# 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

The time complexity of the above naive recursive approach is O(2^{n}).

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 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.

## 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.

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