Print sums of all subsets of a given set

Last Updated : 22 Feb, 2023

Given an array of integers, print sums of all subsets in it. Output sums can be printed in any order.

Examples :

Input : arr[] = {2, 3}
Output: 0 2 3 5

Input : arr[] = {2, 4, 5}
Output : 0 2 4 5 6 7 9 11

Method 1 (Recursive)
We can recursively solve this problem. There are total 2ⁿ subsets. For every element, we consider two choices, we include it in a subset and we don’t include it in a subset. Below is recursive solution based on this idea.

C++

// C++ program to print sums of all possible
// subsets.
#include <bits/stdc++.h>
using namespace std;
 
// Prints sums of all subsets of arr[l..r]
void subsetSums(int arr[], int l, int r, int sum = 0)
{
    // Print current subset
    if (l > r) {
        cout << sum << " ";
        return;
    }
 
    // Subset including arr[l]
    subsetSums(arr, l + 1, r, sum + arr[l]);
 
    // Subset excluding arr[l]
    subsetSums(arr, l + 1, r, sum);
}
 
// Driver code
int main()
{
    int arr[] = { 5, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    subsetSums(arr, 0, n - 1);
    return 0;
}

Java

// Java program to print sums
// of all possible subsets.
import java.io.*;
 
class GFG {
 
    // Prints sums of all
    // subsets of arr[l..r]
    static void subsetSums(int[] arr, int l, int r, int sum)
    {
 
        // Print current subset
        if (l > r) {
            System.out.print(sum + " ");
            return;
        }
 
        // Subset including arr[l]
        subsetSums(arr, l + 1, r, sum + arr[l]);
 
        // Subset excluding arr[l]
        subsetSums(arr, l + 1, r, sum);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 5, 4, 3 };
        int n = arr.length;
 
        subsetSums(arr, 0, n - 1, 0);
    }
}
 
// This code is contributed by anuj_67

Python3

# Python3 program to print sums of
# all possible subsets.
 
# Prints sums of all subsets of arr[l..r]
 
 
def subsetSums(arr, l, r, sum=0):
 
    # Print current subset
    if l > r:
        print(sum, end=" ")
        return
 
    # Subset including arr[l]
    subsetSums(arr, l + 1, r, sum + arr[l])
 
    # Subset excluding arr[l]
    subsetSums(arr, l + 1, r, sum)
 
 
# Driver code
arr = [5, 4, 3]
n = len(arr)
subsetSums(arr, 0, n - 1)
 
# This code is contributed by Shreyanshi Arun.

C#

// C# program to print sums of all possible
// subsets.
using System;
 
class GFG {
 
    // Prints sums of all subsets of
    // arr[l..r]
    static void subsetSums(int[] arr, int l, int r, int sum)
    {
 
        // Print current subset
        if (l > r) {
            Console.Write(sum + " ");
            return;
        }
 
        // Subset including arr[l]
        subsetSums(arr, l + 1, r, sum + arr[l]);
 
        // Subset excluding arr[l]
        subsetSums(arr, l + 1, r, sum);
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 5, 4, 3 };
        int n = arr.Length;
 
        subsetSums(arr, 0, n - 1, 0);
    }
}
 
// This code is contributed by anuj_67

PHP

<?php
// PHP program to print sums 
// of all possible subsets.
 
// Prints sums of all 
// subsets of arr[l..r]
function subsetSums($arr, $l, 
                    $r, $sum = 0)
{
    // Print current subset
    if ($l > $r)
    {
        echo $sum , " ";
        return;
    }
 
    // Subset including arr[l]
    subsetSums($arr, $l + 1, $r, 
               $sum + $arr[$l]);
 
    // Subset excluding arr[l]
    subsetSums($arr, $l + 1, $r, $sum);
}
 
// Driver code
$arr = array(5, 4, 3);
$n = count($arr);
 
subsetSums($arr, 0, $n - 1);
 
// This code is contributed by anuj_67.
?>

Javascript

<script>
// Javascript program to program to print
// sums of all possible subsets.
 
// Prints sums of all 
// subsets of arr[l..r]
function subsetSums(arr, l, r, sum)
{
     
    // Print current subset
    if (l > r)
    {
        document.write(sum + " ");
        return;
    }
   
    // Subset including arr[l]
    subsetSums(arr, l + 1, r, 
               sum + arr[l]);
                
    // Subset excluding arr[l]
    subsetSums(arr, l + 1, r, sum);
}
     
// Driver code
let arr = [5, 4, 3];
let n = arr.length;
 
subsetSums(arr, 0, n - 1, 0);
 
// This code is contributed by code_hunt
     
</script>

Output :

12 9 8 5 7 4 3 0

Time complexity : O(2^n)

Auxiliary Space : O(n)

Method 2 (Iterative)
As discussed above, there are total 2ⁿ subsets. The idea is to generate a loop from 0 to 2ⁿ – 1. For every number, pick all array elements corresponding to 1s in the binary representation of the current number.

C++

// Iterative C++ program to print sums of all
// possible subsets.
#include <bits/stdc++.h>
using namespace std;
 
// Prints sums of all subsets of array
void subsetSums(int arr[], int n)
{
    // There are total 2^n subsets
    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 binary representation of
        // current i to decide which elements
        // to pick.
        for (int j = 0; j < n; j++)
            if (i & (1 << j))
                sum += arr[j];
 
        // Print sum of picked elements.
        cout << sum << " ";
    }
}
 
// Driver code
int main()
{
    int arr[] = { 5, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    subsetSums(arr, n);
    return 0;
}

Java

// Iterative Java program to print sums of all
// possible subsets.
import java.util.*;
 
class GFG {
 
    // Prints sums of all subsets of array
    static void subsetSums(int arr[], int n)
    {
 
        // There are total 2^n subsets
        int total = 1 << n;
 
        // Consider all numbers from 0 to 2^n - 1
        for (int i = 0; i < total; i++) {
            int sum = 0;
 
            // Consider binary representation of
            // current i to decide which elements
            // to pick.
            for (int j = 0; j < n; j++)
                if ((i & (1 << j)) != 0)
                    sum += arr[j];
 
            // Print sum of picked elements.
            System.out.print(sum + " ");
        }
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = new int[] { 5, 4, 3 };
        int n = arr.length;
 
        subsetSums(arr, n);
    }
}
 
// This code is contributed by spp____

Python3

# Iterative Python3 program to print sums of all possible subsets
 
# Prints sums of all subsets of array
def subsetSums(arr, n):
    # There are total 2^n subsets
    total = 1 << n
 
    # Consider all numbers from 0 to 2^n - 1
    for i in range(total):
       Sum = 0
 
       # Consider binary representation of
       # current i to decide which elements
       # to pick.
       for j in range(n):
          if ((i & (1 << j)) != 0):
              Sum += arr[j]
 
       # Print sum of picked elements.
       print(Sum, "", end = "")
 
arr = [ 5, 4, 3 ]
n = len(arr)
 
subsetSums(arr, n);
 
# This code is contributed by mukesh07.

C#

// Iterative C# program to print sums of all
// possible subsets.
using System;
class GFG {
     
    // Prints sums of all subsets of array
    static void subsetSums(int[] arr, int n)
    {
  
        // There are total 2^n subsets
        int total = 1 << n;
  
        // Consider all numbers from 0 to 2^n - 1
        for (int i = 0; i < total; i++) {
            int sum = 0;
  
            // Consider binary representation of
            // current i to decide which elements
            // to pick.
            for (int j = 0; j < n; j++)
                if ((i & (1 << j)) != 0)
                    sum += arr[j];
  
            // Print sum of picked elements.
            Console.Write(sum + " ");
        }
    }
     
  static void Main() {
    int[] arr = { 5, 4, 3 };
    int n = arr.Length;
     
    subsetSums(arr, n);
  }
}
 
// This code is contributed by divyesh072019.

PHP

<?php
// Iterative PHP program to print 
// sums of all possible subsets. 
 
// Prints sums of all subsets of array 
function subsetSums($arr, $n) 
{ 
     
        // There are total 2^n subsets 
        $total = 1 << $n; 
 
    // Consider all numbers
    // from 0 to 2^n - 1 
    for ($i = 0; $i < $total; $i++) 
    { 
        $sum = 0; 
 
        // Consider binary representation of 
        // current i to decide which elements 
        // to pick. 
        for ($j = 0; $j < $n; $j++) 
            if ($i & (1 << $j)) 
                $sum += $arr[$j]; 
 
        // Print sum of picked elements. 
        echo $sum , " "; 
    } 
} 
 
    // Driver code 
    $arr = array(5, 4, 3); 
    $n = sizeof($arr);
    subsetSums($arr, $n); 
     
// This Code is Contributed by ajit
?>

Javascript

<script>
 
    // Iterative Javascript program to print sums of all
    // possible subsets.
     
    // Prints sums of all subsets of array
    function subsetSums(arr, n)
    {
 
        // There are total 2^n subsets
        let total = 1 << n;
 
        // Consider all numbers from 0 to 2^n - 1
        for(let i = 0; i < total; i++)
        {
           let sum = 0;
 
           // Consider binary representation of
           // current i to decide which elements
           // to pick.
           for(let j = 0; j < n; j++)
              if ((i & (1 << j)) != 0)
                  sum += arr[j];
 
           // Print sum of picked elements.
           document.write(sum + " ");
        }
    }
     
    let arr = [ 5, 4, 3 ];
    let n = arr.length;
  
    subsetSums(arr, n);
     
</script>

Output :

0 5 4 9 3 8 7 12

Time Complexity: O( $N * 2^N$ )
Auxiliary Space: O(1)
Thanks to cfh for suggesting above iterative solution in a comment.
Note: We haven’t actually created sub-sets to find their sums rather we have just used recursion to find sum of non-contiguous sub-sets of the given set.

A more Efficient Iterative method:

In this method, while visiting a new element, we take its sum with all previously stored sums. This method stores the sums of all subsets and hence it is valid for smaller inputs.

C++

// Iterative C++ program to print sums of all
// possible subsets.
#include <bits/stdc++.h>
using namespace std;
 
// Prints sums of all subsets of array
void subsetSums(int nums[], int n)
{
    // There are total 2^n subsets
    vector<int> s = {0};//store the sums
         
        for (int i = 0; i <n; i++) {
            const int v = s.size();
            for (int t = 0; t < v; t++) {
                s.push_back(s[t] + nums[i]); //add this element with previous subsets
            }
        }
        // Print
          for(int i=0;i<s.size();i++)
        cout << s[i] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 5, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    subsetSums(arr, n);
    return 0;
}

Java

import java.util.ArrayList;
 
public class Main {
    public static void subsetSums(int[] nums, int n) {
        ArrayList<Integer> s = new ArrayList<Integer>();
        s.add(0);
 
        for (int i = 0; i < n; i++) {
            int v = s.size();
            for (int t = 0; t < v; t++) {
                s.add(s.get(t) + nums[i]);
            }
        }
 
        for (int i = 0; i < s.size(); i++) {
            System.out.print(s.get(i) + " ");
        }
    }
 
    public static void main(String[] args) {
        int[] arr = { 5, 4, 3 };
        int n = arr.length;
 
        subsetSums(arr, n);
    }
}

Python3

# Iterative Python program to print sums of all
# possible subsets.
 
# Prints sums of all subsets of array
def subsetSums(nums,n):
    # There are total 2^n subsets
    s = [0]
    for i in range(n):
        v = len(s)
        for t in range(v):
            s.append(s[t] + nums[i]) # add this element with previous subsets
    # Print
    print(s,end=' ')
 
# Driver code
arr = [5, 4, 3 ]
n = len(arr)
subsetSums(arr, n)

C#

// C# program to print sums of all possible subsets
using System;
 
public class Subsets
{
  static void subsetSums(int[] nums, int n)
  {
     
    // There are total 2^n subsets
    int[] s = new int[1];
    s[0] = 0;
    for (int i = 0; i < n; i++)
    {
      int v = s.Length;
      for (int t = 0; t < v; t++)
      {
         
        // add the current element with all previous subsets
        Array.Resize(ref s, s.Length + 1); // increase the size of the array
        s[s.Length - 1] = s[t] + nums[i]; // add the current element with previous subsets
      }
    }
     
    // Print
    Console.Write(string.Join(" ", s));
  }public static void Main()
  {
    int[] arr = { 5, 4, 3 };
    int n = arr.Length;
    subsetSums(arr, n);
  }
}

Javascript

// Iterative JavaScript program to print sums of all possible subsets
function subsetSums(nums, n) {
  // There are total 2^n subsets
  let s = [0];
  for (let i = 0; i < n; i++) {
    let v = s.length;
    for (let t = 0; t < v; t++) {
      s.push(s[t] + nums[i]); // add this element with previous subsets
    }
  }
  // Print
  for (let i=0; i<s.length; i++) {
    process.stdout.write(s[i]+" ");
  }
}
 
// Driver code
let arr = [5, 4, 3];
let n = arr.length;
subsetSums(arr, n);

Output:

0 5 4 9 3 8 7 12

Time Complexity: O(2^N)

Auxiliary Space: O(N) for storing the ans

Thanks to shivampatidar6116 for suggesting the above iterative solution

The above-mentioned techniques can be used to perform various operations on sub-sets like multiplication, division, XOR, OR, etc, without actually creating and storing the sub-sets and thus making the program memory efficient.

Suggest improvement

Decimal to Binary using recursion and without using power operator

All ways to add parenthesis for evaluation

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Print sums of all subsets of a given set

C++

Java

Python3

C#

PHP

Javascript

C++

Java

Python3

C#

PHP

Javascript

C++

Java

Python3

C#

Javascript

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