Find all distinct subset (or subsequence) sums of an array
Last Updated : 11 Nov, 2024
Given an array arr[] of size n, the task is to find a distinct sum that can be generated from the subsets of the given sets and return them in increasing order. It is given that the sum of array elements is small.
Examples:
Input: arr[] = [1, 2]
Output: [0, 1, 2, 3]
Explanation: Four distinct sums can be calculated which are 0, 1, 2 and 3.
- 0 if we do not choose any number.
- 1 if we choose only 1.
- 2 if we choose only 2.
- 3 if we choose 1 and 2.
Input: arr[] = [1, 2, 3]
Output: [0, 1, 2, 3, 4, 5, 6]
Explanation: Seven distinct sums can be calculated which are 0, 1, 2, 3, 4, 5 and 6.
- 0 if we do not choose any number.
- 1 if we choose only 1.
- 2 if we choose only 2.
- 3 if we choose only 3.
- 4 if we choose 1 and 3.
- 5 if we choose 2 and 3.
- 6 if we choose 1, 2 and 3.
Using Recursion – O(2^n) Time and O(n) Space
The naive solution for this problem is to generate all the subsets, store their sums in a hash set and finally return all keys from the hash set. recursively generates all possible subset sums by considering each element twice – once including it in the sum and once excluding it. When index is reached at n store the current sum to hashSet and return.
Mathematically:
- Include the current element (arr[i]) in the subset sum: distSumRec(arr, n, sum + arr[i], i + 1, s)
- Exclude the current element from the subset sum: distSumRec(arr, n, sum, i + 1, s)
C++ // C++ program to print distinct subset sums of // a given array. #include<bits/stdc++.h> using namespace std; // Recursive function to calculate distinct subset sums // sum: the current sum of the subset void distSumRec(vector<int> &arr, int n, int sum, int i, set<int> &s) { if (i > n) return; // If we have considered all elements in the array, // insert the current sum into the set if (i == n) { s.insert(sum); return; } // Include the current element (arr[i]) in the subset sum distSumRec(arr, n, sum + arr[i], i + 1, s); // Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, s); } // This function calls distSumRec() to generate // distinct sum subsets and return it. vector<int> DistinctSum(vector<int> &arr) { // Set to store distinct sums set<int> s; int n = arr.size(); // Start the recursive process with an // initial sum of 0 and index 0 distSumRec(arr, n, 0, 0, s); vector<int> result; for(int i : s) result.push_back(i); return result; } int main() { vector<int> arr = {2, 3, 4, 5, 6}; vector<int> result = DistinctSum(arr); for(int i : result) { cout << i << " "; } return 0; } // Java program to print distinct subset sums of a given // array. import java.util.*; class GfG { // Recursive function to calculate distinct subset sums // sum: the current sum of the subset static void distSumRec(int[] arr, int n, int sum, int i, Set<Integer> s) { if (i > n) return; // If we have considered all elements in the array, // insert the current sum into the set if (i == n) { s.add(sum); return; } // Include the current element (arr[i]) in the // subset sum distSumRec(arr, n, sum + arr[i], i + 1, s); // Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, s); } // This function calls distSumRec() to generate // distinct sum subsets and return it. static List<Integer> DistinctSum(int[] arr) { // Set to store distinct sums Set<Integer> s = new HashSet<>(); int n = arr.length; // Start the recursive process with an initial sum // of 0 and index 0 distSumRec(arr, n, 0, 0, s); List<Integer> result = new ArrayList<>(s); return result; } public static void main(String[] args) { int[] arr = { 2, 3, 4, 5, 6 }; List<Integer> result = DistinctSum(arr); for (int i : result) { System.out.print(i + " "); } } } # Python program to print distinct subset sums of a given array. # Recursive function to calculate distinct subset sums # sum: the current sum of the subset def distSumRec(arr, n, sum, i, s): if i > n: return # If we have considered all elements in the array, # insert the current sum into the set if i == n: s.add(sum) return # Include the current element (arr[i]) # in the subset sum distSumRec(arr, n, sum + arr[i], i + 1, s) # Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, s) # This function calls distSumRec() to generate # distinct sum subsets and return it. def DistinctSum(arr): # Set to store distinct sums s = set() n = len(arr) # Start the recursive process with an initial # sum of 0 and index 0 distSumRec(arr, n, 0, 0, s) return list(s) arr = [2, 3, 4, 5, 6] result = DistinctSum(arr) print(" ".join(map(str, result))) // C# program to print distinct subset sums of a given // array. using System; using System.Collections.Generic; class GfG { // Recursive function to calculate distinct subset sums // sum: the current sum of the subset static void distSumRec(int[] arr, int n, int sum, int i, HashSet<int> s) { if (i > n) return; // If we have considered all elements in the array, // insert the current sum into the set if (i == n) { s.Add(sum); return; } // Include the current element (arr[i]) in the // subset sum distSumRec(arr, n, sum + arr[i], i + 1, s); // Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, s); } // This function calls distSumRec() to generate // distinct sum subsets and return it. static List<int> DistinctSum(int[] arr) { // Set to store distinct sums HashSet<int> s = new HashSet<int>(); int n = arr.Length; // Start the recursive process with an // initial sum of 0 and index 0 distSumRec(arr, n, 0, 0, s); List<int> result = new List<int>(s); result.Sort(); return result; } static void Main() { int[] arr = { 2, 3, 4, 5, 6 }; List<int> result = DistinctSum(arr); foreach(int i in result) { Console.Write(i + " "); } } } // JavaScript program to print distinct subset sums of a // given array. // Recursive function to calculate distinct subset sums // sum: the current sum of the subset function distSumRec(arr, n, sum, i, s) { if (i > n) return; // If we have considered all elements in the array, // insert the current sum into the set if (i === n) { s.add(sum); return; } // Include the current element (arr[i]) in the subset // sum distSumRec(arr, n, sum + arr[i], i + 1, s); // Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, s); } // This function calls distSumRec() to generate // distinct sum subsets and return it. function DistinctSum(arr) { // Set to store distinct sums let s = new Set(); let n = arr.length; // Start the recursive process with an initial sum of 0 // and index 0 distSumRec(arr, n, 0, 0, s); return Array.from(s).sort((a, b) => a - b); } let arr = [ 2, 3, 4, 5, 6 ]; let result = DistinctSum(arr); console.log(result.join(" ")); Output0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20
Using Top-Down DP (Memoization) – O(n * sum) Time and O(n * sum) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming.
1. Optimal Substructure: The solution to the Distinct Subset Sum Problem can be derived from the optimal solutions of smaller subproblems.
- Include the current element in the subset: If the current element (arr[i]) is included in the subset, the new required sum becomes sum + arr[i].
- Exclude the current element from the subset: If the current element is excluded, the required sum remains the same.
2. Overlapping Subproblems: In this case, the subproblems overlap because the same subset sums are computed multiple times during recursion. For example, when considering an element in the set, the same sum can be encountered in different recursive calls.
- We create a 2D memoization table memo[n+1][totalSum+1] where n is the number of elements in the array and totalSum is the sum of all the elements in the array. Each entry memo[i][sum] will store whether the sum sum can be formed by the first i elements of the array.
- Initially, all entries are set to -1 to indicate that no subproblems have been computed yet.
- Before computing memo[i][sum], we check if it is already computed by checking if memo[i][sum] != -1. If it’s already computed, we return the stored result; otherwise, we calculate it recursively using the inclusion/exclusion approach.
C++ // C++ program to print distinct subset sums of // a given array. #include <bits/stdc++.h> using namespace std; // Recursive function to calculate distinct subset sums void distSumRec(vector<int> &arr, int n, int sum, int i, vector<vector<int>> &memo) { // If we have considered all elements in // the array, mark this sum if (i == n) { memo[i][sum] = 1; return; } // If this state has already been computed, // skip further processing if (memo[i][sum] != -1) return; // Mark the current state as visited memo[i][sum] = 1; // Include the current element (arr[i]) in the subset sum distSumRec(arr, n, sum + arr[i], i + 1, memo); // Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, memo); } // This function calls distSumRec() to generate // distinct sum subsets and return them vector<int> DistinctSum(vector<int> &arr) { int n = arr.size(); // Calculate the maximum possible sum int totalSum = accumulate(arr.begin(), arr.end(), 0); // Memoization table initialized with -1 vector<vector<int>> memo(n + 1, vector<int>(totalSum + 1, -1)); // Start the recursive process with an // initial sum of 0 and index 0 distSumRec(arr, n, 0, 0, memo); // Collect all distinct sums from the memo table vector<int> result; for (int i = 0; i <= totalSum; i++) { if (memo[n][i] == 1) { result.push_back(i); } } return result; } int main() { vector<int> arr = {2, 3, 4, 5, 6}; vector<int> result = DistinctSum(arr); for (int i : result) { cout << i << " "; } return 0; } // Java program to print distinct subset sums of // a given array. import java.util.*; class GfG { // Recursive function to calculate distinct subset sums static void distSumRec(int[] arr, int n, int sum, int i, int[][] memo) { // If we have considered all elements in // the array, mark this sum if (i == n) { memo[i][sum] = 1; return; } // If this state has already been computed, // skip further processing if (memo[i][sum] != -1) return; // Mark the current state as visited memo[i][sum] = 1; // Include the current element (arr[i]) in the subset sum distSumRec(arr, n, sum + arr[i], i + 1, memo); // Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, memo); } // This function calls distSumRec() to generate // distinct sum subsets and return them static List<Integer> DistinctSum(int[] arr) { int n = arr.length; // Calculate the maximum possible sum int totalSum = Arrays.stream(arr).sum(); // Memoization table initialized with -1 int[][] memo = new int[n + 1][totalSum + 1]; for (int[] row : memo) { Arrays.fill(row, -1); } // Start the recursive process with an // initial sum of 0 and index 0 distSumRec(arr, n, 0, 0, memo); // Collect all distinct sums from the memo table List<Integer> result = new ArrayList<>(); for (int i = 0; i <= totalSum; i++) { if (memo[n][i] == 1) { result.add(i); } } return result; } public static void main(String[] args) { int[] arr = {2, 3, 4, 5, 6}; List<Integer> result = DistinctSum(arr); for (int i : result) { System.out.print(i + " "); } } } # Python program to print distinct subset sums of # a given array. # Recursive function to calculate distinct subset sums def distSumRec(arr, n, sum, i, memo): # If we have considered all elements in # the array, mark this sum if i == n: memo[i][sum] = 1 return # If this state has already been computed, # skip further processing if memo[i][sum] != -1: return # Mark the current state as visited memo[i][sum] = 1 # Include the current element (arr[i]) in the subset sum distSumRec(arr, n, sum + arr[i], i + 1, memo) # Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, memo) # This function calls distSumRec() to generate # distinct sum subsets and return them def DistinctSum(arr): n = len(arr) # Calculate the maximum possible sum totalSum = sum(arr) # Memoization table initialized with -1 memo = [[-1 for _ in range(totalSum + 1)] for _ in range(n + 1)] # Start the recursive process with an # initial sum of 0 and index 0 distSumRec(arr, n, 0, 0, memo) # Collect all distinct sums from the memo table result = [] for i in range(totalSum + 1): if memo[n][i] == 1: result.append(i) return result if __name__ == "__main__": arr = [2, 3, 4, 5, 6] result = DistinctSum(arr) for i in result: print(i, end=" ")
// C# program to print distinct subset sums of // a given array. using System; using System.Collections.Generic; class GfG { // Recursive function to calculate distinct subset sums static void distSumRec(int[] arr, int n, int sum, int i, int[,] memo) { // If we have considered all elements in // the array, mark this sum if (i == n) { memo[i, sum] = 1; return; } // If this state has already been computed, // skip further processing if (memo[i, sum] != -1) return; // Mark the current state as visited memo[i, sum] = 1; // Include the current element (arr[i]) // in the subset sum distSumRec(arr, n, sum + arr[i], i + 1, memo); // Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, memo); } // This function calls distSumRec() to generate // distinct sum subsets and return them static List<int> DistinctSum(int[] arr) { int n = arr.Length; // Calculate the maximum possible sum int totalSum = 0; foreach (var num in arr) { totalSum += num; } // Memoization table initialized with -1 int[,] memo = new int[n + 1, totalSum + 1]; for (int i = 0; i <= n; i++) { for (int j = 0; j <= totalSum; j++) { memo[i, j] = -1; } } // Start the recursive process with an // initial sum of 0 and index 0 distSumRec(arr, n, 0, 0, memo); // Collect all distinct sums from // the memo table List<int> result = new List<int>(); for (int i = 0; i <= totalSum; i++) { if (memo[n, i] == 1) { result.Add(i); } } return result; } static void Main() { int[] arr = {2, 3, 4, 5, 6}; List<int> result = DistinctSum(arr); foreach (int sum in result) { Console.Write(sum + " "); } } } // JavaScript program to print distinct subset sums of // a given array. // Recursive function to calculate distinct subset sums function distSumRec(arr, n, sum, i, memo) { // If we have considered all elements in // the array, mark this sum if (i === n) { memo[i][sum] = 1; return; } // If this state has already been computed, // skip further processing if (memo[i][sum] !== -1) { return; } // Mark the current state as visited memo[i][sum] = 1; // Include the current element (arr[i]) in the subset // sum distSumRec(arr, n, sum + arr[i], i + 1, memo); // Exclude the current element from the subset sum distSumRec(arr, n, sum, i + 1, memo); } // This function calls distSumRec() to generate // distinct sum subsets and return them function DistinctSum(arr) { const n = arr.length; // Calculate the maximum possible sum const totalSum = arr.reduce((acc, num) => acc + num, 0); // Memoization table initialized with -1 const memo = Array.from({length : n + 1}, () => Array(totalSum + 1).fill(-1)); // Start the recursive process with an // initial sum of 0 and index 0 distSumRec(arr, n, 0, 0, memo); // Collect all distinct sums from the memo table const result = []; for (let i = 0; i <= totalSum; i++) { if (memo[n][i] === 1) { result.push(i); } } return result; } const arr = [ 2, 3, 4, 5, 6 ]; const result = DistinctSum(arr); console.log(result.join(" ")); Output0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20
Using Bottom-Up DP (Tabulation) – O(n * sum) Time and O(n * sum) Space
The approach is similar to the previous one. just instead of breaking down the problem recursively, we iteratively build up the solution by calculating in bottom-up manner.
We will create a 2D array dp[][] of size (n + 1) x (sum + 1) where sum is the sum of all elements in the array. Each dp[i][j] represents whether a subset of the first i elements of the array can sum to j. dp[i][j] = true means that there is a subset of elements from arr[0..i-1] that sums to j.
Base case:
- We always have a subset with sum 0, which is the empty subset. Therefore, we initialize the first column dp[i][0] = true for all i, as the sum of 0 is always achievable with the empty subset.
For every element arr[i-1], we will either include or exclude it in the subset sum, and update the table accordingly.
- Including the Element: If we include arr[i-1], then the new sum becomes j + arr[i-1]. So, we will check the previous state dp[i-1][j-arr[i-1]] to see if that sum was possible.
- Excluding the Element: If we exclude arr[i-1], then the sum remains j, and we will check the previous state dp[i-1][j] to see if that sum was possible without including the element.
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 vector<int> DistinctSum(vector<int> &arr) { int n = arr.size(); 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. vector<vector<bool>> dp(n + 1, vector<bool>(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; } } } vector<int> result; for (int j = 0; j <= sum; j++) if (dp[n][j] == true) result.push_back(j); return result; } int main() { vector<int> arr = {2, 3, 4, 5, 6}; vector<int> res = DistinctSum(arr); for (int i : res) cout << i << " "; return 0; } // Java program to print distinct subset sums of // a given array using dynamic programming. import java.util.*; class GfG { // Uses Dynamic Programming to find // distinct subset sums static List<Integer> DistinctSum(int[] arr) { int n = arr.length; 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 the // current array element if (dp[i - 1][j] == true) { dp[i][j] = true; dp[i][j + arr[i - 1]] = true; } } } List<Integer> result = new ArrayList<>(); for (int j = 0; j <= sum; j++) { if (dp[n][j]) { result.add(j); } } return result; } public static void main(String[] args) { int[] arr = { 2, 3, 4, 5, 6 }; List<Integer> result = DistinctSum(arr); for (int i : result) { System.out.print(i + " "); } } } # Python program to print distinct subset sums of # a given array using dynamic programming. # Uses Dynamic Programming to find distinct # subset sums def DistinctSum(arr): n = len(arr) total_sum = sum(arr) # dp[i][j] would be true if arr[0..i-1] has a # subset with sum equal to j. dp = [[False] * (total_sum + 1) for _ 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, total_sum + 1): # Sums that were achievable without the # current array element if dp[i - 1][j]: dp[i][j] = True dp[i][j + arr[i - 1]] = True result = [j for j in range(total_sum + 1) if dp[n][j]] return result arr = [2, 3, 4, 5, 6] result = DistinctSum(arr) print(" ".join(map(str, result))) // C# program to print distinct subset sums of // a given array using dynamic programming. using System; using System.Collections.Generic; class GfG { // Uses Dynamic Programming to find // distinct subset sums static List<int> DistinctSum(int[] arr) { int n = arr.Length; int sum = 0; foreach(int num in arr) sum += num; // 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 the // current array element if (dp[i - 1, j]) { dp[i, j] = true; dp[i, j + arr[i - 1]] = true; } } } List<int> result = new List<int>(); for (int j = 0; j <= sum; j++) { if (dp[n, j]) { result.Add(j); } } return result; } static void Main() { int[] arr = { 2, 3, 4, 5, 6 }; List<int> result = DistinctSum(arr); foreach(int i in result) { Console.Write(i + " "); } } } // JavaScript program to print distinct subset sums // of a given array using dynamic programming. // Uses Dynamic Programming to find distinct subset sums function DistinctSum(arr) { const n = arr.length; const totalSum = arr.reduce((acc, num) => acc + num, 0); // dp[i][j] would be true if arr[0..i-1] has a // subset with sum equal to j. let dp = Array.from({length : n + 1}, () => Array(totalSum + 1).fill(false)); // There is always a subset with 0 sum for (let i = 0; i <= n; i++) { dp[i][0] = true; } // Fill dp[][] in bottom up manner for (let i = 1; i <= n; i++) { dp[i][arr[i - 1]] = true; for (let j = 1; j <= totalSum; j++) { // Sums that were achievable without the current // array element if (dp[i - 1][j]) { dp[i][j] = true; dp[i][j + arr[i - 1]] = true; } } } let result = []; for (let j = 0; j <= totalSum; j++) { if (dp[n][j]) { result.push(j); } } return result; } let arr = [ 2, 3, 4, 5, 6 ]; let result = DistinctSum(arr); console.log(result.join(" ")); Output0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20
Optimized Bit-set Approach:
Above Code snippet does the same as naive solution, where dp is a bit mask (we’ll use bit-set). Lets see how:
- dp ? all the sums which were produced before element a[i]
- dp << a[i] ? shifting all the sums by a[i], i.e. adding a[i] to all the sums.
- For example, Suppose initially the bit-mask was 000010100 meaning we could generate only 2 and 4 (count from right).
- Now if we get a element 3, we could make 5 and 7 as well by adding to 2 and 4 respectively.
- This can be denoted by 010100000 which is equivalent to (000010100) << 3
- dp | (dp << a[i]) ? 000010100 | 010100000 = 010110100 This is union of above two sums representing which sums are possible, namely 2, 4, 5 and 7.
C++ // C++ program to compute all possible // distinct subset sums using BitSet #include <bits/stdc++.h> using namespace std; // Function to compute all possible // distinct subset sums using bitset vector<int> DistinctSum(vector<int> &arr) { int n = arr.size(); // Calculate total sum of the array int sum = accumulate(arr.begin(), arr.end(), 0); // Bitset of size sum+1, dp[i] is 1 // if sum i is possible, 0 otherwise bitset<100001> dp; // Sum 0 is always possible (empty subset) dp[0] = 1; for (int i = 0; i < n; ++i) { // Shift the current possible sums by a[i] dp |= dp << arr[i]; } // Collect all the sums that are possible vector<int> res; for (int i = 0; i <= sum; ++i) { if (dp[i]) { // If dp[i] is 1, it means sum i is possible res.push_back(i); } } return res; } int main() { vector<int> arr = {2, 3, 4, 5, 6}; vector<int> result = DistinctSum(arr); for (int sum : result) { cout << sum << " "; } return 0; } // Java program to compute all possible // distinct subset sums using BitSet import java.util.*; class GfG { // Function to compute all possible // distinct subset sums using BitSet static List<Integer> DistinctSum(int[] arr) { int n = arr.length; // Calculate total sum of the array int sum = Arrays.stream(arr).sum(); // BitSet of size sum+1, dp[i] is 1 // if sum i is possible, 0 otherwise BitSet dp = new BitSet(sum + 1); // Sum 0 is always possible (empty subset) dp.set(0); for (int i = 0; i < n; ++i) { // Create a copy of the current dp state BitSet temp = (BitSet) dp.clone(); // Add the current element to the previously // possible sums for (int j = 0; j <= sum; ++j) { if (dp.get(j)) { temp.set(j + arr[i]); } } // Update dp with the new sums dp.or(temp); } // Collect all the sums that are possible List<Integer> res = new ArrayList<>(); for (int i = 0; i <= sum; ++i) { if (dp.get(i)) { // If dp[i] is 1, it means sum i is possible res.add(i); } } return res; } public static void main(String[] args) { int[] arr = {2, 3, 4, 5, 6}; List<Integer> result = DistinctSum(arr); for (int sum : result) { System.out.print(sum + " "); } } } # Python program to compute all possible # distinct subset sums using bitset # Function to compute all possible # distinct subset sums using bitset def DistinctSum(arr): n = len(arr) # Calculate total sum of the array total_sum = sum(arr) # Bitset of size total_sum+1, dp[i] is 1 if # sum i is possible, 0 otherwise dp = [False] * (total_sum + 1) # Sum 0 is always possible (empty subset) dp[0] = True for num in arr: # Shift the current possible sums by num for j in range(total_sum, num - 1, -1): if dp[j - num]: dp[j] = True # Collect all the sums that are possible result = [i for i in range(total_sum + 1) if dp[i]] return result arr = [2, 3, 4, 5, 6] result = DistinctSum(arr) print(" ".join(map(str, result))) // C# program to compute all possible // distinct subset sums using BitArray using System; using System.Collections; using System.Collections.Generic; class GfG { // Function to compute all possible // distinct subset sums using BitArray static List<int> DistinctSum(int[] arr) { int n = arr.Length; // Calculate total sum of the array int sum = 0; foreach(var num in arr) sum += num; // BitArray of size sum+1, dp[i] is 1 if // sum i is possible, 0 otherwise BitArray dp = new BitArray(sum + 1); // Sum 0 is always possible (empty subset) dp[0] = true; for (int i = 0; i < n; ++i) { // Create a clone of the current dp state BitArray temp = (BitArray)dp.Clone(); for (int j = 0; j <= sum - arr[i]; ++j) { if (dp[j]) { temp[j + arr[i]] = true; } } // Update dp with the new sums dp = temp; } // Collect all the sums that are possible List<int> result = new List<int>(); for (int i = 0; i <= sum; ++i) { if (dp[i]) { // If dp[i] is true, it means sum i is // possible result.Add(i); } } return result; } static void Main() { int[] arr = { 2, 3, 4, 5, 6 }; List<int> result = DistinctSum(arr); foreach(int sum in result) { Console.Write(sum + " "); } } } // JavaScript program to compute all possible // distinct subset sums using bitset // Function to compute all possible // distinct subset sums using bitset function DistinctSum(arr) { const n = arr.length; // Calculate total sum of the array const totalSum = arr.reduce((acc, num) => acc + num, 0); // Bitset of size totalSum+1, dp[i] is 1 // if sum i is possible, 0 otherwise let dp = Array(totalSum + 1).fill(false); // Sum 0 is always possible (empty subset) dp[0] = true; for (let i = 0; i < n; ++i) { // Shift the current possible sums by a[i] for (let j = totalSum; j >= arr[i]; --j) { if (dp[j - arr[i]]) { dp[j] = true; } } } // Collect all the sums that are possible let result = []; for (let i = 0; i <= totalSum; ++i) { if (dp[i]) { result.push(i); } } return result; } let arr = [ 2, 3, 4, 5, 6 ]; let result = DistinctSum(arr); console.log(result.join(" ")); Output0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20
Time Complexity: also seems to be O(n * s). Because if we would have used a array instead of bitset the shifting would have taken linear time O(S). However the shift (and almost all) operation on bitset takes O(s / w) time. Where w is the word size of the system, Usually its 32 bit or 64 bit. Thus the final time complexity becomes O(n * s / w)
Auxiliary Space:O(m), where m is the maximum value of the input array.
Some Important Points:
- The size of bitset must be a constant, this sometimes is a drawback as we might waste some space.
- Bitset can be thought of a array where every element takes care of W elements. For example 010110100 is equivalent to {2, 6, 4} in a hypothetical system with word size w = 3.
- Bitset optimized knapsack solution reduced the time complexity by a factor of w which sometimes is just enough to get AC.
Similar Reads
Subset Sum Problem
Given an array arr[] of non-negative integers and a value sum, the task is to check if there is a subset of the given array whose sum is equal to the given sum. Examples: Input: arr[] = [3, 34, 4, 12, 5, 2], sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9. Input: arr[] = [3, 34,
15+ min read
Subset sum in Different languages
Python Program for Subset Sum Problem | DP-25
Write a Python program for a given set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum. Examples: Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9. Input:
7 min read
Java Program for Subset Sum Problem | DP-25
Write a Java program for a given set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum. Examples: Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9. Input: se
8 min read
C Program for Subset Sum Problem | DP-25
Write a C program for a given set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum. Examples: Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9. Input: set[]
8 min read
PHP Program for Subset Sum Problem | DP-25
Write a PHP program for a given set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum. Examples: Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9. Input: set
7 min read
C# Program for Subset Sum Problem | DP-25
Write a C# program for a given set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum. Examples: Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9. Input: set[
8 min read
Subset Sum Problem using Backtracking
Given a set[] of non-negative integers and a value sum, the task is to print the subset of the given set whose sum is equal to the given sum. Examples:Â Input:Â set[] = {1,2,1}, sum = 3Output:Â [1,2],[2,1]Explanation:Â There are subsets [1,2],[2,1] with sum 3. Input:Â set[] = {3, 34, 4, 12, 5, 2}, sum =
8 min read
Print all subsets with given sum
Given an array arr[] of non-negative integers and an integer target. The task is to print all subsets of the array whose sum is equal to the given target. Note: If no subset has a sum equal to target, print -1. Examples: Input: arr[] = [5, 2, 3, 10, 6, 8], target = 10Output: [ [5, 2, 3], [2, 8], [10
15+ min read
Subset Sum Problem in O(sum) space
Given an array of non-negative integers and a value sum, determine if there is a subset of the given set with sum equal to given sum. Examples: Input: arr[] = {4, 1, 10, 12, 5, 2}, sum = 9Output: TRUEExplanation: {4, 5} is a subset with sum 9. Input: arr[] = {1, 8, 2, 5}, sum = 4Output: FALSE Explan
13 min read
Subset Sum is NP Complete
Prerequisite: NP-Completeness, Subset Sum Problem Subset Sum Problem: Given N non-negative integers a1...aN and a target sum K, the task is to decide if there is a subset having a sum equal to K. Explanation: An instance of the problem is an input specified to the problem. An instance of the subset
5 min read
Minimum Subset sum difference problem with Subset partitioning
Given a set of N integers with up to 40 elements, the task is to partition the set into two subsets of equal size (or the closest possible), such that the difference between the sums of the subsets is minimized. If the size of the set is odd, one subset will have one more element than the other. If
13 min read
Maximum subset sum such that no two elements in set have same digit in them
Given an array of N elements. Find the subset of elements which has maximum sum such that no two elements in the subset has common digit present in them.Examples: Input : array[] = {22, 132, 4, 45, 12, 223} Output : 268 Maximum Sum Subset will be = {45, 223} . All possible digits are present except
12 min read
Find all distinct subset (or subsequence) sums of an array
Given an array arr[] of size n, the task is to find a distinct sum that can be generated from the subsets of the given sets and return them in increasing order. It is given that the sum of array elements is small. Examples: Input: arr[] = [1, 2]Output: [0, 1, 2, 3]Explanation: Four distinct sums can
15+ min read
Subset sum problem where Array sum is at most N
Given an array arr[] of size N such that the sum of all the array elements does not exceed N, and array queries[] containing Q queries. For each query, the task is to find if there is a subset of the array whose sum is the same as queries[i]. Examples: Input: arr[] = {1, 0, 0, 0, 0, 2, 3}, queries[]
10 min read