Subset Sum Problem in O(sum) space

3.2

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 cab 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 = 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


Output:

There exists a subset with given sum

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