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, 6Input: 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 20Input: 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 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[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 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[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 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[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 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 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 sumfunction 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 sumfunction 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 Codevar 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)
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