Print sums of all subsets of a given set (original) (raw)
Last Updated : 27 Feb, 2025
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
**Explanation: All subsets of this array are - {{}, {2}, {3}, {2, 3}}, having sums - 0, 2, 3 and 5 respectively.**Input: arr[] = {2, 4, 5}
**Output: 0 2 4 5 6 7 9 11
**[Naive Approach] Using Iterative Method - O(N * 2^N) Time and O(1) Space
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(vector &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() { vector arr = { 5, 4, 3 }; int n = arr.size();
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____
Python
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.
JavaScript
// 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.
process.stdout.write(sum + " ");
}
}
let arr = [ 5, 4, 3 ];
let n = arr.length;
subsetSums(arr, n);`
**Output :
0 5 4 9 3 8 7 12
**[Expected Approach] Using Recursion - O(2^N) Time and O(N) Space
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(vector &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() { vector arr = { 5, 4, 3 }; int n = arr.size();
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
Python
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
JavaScript
// 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, result) { // Print current subset if (l > r) { result.push(sum); return; }
// Subset including arr[l]
subsetSums(arr, l + 1, r,
sum + arr[l],result);
// Subset excluding arr[l]
subsetSums(arr, l + 1, r, sum,result);}
// Driver code let arr = [5, 4, 3]; let n = arr.length; let result = []; subsetSums(arr, 0, n - 1, 0,result); console.log(result.join(" ")); // This code is contributed by code_hunt
`
**Output :
12 9 8 5 7 4 3 0
[Alternate Approach**] Using Iterative method -** O(2^N) Time and O(N) Space
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(vector &nums, int n) { // There are total 2^n subsets vector 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() { vector arr = { 5, 4, 3 }; int n = arr.size();
subsetSums(arr, n);
return 0;}
Java
import java.util.ArrayList;
public class Main { public static void subsetSums(int[] nums, int n) { ArrayList s = new ArrayList(); 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);
}}
Python
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 for i in s: print(i, 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
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.