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
{4, 5} is a subset with sum 9.
Input : arr[] = {1, 8, 2, 5},
sum = 4
Output : FALSE
There exists no subset with sum 4.
We have discussed a Dynamic Programming based solution in below post.
Dynamic Programming | Set 25 (Subset Sum Problem)
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.
C++
// Returns true if there exists a subset // with given sum in arr[] #include <iostream> using namespace std; 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) cout <<"There exists a subset with given sum"; else cout <<"No subset exists with given sum"; return 0; } // This code is contributed by shivanisinghss2110 |
C
// 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
// 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 |
Python
# 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 = 7n = 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#
// 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 |
PHP
<?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 ?> |
Javascript
<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> |
Output
There exists a subset with given sum
Another Approach: 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.
Here’s the code for the given approach:
C++
#include <iostream> using namespace std; bool isPossible(int elements[], int sum, int n) { int dp[sum + 1]; // 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 |
Java
import java.io.*; import java.util.*; class GFG { static boolean isPossible(int elements[], int sum) { int dp[] = new int[sum + 1]; // 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"); } } |
Python3
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. |
C#
using System; class GFG { static Boolean isPossible(int []elements, int sum) { int []dp = new int[sum + 1]; // 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 |
Javascript
<script> function isPossible(elements, sum) { var dp = [sum + 1]; // 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 </script> |
Output
YES
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.
This article is contributed by Neelesh (Neelesh_Sinha). If you like neveropen and would like to contribute, you can also write an article using write.neveropen.co.za or mail your article to review-team@neveropen.co.za. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
