Subset Sum Problem in O(sum) space
Last Updated : 16 May, 2024
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 = 9
Output: TRUE
Explanation: {4, 5} is a subset with sum 9.
Input: arr[] = {1, 8, 2, 5}, sum = 4
Output: FALSE
Explanation: There exists no subset with sum 4.
We have discussed a Dynamic Programming based solution in the post "Dynamic Programming | Set 25 (Subset Sum Problem)".
Subset Sum Problem in O(sum) space using 2D array:
The solution discussed above requires O(n * sum) space and O(n * sum) time. We can optimize space. We create a boolean 2D array subset[2][sum+1]. Using bottom-up manner we can fill up this table. The idea behind using 2 in "subset[2][sum+1]" is that for filling a row only the values from previous row are required. So alternate rows are used either making the first one as current and second as previous or the first as previous and second as current.
Below is the implementation of the above approach:
C++ #include <iostream> using namespace std; bool isSubsetSum(int arr[], int n, int sum) { // The value of subset[i][j] will be true // if there exists a subset of sum j in // arr[0, 1, ...., i-1] bool subset[n+1][sum + 1]; for (int i = 0; i <= n; i++) { for (int j = 0; j <= sum; j++) { // A subset with sum 0 is always possible if (j == 0) subset[i][j] = true; // If there exists no element no sum // is possible else if (i == 0) subset[i][j] = false; else if (arr[i - 1] <= j) subset[i][j] = subset[i - 1][j - arr[i - 1]] || subset[i - 1][j]; else subset[i][j] = subset[i - 1][j]; } } return subset[n][sum]; } // Driver code int main() { int arr[] = { 6, 2, 5 }; int sum = 7; int n = sizeof(arr) / sizeof(arr[0]); if (isSubsetSum(arr, n, sum) == true) cout <<"There exists a subset with given sum"; else cout <<"No subset exists with given sum"; return 0; } // Returns true if there exists a subset // with given sum in arr[] #include <stdio.h> #include <stdbool.h> bool isSubsetSum(int arr[], int n, int sum) { // The value of subset[i%2][j] will be true // if there exists a subset of sum j in // arr[0, 1, ...., i-1] bool subset[2][sum + 1]; for (int i = 0; i <= n; i++) { for (int j = 0; j <= sum; j++) { // A subset with sum 0 is always possible if (j == 0) subset[i % 2][j] = true; // If there exists no element no sum // is possible else if (i == 0) subset[i % 2][j] = false; else if (arr[i - 1] <= j) subset[i % 2][j] = subset[(i + 1) % 2] [j - arr[i - 1]] || subset[(i + 1) % 2][j]; else subset[i % 2][j] = subset[(i + 1) % 2][j]; } } return subset[n % 2][sum]; } // Driver code int main() { int arr[] = { 6, 2, 5 }; int sum = 7; int n = sizeof(arr) / sizeof(arr[0]); if (isSubsetSum(arr, n, sum) == true) printf("There exists a subset with given sum"); else printf("No subset exists with given sum"); return 0; } // Java Program to get a subset with a // with a sum provided by the user public class Subset_sum { // Returns true if there exists a subset // with given sum in arr[] static boolean isSubsetSum(int arr[], int n, int sum) { // The value of subset[i%2][j] will be true // if there exists a subset of sum j in // arr[0, 1, ...., i-1] boolean subset[][] = new boolean[2][sum + 1]; for (int i = 0; i <= n; i++) { for (int j = 0; j <= sum; j++) { // A subset with sum 0 is always possible if (j == 0) subset[i % 2][j] = true; // If there exists no element no sum // is possible else if (i == 0) subset[i % 2][j] = false; else if (arr[i - 1] <= j) subset[i % 2][j] = subset[(i + 1) % 2] [j - arr[i - 1]] || subset[(i + 1) % 2][j]; else subset[i % 2][j] = subset[(i + 1) % 2][j]; } } return subset[n % 2][sum]; } // Driver code public static void main(String args[]) { int arr[] = { 1, 2, 5 }; int sum = 7; int n = arr.length; if (isSubsetSum(arr, n, sum) == true) System.out.println("There exists a subset with" + " given sum"); else System.out.println("No subset exists with" + " given sum"); } } // This code is contributed by Sumit Ghosh # Returns true if there exists a subset # with given sum in arr[] def isSubsetSum(arr, n, sum): # The value of subset[i%2][j] will be true # if there exists a subset of sum j in # arr[0, 1, ...., i-1] subset = [[False for j in range(sum + 1)] for i in range(3)] for i in range(n + 1): for j in range(sum + 1): # A subset with sum 0 is always possible if (j == 0): subset[i % 2][j] = True # If there exists no element no sum # is possible elif (i == 0): subset[i % 2][j] = False elif (arr[i - 1] <= j): subset[i % 2][j] = subset[(i + 1) % 2][j - arr[i - 1]] or subset[(i + 1) % 2][j] else: subset[i % 2][j] = subset[(i + 1) % 2][j] return subset[n % 2][sum] # Driver code arr = [6, 2, 5] sum = 7 n = len(arr) if (isSubsetSum(arr, n, sum) == True): print("There exists a subset with given sum") else: print("No subset exists with given sum") # This code is contributed by Sachin Bisht // C# Program to get a subset with a // with a sum provided by the user using System; public class Subset_sum { // Returns true if there exists a subset // with given sum in arr[] static bool isSubsetSum(int []arr, int n, int sum) { // The value of subset[i%2][j] will be true // if there exists a subset of sum j in // arr[0, 1, ...., i-1] bool [,]subset = new bool[2,sum + 1]; for (int i = 0; i <= n; i++) { for (int j = 0; j <= sum; j++) { // A subset with sum 0 is always possible if (j == 0) subset[i % 2,j] = true; // If there exists no element no sum // is possible else if (i == 0) subset[i % 2,j] = false; else if (arr[i - 1] <= j) subset[i % 2,j] = subset[(i + 1) % 2,j - arr[i - 1]] || subset[(i + 1) % 2,j]; else subset[i % 2,j] = subset[(i + 1) % 2,j]; } } return subset[n % 2,sum]; } // Driver code public static void Main() { int []arr = { 1, 2, 5 }; int sum = 7; int n = arr.Length; if (isSubsetSum(arr, n, sum) == true) Console.WriteLine("There exists a subset with" + "given sum"); else Console.WriteLine("No subset exists with" + "given sum"); } } // This code is contributed by Ryuga <script> // Javascript Program to get a subset with a // with a sum provided by the user // Returns true if there exists a subset // with given sum in arr[] function isSubsetSum(arr, n, sum) { // The value of subset[i%2][j] will be true // if there exists a subset of sum j in // arr[0, 1, ...., i-1] let subset = new Array(2); // Loop to create 2D array using 1D array for (var i = 0; i < subset.length; i++) { subset[i] = new Array(2); } for (let i = 0; i <= n; i++) { for (let j = 0; j <= sum; j++) { // A subset with sum 0 is always possible if (j == 0) subset[i % 2][j] = true; // If there exists no element no sum // is possible else if (i == 0) subset[i % 2][j] = false; else if (arr[i - 1] <= j) subset[i % 2][j] = subset[(i + 1) % 2] [j - arr[i - 1]] || subset[(i + 1) % 2][j]; else subset[i % 2][j] = subset[(i + 1) % 2][j]; } } return subset[n % 2][sum]; } // driver program let arr = [ 1, 2, 5 ]; let sum = 7; let n = arr.length; if (isSubsetSum(arr, n, sum) == true) document.write("There exists a subset with" + "given sum"); else document.write("No subset exists with" + "given sum"); // This code is contributed by code_hunt. </script> <?php // Returns true if there exists a subset // with given sum in arr[] function isSubsetSum($arr, $n, $sum) { // The value of subset[i%2][j] will be // true if there exists a subset of // sum j in arr[0, 1, ...., i-1] $subset[2][$sum + 1] = array(); for ($i = 0; $i <= $n; $i++) { for ($j = 0; $j <= $sum; $j++) { // A subset with sum 0 is // always possible if ($j == 0) $subset[$i % 2][$j] = true; // If there exists no element no // sum is possible else if ($i == 0) $subset[$i % 2][$j] = false; else if ($arr[$i - 1] <= $j) $subset[$i % 2][$j] = $subset[($i + 1) % 2] [$j - $arr[$i - 1]] || $subset[($i + 1) % 2][$j]; else $subset[$i % 2][$j] = $subset[($i + 1) % 2][$j]; } } return $subset[$n % 2][$sum]; } // Driver code $arr = array( 6, 2, 5 ); $sum = 7; $n = sizeof($arr); if (isSubsetSum($arr, $n, $sum) == true) echo ("There exists a subset with given sum"); else echo ("No subset exists with given sum"); // This code is contributed by Sach_Code ?> OutputThere exists a subset with given sum
Subset Sum Problem in O(sum) space using 1D array:
To further reduce space complexity, we create a boolean 1D array subset[sum+1]. Using bottom-up manner we can fill up this table. The idea is that we can check if the sum till position "i" is possible then if the current element in the array at position j is x, then sum i+x is also possible. We traverse the sum array from back to front so that we don't count any element twice.
Below is the implementation of the above approach:
C++ #include <iostream> using namespace std; bool isPossible(int elements[], int sum, int n) { int dp[sum + 1] = { 0 }; // Initializing with 1 as sum 0 is // always possible dp[0] = 1; // Loop to go through every element of // the elements array for (int i = 0; i < n; i++) { // To change the values of all possible sum // values to 1 for (int j = sum; j >= elements[i]; j--) { if (dp[j - elements[i]] == 1) dp[j] = 1; } } // If sum is possible then return 1 if (dp[sum] == 1) return true; return false; } // Driver code int main() { int elements[] = { 6, 2, 5 }; int n = sizeof(elements) / sizeof(elements[0]); int sum = 7; if (isPossible(elements, sum, n)) cout << ("YES"); else cout << ("NO"); return 0; } // This code is contributed by Potta Lokesh // This code is modified by Susobhan Akhuli import java.io.*; import java.util.*; class GFG { static boolean isPossible(int elements[], int sum) { int dp[] = new int[sum + 1]; Arrays.fill(dp, 0); // initializing with 1 as sum 0 is always possible dp[0] = 1; // loop to go through every element of the elements // array for (int i = 0; i < elements.length; i++) { // to change the values of all possible sum // values to 1 for (int j = sum; j >= elements[i]; j--) { if (dp[j - elements[i]] == 1) dp[j] = 1; } } // if sum is possible then return 1 if (dp[sum] == 1) return true; return false; } public static void main(String[] args) throws Exception { int elements[] = { 6, 2, 5 }; int sum = 7; if (isPossible(elements, sum)) System.out.println("YES"); else System.out.println("NO"); } } // This code is modified by Susobhan Akhuli def isPossible(elements, target): dp = [False]*(target+1) # initializing with 1 as sum 0 is always possible dp[0] = True # loop to go through every element of the elements array for ele in elements: # to change the value o all possible sum values to True for j in range(target, ele - 1, -1): if dp[j - ele]: dp[j] = True # If target is possible return True else False return dp[target] # Driver code arr = [6, 2, 5] target = 7 if isPossible(arr, target): print("YES") else: print("NO") # The code is contributed by Arpan. using System; class GFG { static Boolean isPossible(int[] elements, int sum) { int[] dp = new int[sum + 1]; Array.Fill(dp, 0); // initializing with 1 as sum 0 is always possible dp[0] = 1; // loop to go through every element of the elements // array for (int i = 0; i < elements.Length; i++) { // to change the values of all possible sum // values to 1 for (int j = sum; j >= elements[i]; j--) { if (dp[j - elements[i]] == 1) dp[j] = 1; } } // if sum is possible then return 1 if (dp[sum] == 1) return true; return false; } // Driver code public static void Main(String[] args) { int[] elements = { 6, 2, 5 }; int sum = 7; if (isPossible(elements, sum)) Console.Write("YES"); else Console.Write("NO"); } } // This code is contributed by shivanisinghss2110 // This code is modified by Susobhan Akhuli <script> function isPossible(elements, sum) { var dp = Array(sum+1).fill(0); // initializing with 1 as sum 0 is always possible dp[0] = 1; // loop to go through every element of the elements // array for (var i = 0; i < elements.length; i++) { // to change the values of all possible sum // values to 1 for (var j = sum; j >= elements[i]; j--) { if (dp[j - elements[i]] == 1) dp[j] = 1; } } // if sum is possible then return 1 if (dp[sum] == 1) return true; return false; } var elements = [ 6, 2, 5 ]; var sum = 7; if (isPossible(elements, sum)) document.write("YES"); else document.write("NO"); // This code is contributed by shivanisinghss2110 // This code is modified by Susobhan Akhuli </script> Time Complexity: O(N*K) where N is the number of elements in the array and K is total sum.
Auxiliary Space: O(K), since K extra space has been taken.
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