# 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++ 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; ` `} ` |

*chevron_right*

*filter_none*

**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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