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

• Difficulty Level : Medium
• Last Updated : 02 Jun, 2021

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] 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 using namespace std; // Function to print all th distinct sumvoid 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] = 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 sumvoid 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 Codeint main(){    int arr[] = { 2, 3, 4, 5, 6 };    int n = sizeof(arr) / sizeof(arr);    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 th 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] = 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 th distinct sumdef 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] = 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 sumdef 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 Codeif __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 th 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] = 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


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)

My Personal Notes arrow_drop_up