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 = 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 is 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 <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 ?> |
Output:
There exists a subset with given sum
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