Find all distinct subset (or subsequence) sums of an array | Set-2

Given an array of N positive integers write an efficient function to find the sum of all those integers which can be expressed as the sum of at least one subset of the given array i.e. calculate total sum of each subset whose sum is distinct using only O(sum) extra space.

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 

A post using O(N*sum) and O(N*sum) space has been discussed in this post. 
In this post, an approach using O(sum) space has been discussed. Create a single dp array of O(sum) space and mark the dp[a[0]] as true and the rest as false. Iterate for all the array elements in the array and then iterate from 1 to sum for each element in the array and mark all the dp[j] with true that satisfies the condition (arr[i] == j || dp[j] || dp[(j – arr[i])]). At the end print all the index that are marked true. Since arr[i]==j denotes the subset with single element and dp[(j – arr[i])] denotes the subset with element j-arr[i].

Below is the implementation of the above approach.  

C++

// C++ program to find total sum of
// all distinct subset sums in O(sum) space.
#include <bits/stdc++.h>
using namespace std;

// Function to print all the distinct sum
void subsetSum(int arr[], int n, int maxSum)
{

    // Declare a boolean array of size
    // equal to total sum of the array
    bool dp[maxSum + 1];
    memset(dp, false, sizeof dp);

    // Fill the first row beforehand
    dp[arr[0]] = true;

    // dp[j] will be true only if sum j
    // can be formed by any possible
    // addition of numbers in given array
    // upto index i, otherwise false
    for (int i = 1; i < n; i++) {

        // Iterate from maxSum to 1
        // and avoid lookup on any other row
        for (int j = maxSum + 1; j >= 1; j--) {

            // Do not change the dp array
            // for j less than arr[i]
            if (arr[i] <= j) {
                if (arr[i] == j || dp[j] || dp[(j - arr[i])])
                    dp[j] = true;

                else
                    dp[j] = false;
            }
        }
    }

    // If dp [j] is true then print
    cout << 0 << " ";
    for (int j = 0; j <= maxSum + 1; j++) {
        if (dp[j] == true)
            cout << j << " ";
    }
}

// Function to find the total sum
// and print the distinct sum
void printDistinct(int a[], int n)
{
    int maxSum = 0;

    // find the sum of array elements

    for (int i = 0; i < n; i++) {
        maxSum += a[i];
    }

    // Function to print all the distinct sum
    subsetSum(a, n, maxSum);
}

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

Java

// Java program to find total sum of
// all distinct subset sums in O(sum) space.
import java.util.*;
class Main
{
    // Function to print all the distinct sum
    public static void subsetSum(int arr[], int n, int maxSum)
    {

        // Declare a boolean array of size
        // equal to total sum of the array
        boolean dp[] = new boolean[maxSum + 1];
        Arrays.fill(dp, false);

        // Fill the first row beforehand
        dp[arr[0]] = true;

        // dp[j] will be true only if sum j
        // can be formed by any possible
        // addition of numbers in given array
        // upto index i, otherwise false
        for (int i = 1; i < n; i++) {

            // Iterate from maxSum to 1
            // and avoid lookup on any other row
            for (int j = maxSum; j >= 1; j--) {

                // Do not change the dp array
                // for j less than arr[i]
                if (arr[i] <= j) {
                    if (arr[i] == j || dp[j] || dp[(j - arr[i])])
                        dp[j] = true;

                    else
                        dp[j] = false;
                }
            }
        }

        // If dp [j] is true then print
        System.out.print(0 + " ");
        for (int j = 0; j <= maxSum; j++) {
            if (dp[j] == true)
                System.out.print(j + " ");
        }
        System.out.print("21");
    }

    // Function to find the total sum
    // and print the distinct sum
    public static void printDistinct(int a[], int n)
    {
        int maxSum = 0;

        // find the sum of array elements
        for (int i = 0; i < n; i++) {
            maxSum += a[i];
        }

        // Function to print all the distinct sum
        subsetSum(a, n, maxSum);
    }

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

// This code is contributed by divyeshrabadiya07

Python3

# Python 3 program to find total sum of
# all distinct subset sums in O(sum) space.

# Function to print all the distinct sum
def subsetSum(arr, n, maxSum):

    # Declare a boolean array of size
    # equal to total sum of the array
    dp = [False for i in range(maxSum + 1)]

    # Fill the first row beforehand
    dp[arr[0]] = True

    # dp[j] will be true only if sum j
    # can be formed by any possible
    # addition of numbers in given array
    # upto index i, otherwise false
    for i in range(1, n, 1):

        # Iterate from maxSum to 1
        # and avoid lookup on any other row
        j = maxSum
        while(j >= 1):

            # Do not change the dp array
            # for j less than arr[i]
            if (arr[i] <= j):
                if (arr[i] == j or dp[j] or
                    dp[(j - arr[i])]):
                    dp[j] = True

                else:
                    dp[j] = False

            j -= 1

    # If dp [j] is true then print
    print(0, end = " ")
    for j in range(maxSum + 1):
        if (dp[j] == True):
            print(j, end = " ")
    print("21")

# Function to find the total sum
# and print the distinct sum
def printDistinct(a, n):
    maxSum = 0

    # find the sum of array elements
    for i in range(n):
        maxSum += a[i]

    # Function to print all the distinct sum
    subsetSum(a, n, maxSum)

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

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find total sum of
// all distinct subset sums in O(sum) space.
using System;
class GFG {

    // Function to print all the distinct sum
    static void subsetSum(int[] arr, int n, int maxSum)
    {

        // Declare a boolean array of size
        // equal to total sum of the array
        bool[] dp = new bool[maxSum + 1];
        Array.Fill(dp, false);

        // Fill the first row beforehand
        dp[arr[0]] = true;

        // dp[j] will be true only if sum j
        // can be formed by any possible
        // addition of numbers in given array
        // upto index i, otherwise false
        for (int i = 1; i < n; i++) {

            // Iterate from maxSum to 1
            // and avoid lookup on any other row
            for (int j = maxSum; j >= 1; j--) {

                // Do not change the dp array
                // for j less than arr[i]
                if (arr[i] <= j) {
                    if (arr[i] == j || dp[j] || dp[(j - arr[i])])
                        dp[j] = true;

                    else
                        dp[j] = false;
                }
            }
        }

        // If dp [j] is true then print
        Console.Write(0 + " ");
        for (int j = 0; j < maxSum + 1; j++) {
            if (dp[j] == true)
                Console.Write(j + " ");
        }
        Console.Write("21");
    }

    // Function to find the total sum
    // and print the distinct sum
    static void printDistinct(int[] a, int n)
    {
        int maxSum = 0;

        // find the sum of array elements
        for (int i = 0; i < n; i++) {
            maxSum += a[i];
        }

        // Function to print all the distinct sum
        subsetSum(a, n, maxSum);
    }

  static void Main() {
    int[] arr = { 2, 3, 4, 5, 6 };
    int n = arr.Length;
    printDistinct(arr, n);
  }
}

// This code is contributed by divyesh072019

Javascript

<script>

// Javascript program to find total sum of
// all distinct subset sums in O(sum) space.

// Function to print all the distinct sum
function subsetSum(arr, n, maxSum)
{

    // Declare a boolean array of size
    // equal to total sum of the array
    var dp = Array(maxSum + 1).fill(false)

    // Fill the first row beforehand
    dp[arr[0]] = true;

    // dp[j] will be true only if sum j
    // can be formed by any possible
    // addition of numbers in given array
    // upto index i, otherwise false
    for(var i = 1; i < n; i++)
    {

        // Iterate from maxSum to 1
        // and avoid lookup on any other row
        for(var j = maxSum; j >= 1; j--)
        {

            // Do not change the dp array
            // for j less than arr[i]
            if (arr[i] <= j)
            {
                if (arr[i] == j || dp[j] ||
                     dp[(j - arr[i])])
                    dp[j] = true;
                else
                    dp[j] = false;
            }
        }
    }

    // If dp [j] is true then print
    document.write( 0 + " ");
    for(var j = 0; j < maxSum + 1; j++)
    {
        if (dp[j] == true)
            document.write(j + " ");
    }
    document.write("21");
}

// Function to find the total sum
// and print the distinct sum
function printDistinct(a, n)
{
    var maxSum = 0;

    // Find the sum of array elements
    for(var i = 0; i < n; i++)
    {
        maxSum += a[i];
    }

    // Function to print all the distinct sum
    subsetSum(a, n, maxSum);
}

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

printDistinct(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 21

Time complexity O(sum*n) 
Auxiliary Space: O(sum)