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

Given a set of integers, find distinct sum that can be generated from the subsets of the given sets and print them in an 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 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[0]); 
    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) { 
            s.add(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. 
    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): 
        s.add(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. 
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)  
        { 
            s.Add(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. 
    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 */

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 naive recursive approach is O(2ⁿ).

Time complexity of the above problem can be improved using Dynamic Programming, especially when sum of given elements is small. We can make a dp table with rows containing the size of the array and size of the column will be 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][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) 
            cout << j << " "; 
} 
  
  
// Driver code 
int main() 
{ 
    int arr[] = {2, 3, 4, 5, 6}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    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][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) 
                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][0] = 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. 

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.

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