0``)``        ``{``            ``sum_div += i;``        ``}``    ``}``  ` `    ``// If the sum of divisors is equal``    ``// to the given number, return true``    ``if` `(sum_div == x)``    ``{``        ``return` `1``;``    ``}``    ` `    ``// Otherwise, return false``    ``else``        ``return` `0``;``}` `// Function to find sum of all``// subsets from an array whose``// sum is a perfect number``static` `void` `subsetSum(``int``[] arr, ``int` `l,``                      ``int` `r, ``int` `sum)``{``    ` `    ``// Print the current subset sum``    ``// if it is a perfect number``    ``if` `(l > r)``    ``{``        ` `        ``// Check if sum is a``        ``// perfect number or not``        ``if` `(isPerfect(sum) != ``0``)``        ``{``            ``System.out.print(sum + ``" "``);``        ``}``        ``return``;``    ``}``  ` `    ``// Calculate sum of the subset``    ``// including arr[l]``    ``subsetSum(arr, l + ``1``, r, sum + arr[l]);``  ` `    ``// Calculate sum of the subset``    ``// excluding arr[l]``    ``subsetSum(arr, l + ``1``, r, sum);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int``[] arr = { ``5``, ``4``, ``6` `};``    ``int` `N = arr.length;``    ` `    ``subsetSum(arr, ``0``, N - ``1``, ``0``);``}``}` `// This code is contributed by code_hunt`

## Python3

 `# Python3 program for the above approach``import` `math` `# Function to check is a given number``# is a  perfect number or not``def` `isPerfect(x) :``    ` `    ``# Stores the sum of its divisors``    ``sum_div ``=` `1` `    ``# Add all divisors of x to sum_div``    ``for` `i ``in` `range``(``2``, (x ``/``/` `2``) ``+` `1``) :``        ``if` `(x ``%` `i ``=``=` `0``) :``            ``sum_div ``+``=` `i``        ` `    ``# If the sum of divisors is equal``    ``# to the given number, return true``    ``if` `(sum_div ``=``=` `x) :``        ``return` `1``    ` `    ``# Otherwise, return false``    ``else` `:``        ``return` `0` `# Function to find sum of all``# subsets from an array whose``# sum is a perfect number``def` `subsetSum(arr, l,``               ``r, ``sum``) :``  ` `    ``# Print current subset sum``    ``# if it is a perfect number``    ``if` `(l > r) :` `        ``# Check if sum is a``        ``# perfect number or not``        ``if` `(isPerfect(``sum``) !``=` `0``) :``            ``print``(``sum``, end ``=` `" "``)``        ` `        ``return``    ` `    ``# Calculate sum of the subset``    ``# including arr[l]``    ``subsetSum(arr, l ``+` `1``, r, ``sum` `+` `arr[l])` `    ``# Calculate sum of the subset``    ``# excluding arr[l]``    ``subsetSum(arr, l ``+` `1``, r, ``sum``)` `# Driver Code``arr ``=` `[ ``5``, ``4``, ``6` `]``N ``=` `len``(arr)``subsetSum(arr, ``0``, N ``-` `1``, ``0``)` `# This code is contributed by sanjoy_62.`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to check is a given number``// is a  perfect number or not``static` `int` `isPerfect(``int` `x)``{``    ` `    ``// Stores the sum of its divisors``    ``int` `sum_div = 1;``  ` `    ``// Add all divisors of x to sum_div``    ``for``(``int` `i = 2; i <= x / 2; ++i)``    ``{``        ``if` `(x % i == 0)``        ``{``            ``sum_div += i;``        ``}``    ``}``  ` `    ``// If the sum of divisors is equal``    ``// to the given number, return true``    ``if` `(sum_div == x)``    ``{``        ``return` `1;``    ``}``  ` `    ``// Otherwise, return false``    ``else``        ``return` `0;``}` `// Function to find sum of all``// subsets from an array whose``// sum is a perfect number``static` `void` `subsetSum(``int``[] arr, ``int` `l,``                      ``int` `r, ``int` `sum = 0)``{``    ` `    ``// Print the current subset sum``    ``// if it is a perfect number``    ``if` `(l > r)``    ``{``        ` `        ``// Check if sum is a``        ``// perfect number or not``        ``if` `(isPerfect(sum) != 0)``        ``{``            ``Console.Write(sum + ``" "``);``        ``}``        ``return``;``    ``}``  ` `    ``// Calculate sum of the subset``    ``// including arr[l]``    ``subsetSum(arr, l + 1, r, sum + arr[l]);``  ` `    ``// Calculate sum of the subset``    ``// excluding arr[l]``    ``subsetSum(arr, l + 1, r, sum);``}` `// Driver code``static` `void` `Main()``{``    ``int``[] arr = { 5, 4, 6 };``    ``int` `N = arr.Length;``    ``subsetSum(arr, 0, N - 1);``}``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output:

`6`

Time Complexity: O(M * 2N), where M is the sum of the elements of the array arr[]
Auxiliary Space: O(1)

Iterative Approach: Since there are 2N possible subsets from an array of size N, the idea is to iterate a loop from 0 to 2N – 1 and for every number, pick all array elements which correspond to 1s in the binary representation of the current number and then check if the sum of the chosen elements is a perfect number or not. Follow the steps below to solve the problem:

• Iterate in the range[0, 2N – 1] using the variable i and perform the following steps:
• Initialize a variable, say S with 0 to store the sum of the current subset.
• Traverse the array arr[] using the variable j and perform the following steps:
• Check if S is a perfect number or not, if found to be true print the value of S 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 is a given``// number is a perfect number or not``int` `isPerfect(``int` `x)``{``    ``// Stores sum of divisors``    ``int` `sum_div = 1;` `    ``// Add all divisors of x to sum_div``    ``for` `(``int` `i = 2; i <= x / 2; ++i) {``        ``if` `(x % i == 0) {``            ``sum_div += i;``        ``}``    ``}` `    ``// If the sum of divisors is equal``    ``// to the given number, return true``    ``if` `(sum_div == x) {``        ``return` `1;``    ``}` `    ``// Otherwise, return false``    ``else``        ``return` `0;``}` `// Function to find the sum of all the``// subsets from an array whose sum is``// a perfect number``void` `subsetSum(``int` `arr[], ``int` `n)``{``    ``// Stores the total number of``    ``// subsets, i.e. 2 ^ n``    ``long` `long` `total = 1 << n;` `    ``// Consider all numbers from 0 to 2 ^ n - 1``    ``for` `(``long` `long` `i = 0; i < total; i++) {``        ``long` `long` `sum = 0;` `        ``// Consider array elements from``        ``// positions of set bits in the``        ``// binary representation of n``        ``for` `(``int` `j = 0; j < n; j++)``            ``if` `(i & (1 << j))``                ``sum += arr[j];` `        ``// If sum of chosen elements``        ``// is a perfect number``        ``if` `(isPerfect(sum)) {``            ``cout << sum << ``" "``;``        ``}``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 5, 4, 6 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``subsetSum(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG``{` `  ``// Function to check is a given``  ``// number is a perfect number or not``  ``static` `int` `isPerfect(``int` `x)``  ``{``    ` `    ``// Stores sum of divisors``    ``int` `sum_div = ``1``;` `    ``// Add all divisors of x to sum_div``    ``for` `(``int` `i = ``2``; i <= x / ``2``; ++i) {``      ``if` `(x % i == ``0``) {``        ``sum_div += i;``      ``}``    ``}` `    ``// If the sum of divisors is equal``    ``// to the given number, return true``    ``if` `(sum_div == x) {``      ``return` `1``;``    ``}` `    ``// Otherwise, return false``    ``else``      ``return` `0``;``  ``}` `  ``// Function to find the sum of all the``  ``// subsets from an array whose sum is``  ``// a perfect number``  ``static` `void` `subsetSum(``int` `arr[], ``int` `n)``  ``{``    ` `    ``// Stores the total number of``    ``// subsets, i.e. 2 ^ n``    ``long` `total = ``1` `<< n;` `    ``// Consider all numbers from 0 to 2 ^ n - 1``    ``for` `(``long` `i = ``0``; i < total; i++) {``      ``int` `sum = ``0``;` `      ``// Consider array elements from``      ``// positions of set bits in the``      ``// binary representation of n``      ``for` `(``int` `j = ``0``; j < n; j++)``        ``if` `((i & (``1` `<< j)) != ``0``)``          ``sum += arr[j];` `      ``// If sum of chosen elements``      ``// is a perfect number``      ``if` `(isPerfect(sum) != ``0``) {``        ``System.out.print(sum + ``" "``);``      ``}``    ``}``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `arr[] = { ``5``, ``4``, ``6` `};``    ``int` `N = arr.length;``    ``subsetSum(arr, N);``  ``}``}` `// This code is contributed by souravghosh0416.`

## Python3

 `# Python3 program for the above approach` `# Function to check is a given``# number is a perfect number or not``def` `isPerfect(x):``    ` `    ``# Stores sum of divisors``    ``sum_div ``=` `1` `    ``# Add all divisors of x to sum_div``    ``for` `i ``in` `range``(``2``, ``int``(x``/``2` `+` `1``)):``        ``if` `(x ``%` `i ``=``=` `0``):``            ``sum_div ``=` `sum_div ``+` `i` `    ``# If the sum of divisors is equal``    ``# to the given number, return true``    ``if` `(sum_div ``=``=` `x):``        ``return` `1` `    ``# Otherwise, return false``    ``else``:``        ``return` `0` `# Function to find the sum of all the``# subsets from an array whose sum is``# a perfect number``def` `subsetSum(arr, n):``    ` `    ``# Stores the total number of``    ``# subsets, i.e. 2 ^ n``    ``total ``=` `1` `<< n` `    ``# Consider all numbers from 0 to 2 ^ n - 1``    ``for` `i ``in` `range``(total):``        ``sum` `=` `0` `        ``# Consider array elements from``        ``# positions of set bits in the``        ``# binary representation of n``        ``for` `j ``in` `range``(n):``            ``if` `(i & (``1` `<< j) !``=` `0``):``                ``sum` `=` `sum` `+` `arr[j]` `        ``# If sum of chosen elements``        ``# is a perfect number``        ``if` `(isPerfect(``sum``)):``            ``print``(``sum``, ``" "``)` `# Driver Code``arr ``=` `[``5``, ``4``, ``6``]``N ``=` `len``(arr)` `subsetSum(arr, N)` `# This code is contributed by Dharanendra L V.`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG{` `  ``// Function to check is a given``  ``// number is a perfect number or not``  ``static` `int` `isPerfect(``int` `x)``  ``{``    ` `    ``// Stores sum of divisors``    ``int` `sum_div = 1;` `    ``// Add all divisors of x to sum_div``    ``for` `(``int` `i = 2; i <= x / 2; ++i) {``      ``if` `(x % i == 0) {``        ``sum_div += i;``      ``}``    ``}` `    ``// If the sum of divisors is equal``    ``// to the given number, return true``    ``if` `(sum_div == x) {``      ``return` `1;``    ``}` `    ``// Otherwise, return false``    ``else``      ``return` `0;``  ``}` `  ``// Function to find the sum of all the``  ``// subsets from an array whose sum is``  ``// a perfect number``  ``static` `void` `subsetSum(``int``[] arr, ``int` `n)``  ``{``    ` `    ``// Stores the total number of``    ``// subsets, i.e. 2 ^ n``    ``long` `total = 1 << n;` `    ``// Consider all numbers from 0 to 2 ^ n - 1``    ``for` `(``long` `i = 0; i < total; i++) {``      ``int` `sum = 0;` `      ``// Consider array elements from``      ``// positions of set bits in the``      ``// binary representation of n``      ``for` `(``int` `j = 0; j < n; j++)``        ``if` `((i & (1 << j)) != 0)``          ``sum += arr[j];` `      ``// If sum of chosen elements``      ``// is a perfect number``      ``if` `(isPerfect(sum) != 0) {``        ``Console.Write(sum + ``" "``);``      ``}``    ``}``  ``}` `// Driver Code``static` `public` `void` `Main()``{``    ``int``[] arr = { 5, 4, 6 };``    ``int` `N = arr.Length;``    ``subsetSum(arr, N);``}``}` `// This code is contributed by splevel62.`

## Javascript

 ``

Output:

`6`

Time Complexity: O((N + M) * 2N), where M is the sum of the elements of the array, arr[]
Auxiliary Space: O(1)

My Personal Notes arrow_drop_up