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Print sums of all subsets of a given set

  • Difficulty Level : Medium
  • Last Updated : 08 Nov, 2021

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

Examples : 

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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 of this solution is O(2^n) and space complexity is O(2^n).

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

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 totoal 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 totoal 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 totoal 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.

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.ncorrect, or you want to share more information about the topic discussed above
 




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