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Sum of all subsets whose sum is a Perfect Number from a given array
  • Difficulty Level : Expert
  • Last Updated : 18 Mar, 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 <bits/stdc++.h>
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) :
   
    # Prthe 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
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 <bits/stdc++.h>
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.
Output: 
6

 

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

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