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Print sums of all subsets of a given set

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  • Difficulty Level : Medium
  • Last Updated : 03 Aug, 2022
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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
Recommended Practice

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;
}

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. 
This article is contributed by Aditya Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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