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.

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// 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 th 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;
}

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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)



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I am a second year BTech (CSE) engineering student at GB Pant Goverment Engineering College Delhi(IPU)

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