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

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

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[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 */

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) {
            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.
    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

Time Complexity: O(2ⁿ).
Auxiliary Space: O(N), due to the use of an unordered_set to store the subset sums.

Dynamic Programming Approach
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][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 print distinct 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.

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][0] = 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.

Space Complexity: O(N * SUM). We are using a two-dimensional array of size N*SUM to store the solution to subproblems.

Optimized Bit-set Approach

dp = dp | dp << a[i]

Above Code snippet does the same as naive solution, where dp is a bit mask (we’ll use bit-set). Lets see how:

1. dp → all the sums which were produced before element a[i]
2. dp << a[i] → shifting all the sums by a[i], i.e. adding a[i] to all the sums.
   1. For example, Suppose initially the bit-mask was 000010100 meaning we could generate only 2 and 4 (count from right).
   2. Now if we get a element 3, we could make 5 and 7 as well by adding to 2 and 4 respectively.
   3. This can be denoted by 010100000 which is equivalent to (000010100) << 3
3. dp | (dp << a[i]) → 000010100 | 010100000 = 010110100 This is union of above two sums representing which sums are possible, namely 2, 4, 5 and 7.

bitset optimized knapsack

C++

// C++ Program to Demonstrate Bitset Optimised Knapsack
// Solution

#include <bits/stdc++.h>
using namespace std;

// Driver Code
int main()
{
    // Input Vector
    vector<int> a = { 2, 3, 4, 5, 6 };

    // we have to make a constant size for bit-set
    // and to be safe keep it significantly high
    int n = a.size();
    const int mx = 40;

    // bitset of size mx, dp[i] is 1 if sum i is possible
    // and 0 otherwise
    bitset<mx> dp;
    // sum 0 is always possible
    dp[0] = 1;

    // dp transitions as explained in article
    for (int i = 0; i < n; ++i) {
        dp |= dp << a[i];
    }

    // print all the  1s in bit-set, this will be the
    // all the unique sums possible
    for (int i = 0; i <= mx; i++) {
        if (dp[i] == 1)
            cout << i << " ";
    }
}

// code is contributed by sarvjot singh

Javascript

// Javascript Program to Demonstrate Bit Optimised Knapsack
      // Solution

      // Driver Code

      // Input Array
      var a = [2, 3, 4, 5, 6];
      var n = a.length;

      // Used a variable "dp" and initialized that with "1"
      // because sum 0 is always possible
      // Since binary of "1" is also "1" which means getting
      // "1" at 0th index and it means sum=0
      var dp = 1;

      // dp transitions as explained in article
      for (var i = 0; i < n; ++i) {
        dp |= dp << a[i];
      }

      //Getting that dp as binary bits of string
      var ans = dp.toString(2);

      // print all the  1s in that binary string, this will be the
      // all the unique sums possible
      for (var j = 0; j <= ans.length; j++) {
        if (ans[j] == "1") {
          console.log(j + " ");
        }
      }

Java

import java.util.*;

public class BitsetKnapsack {

    public static void main(String[] args) {
        // Input Vector
        Integer[] a = {2, 3, 4, 5, 6};

        // we have to make a constant size for bit-set
        // and to be safe keep it significantly high
        int n = a.length;
        final int mx = 40;

        // bitset of size mx, dp[i] is 1 if sum i is possible
        // and 0 otherwise
        BitSet dp = new BitSet(mx);
        // sum 0 is always possible
        dp.set(0);

        // dp transitions as explained in article
        for (int i = 0; i < n; ++i) {
            dp.or(dp.get(0, mx - a[i]));
            dp.set(a[i]);
        }

        // print all the  1s in bit-set, this will be the
        // all the unique sums possible
        for (int i = 0; i <= mx; i++) {
            if (dp.get(i))
                System.out.print(i + " ");
        }
    }
}

Python3

# Input Vector
a = [2, 3, 4, 5, 6]

# We have to make a constant size for bit-set
# and to be safe keep it significantly high
n = len(a)
mx = 40

# bitset of size mx, dp[i] is 1 if sum i is possible
# and 0 otherwise
dp = [0] * mx
# Sum 0 is always possible
dp[0] = 1

# dp transitions as explained in article
for i in range(n):
    for j in range(mx - a[i]):
        dp[j + a[i]] |= dp[j]
    dp[a[i]] = 1

# Print all the  1s in bit-set, this will be the
# all the unique sums possible
for i in range(mx):
    if dp[i] == 1:
        print(i, end=' ')

C#

// C# Program to Demonstrate Bitset Optimised Knapsack
// Solution

using System;
using System.Collections;

class Program {
    static void Main(string[] args)
    {
        // Input Vector
        ArrayList a = new ArrayList() { 2, 3, 4, 5, 6 };

        // we have to make a constant size for bit-set
        // and to be safe keep it significantly high
        int n = a.Count;
        const int mx = 40;

        // bitset of size mx, dp[i] is 1 if sum i is
        // possible and 0 otherwise
        BitArray dp = new BitArray(mx);
        // sum 0 is always possible
        dp[0] = true;

        // dp transitions as explained in article
        for (int i = 0; i < n; ++i) {
            dp = new BitArray(dp).Or(
                new BitArray(dp).LeftShift((int)a[i]));
        }

        // print all the 1s in bit-set, this will be the
        // all the unique sums possible

        for (int i = 0; i < mx; i++) {
            if (dp[i])
                Console.Write(i + " ");
        }
    }
}

// code is contributed by Rohit Yadav

Output

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

Time Complexity also seems to be O(N * S). Because if we would have used a array instead of bitset the shifting would have taken linear time O(S). However the shift (and almost all) operation on bitset takes O(S / W) time. Where W is the word size of the system, Usually its 32 bit or 64 bit. Thus the final time complexity becomes O(N * S / W)

Space Complexity: The space complexity of this approach is O(m) where m is the maximum value of the input array.

Some Important Points:

1. The size of bitset must be a constant, this sometimes is a drawback as we might waste some space.
2. Bitset can be thought of a array where every element takes care of W elements. For example 010110100 is equivalent to {2, 6, 4} in a hypothetical system with word size W = 3.
3. Bitset optimized knapsack solution reduced the time complexity by a factor of W which sometimes is just enough to get AC.

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