Find all distinct subset (or subsequence) sums of an array

• Difficulty Level : Medium
• Last Updated : 02 Jun, 2021

Given a set of integers, find a distinct sum that can be generated from the subsets of the given sets and print them in increasing order. It is given that sum of array elements is small.

Examples:

Input  : arr[] = {1, 2, 3}
Output : 0 1 2 3 4 5 6
Distinct subsets of given set are
{}, {1}, {2}, {3}, {1,2}, {2,3},
{1,3} and {1,2,3}.  Sums of these
subsets are 0, 1, 2, 3, 3, 5, 4, 6
After removing duplicates, we get
0, 1, 2, 3, 4, 5, 6

Input : arr[] = {2, 3, 4, 5, 6}
Output : 0 2 3 4 5 6 7 8 9 10 11 12
13 14 15 16 17 18 20

Input : arr[] = {20, 30, 50}
Output : 0 20 30 50 70 80 100

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The naive solution for this problem is to generate all the subsets, store their sums in a hash set and finally print all keys from the hash set.

C++

// C++ program to print distinct subset sums of
// a given array.
#include<bits/stdc++.h>
using namespace std;

// sum denotes the current sum of the subset
// currindex denotes the index we have reached in
// the given array
void distSumRec(int arr[], int n, int sum,
int currindex, unordered_set<int> &s)
{
if (currindex > n)
return;

if (currindex == n)
{
s.insert(sum);
return;
}

distSumRec(arr, n, sum + arr[currindex],
currindex+1, s);
distSumRec(arr, n, sum, currindex+1, s);
}

// This function mainly calls recursive function
// distSumRec() to generate distinct sum subsets.
// And finally prints the generated subsets.
void printDistSum(int arr[], int n)
{
unordered_set<int> s;
distSumRec(arr, n, 0, 0, s);

// Print the result
for (auto i=s.begin(); i!=s.end(); i++)
cout << *i << " ";
}

// Driver code
int main()
{
int arr[] = {2, 3, 4, 5, 6};
int n = sizeof(arr)/sizeof(arr);
printDistSum(arr, n);
return 0;
}

Java

// Java program to print distinct
// subset sums of a given array.
import java.io.*;
import java.util.*;

class GFG
{
// sum denotes the current sum
// of the subset currindex denotes
// the index we have reached in
// the given array
static void distSumRec(int arr[], int n, int sum,
int currindex, HashSet<Integer> s)
{
if (currindex > n)
return;

if (currindex == n) {
return;
}

distSumRec(arr, n, sum + arr[currindex],
currindex + 1, s);
distSumRec(arr, n, sum, currindex + 1, s);
}

// This function mainly calls
// recursive function distSumRec()
// to generate distinct sum subsets.
// And finally prints the generated subsets.
static void printDistSum(int arr[], int n)
{
HashSet<Integer> s = new HashSet<>();
distSumRec(arr, n, 0, 0, s);

// Print the result
for (int i : s)
System.out.print(i + " ");
}

//Driver code
public static void main(String[] args)
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = arr.length;
printDistSum(arr, n);
}
}

// This code is contributed by Gitanjali.

Python3

# Python 3 program to print distinct subset sums of
# a given array.

# sum denotes the current sum of the subset
# currindex denotes the index we have reached in
# the given array
def distSumRec(arr, n, sum, currindex, s):
if (currindex > n):
return

if (currindex == n):
return

distSumRec(arr, n, sum + arr[currindex], currindex+1, s)
distSumRec(arr, n, sum, currindex+1, s)

# This function mainly calls recursive function
# distSumRec() to generate distinct sum subsets.
# And finally prints the generated subsets.
def printDistSum(arr,n):
s = set()
distSumRec(arr, n, 0, 0, s)

# Print the result
for i in s:
print(i,end = " ")

# Driver code
if __name__ == '__main__':
arr = [2, 3, 4, 5, 6]
n = len(arr)
printDistSum(arr, n)

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to print distinct
// subset sums of a given array.
using System;
using System.Collections.Generic;

class GFG
{
// sum denotes the current sum
// of the subset currindex denotes
// the index we have reached in
// the given array
static void distSumRec(int []arr, int n, int sum,
int currindex, HashSet<int> s)
{
if (currindex > n)
return;

if (currindex == n)
{
return;
}

distSumRec(arr, n, sum + arr[currindex],
currindex + 1, s);
distSumRec(arr, n, sum, currindex + 1, s);
}

// This function mainly calls
// recursive function distSumRec()
// to generate distinct sum subsets.
// And finally prints the generated subsets.
static void printDistSum(int []arr, int n)
{
HashSet<int> s = new HashSet<int>();
distSumRec(arr, n, 0, 0, s);

// Print the result
foreach (int i in s)
Console.Write(i + " ");
}

// Driver code
public static void Main()
{
int []arr = { 2, 3, 4, 5, 6 };
int n = arr.Length;
printDistSum(arr, n);
}
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>
// Javascript program to print distinct
// subset sums of a given array.

// sum denotes the current sum
// of the subset currindex denotes
// the index we have reached in
// the given array
function distSumRec(arr,n,sum,currindex,s)
{
if (currindex > n)
return;

if (currindex == n) {
return;
}

distSumRec(arr, n, sum + arr[currindex],
currindex + 1, s);
distSumRec(arr, n, sum, currindex + 1, s);
}

// This function mainly calls
// recursive function distSumRec()
// to generate distinct sum subsets.
// And finally prints the generated subsets.
function printDistSum(arr,n)
{
let s=new Set();
distSumRec(arr, n, 0, 0, s);
let s1=[...s]
s1.sort(function(a,b){return a-b;})
// Print the result
for (let [key, value] of s1.entries())
document.write(value + " ");
}

//Driver code
let arr=[2, 3, 4, 5, 6 ];
let n = arr.length;
printDistSum(arr, n);

// This code is contributed by unknown2108
</script>

Output:

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20

The time complexity of the above naive recursive approach is O(2n).

The time complexity of the above problem can be improved using Dynamic Programming, especially when the sum of given elements is small. We can make a dp table with rows containing the size of the array and the size of the column will be the sum of all the elements in the array.

C++

// C++ program to print distinct subset sums of
// a given array.
#include<bits/stdc++.h>
using namespace std;

// Uses Dynamic Programming to find distinct
// subset sums
void printDistSum(int arr[], int n)
{
int sum = 0;
for (int i=0; i<n; i++)
sum += arr[i];

// dp[i][j] would be true if arr[0..i-1] has
// a subset with sum equal to j.
bool dp[n+1][sum+1];
memset(dp, 0, sizeof(dp));

// There is always a subset with 0 sum
for (int i=0; i<=n; i++)
dp[i] = true;

// Fill dp[][] in bottom up manner
for (int i=1; i<=n; i++)
{
dp[i][arr[i-1]] = true;
for (int j=1; j<=sum; j++)
{
// Sums that were achievable
// without current array element
if (dp[i-1][j] == true)
{
dp[i][j] = true;
dp[i][j + arr[i-1]] = true;
}
}
}

// Print last row elements
for (int j=0; j<=sum; j++)
if (dp[n][j]==true)
cout << j << " ";
}

// Driver code
int main()
{
int arr[] = {2, 3, 4, 5, 6};
int n = sizeof(arr)/sizeof(arr);
printDistSum(arr, n);
return 0;
}

Java

// Java program to print distinct
// subset sums of a given array.
import java.io.*;
import java.util.*;

class GFG {

// Uses Dynamic Programming to
// find distinct subset sums
static void printDistSum(int arr[], int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];

// dp[i][j] would be true if arr[0..i-1]
// has a subset with sum equal to j.
boolean[][] dp = new boolean[n + 1][sum + 1];

// There is always a subset with 0 sum
for (int i = 0; i <= n; i++)
dp[i] = true;

// Fill dp[][] in bottom up manner
for (int i = 1; i <= n; i++)
{
dp[i][arr[i - 1]] = true;
for (int j = 1; j <= sum; j++)
{
// Sums that were achievable
// without current array element
if (dp[i - 1][j] == true)
{
dp[i][j] = true;
dp[i][j + arr[i - 1]] = true;
}
}
}

// Print last row elements
for (int j = 0; j <= sum; j++)
if (dp[n][j] == true)
System.out.print(j + " ");
}

// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = arr.length;
printDistSum(arr, n);
}
}

// This code is contributed by Gitanjali.

Python3

# Python3 program to prdistinct subset
# Sums of a given array.

# Uses Dynamic Programming to find
# distinct subset Sums
def printDistSum(arr, n):

Sum = sum(arr)

# dp[i][j] would be true if arr[0..i-1]
# has a subset with Sum equal to j.
dp = [[False for i in range(Sum + 1)]
for i in range(n + 1)]

# There is always a subset with 0 Sum
for i in range(n + 1):
dp[i] = True

# Fill dp[][] in bottom up manner
for i in range(1, n + 1):

dp[i][arr[i - 1]] = True

for j in range(1, Sum + 1):

# Sums that were achievable
# without current array element
if (dp[i - 1][j] == True):
dp[i][j] = True
dp[i][j + arr[i - 1]] = True

# Print last row elements
for j in range(Sum + 1):
if (dp[n][j] == True):
print(j, end = " ")

# Driver code
arr = [2, 3, 4, 5, 6]
n = len(arr)
printDistSum(arr, n)

# This code is contributed
# by mohit kumar

C#

// C# program to print distinct
// subset sums of a given array.
using System;

class GFG {

// Uses Dynamic Programming to
// find distinct subset sums
static void printDistSum(int []arr, int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];

// dp[i][j] would be true if arr[0..i-1]
// has a subset with sum equal to j.
bool [,]dp = new bool[n + 1,sum + 1];

// There is always a subset with 0 sum
for (int i = 0; i <= n; i++)
dp[i,0] = true;

// Fill dp[][] in bottom up manner
for (int i = 1; i <= n; i++)
{
dp[i,arr[i - 1]] = true;
for (int j = 1; j <= sum; j++)
{
// Sums that were achievable
// without current array element
if (dp[i - 1,j] == true)
{
dp[i,j] = true;
dp[i,j + arr[i - 1]] = true;
}
}
}

// Print last row elements
for (int j = 0; j <= sum; j++)
if (dp[n,j] == true)
Console.Write(j + " ");
}

// Driver code
public static void Main()
{
int []arr = { 2, 3, 4, 5, 6 };
int n = arr.Length;
printDistSum(arr, n);
}
}

// This code is contributed by nitin mittal.

Javascript

<script>

// Javascript program to print distinct
// subset sums of a given array.

// Uses Dynamic Programming to find
// distinct subset sums
function printDistSum(arr, n)
{
var sum = 0;
for(var i = 0; i < n; i++)
sum += arr[i];

// dp[i][j] would be true if arr[0..i-1] has
// a subset with sum equal to j.
var dp = Array.from(
Array(n + 1), () => Array(sum + 1).fill(0));

// There is always a subset with 0 sum
for(var i = 0; i <= n; i++)
dp[i] = true;

// Fill dp[][] in bottom up manner
for(var i = 1; i <= n; i++)
{
dp[i][arr[i - 1]] = true;
for(var j = 1; j <= sum; j++)
{

// Sums that were achievable
// without current array element
if (dp[i - 1][j] == true)
{
dp[i][j] = true;
dp[i][j + arr[i - 1]] = true;
}
}
}

// Print last row elements
for(var j = 0; j <= sum; j++)
if (dp[n][j] == true)
document.write(j + " ");
}

// Driver code
var arr = [ 2, 3, 4, 5, 6 ];
var n = arr.length;

printDistSum(arr, n);

// This code is contributed by importantly

</script>

Output:

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20

Time complexity of the above approach is O(n*sum) where n is the size of the array and sum is the sum of all the integers in the array.

This article is contributed by Karan Goyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.