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Java Program for Subset Sum Problem | DP-25

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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 = 9
Output: True
Explanation: There is a subset (4, 5) with sum 9.

Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30
Output: False
Explanation: There is no subset that adds up to 30.

Java Program for Subset Sum Problem using Recursion:

For the recursive approach, there will be two cases.

  • Consider the ‘last’ element to be a part of the subset. Now the new required sum = required sum – value of ‘last’ element.
  • Don’t include the ‘last’ element in the subset. Then the new required sum = old required sum.

In both cases, the number of available elements decreases by 1.

Step-by-step approach:

  • Build a recursive function and pass the index to be considered (here gradually moving from the last end) and the remaining sum amount.
  • For each index check the base cases and utilize the above recursive call.
  • If the answer is true for any recursion call, then there exists such a subset. Otherwise, no such subset exists.

Below is the implementation of the above approach.

Java




// A recursive solution for subset sum
 
import java.io.*;
class GFG {
 
    // Returns true if there is a subset
    // of set[] with sum equal to given sum
    static boolean isSubsetSum(int set[], int n, int sum)
    {
        // Base Cases
        if (sum == 0)
            return true;
        if (n == 0)
            return false;
 
        // If last element is greater than
        // sum, then ignore it
        if (set[n - 1] > sum)
            return isSubsetSum(set, n - 1, sum);
 
        // Else, check if sum can be obtained
        // by any of the following
        // (a) including the last element
        // (b) excluding the last element
        return isSubsetSum(set, n - 1, sum)
            || isSubsetSum(set, n - 1, sum - set[n - 1]);
    }
 
    // Driver code
    public static void main(String args[])
    {
        int set[] = { 3, 34, 4, 12, 5, 2 };
        int sum = 9;
        int n = set.length;
        if (isSubsetSum(set, n, sum) == true)
            System.out.println("Found a subset"
                            + " with given sum");
        else
            System.out.println("No subset with"
                            + " given sum");
    }
}
 
/* This code is contributed by Rajat Mishra */


Output

Found a subset with given sum


Time Complexity: O(2n)
Auxiliary space: O(n)

Java Program for Subset Sum Problem using Memoization:

As seen in the previous recursion method, each state of the solution can be uniquely identified using two variables – the index and the remaining sum. So create a 2D array to store the value of each state to avoid recalculation of the same state.

Below is the implementation of the above approach:

Java




// Java program for the above approach
import java.io.*;
class GFG {
 
    // Check if possible subset with
    // given sum is possible or not
    static int subsetSum(int a[], int n, int sum)
    {
        // Storing the value -1 to the matrix
        int tab[][] = new int[n + 1][sum + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= sum; j++) {
                tab[i][j] = -1;
            }
        }
 
        // If the sum is zero it means
        // we got our expected sum
        if (sum == 0)
            return 1;
 
        if (n <= 0)
            return 0;
 
        // If the value is not -1 it means it
        // already call the function
        // with the same value.
        // it will save our from the repetition.
        if (tab[n - 1][sum] != -1)
            return tab[n - 1][sum];
 
        // If the value of a[n-1] is
        // greater than the sum.
        // we call for the next value
        if (a[n - 1] > sum)
            return tab[n - 1][sum]
                = subsetSum(a, n - 1, sum);
        else {
 
            // Here we do two calls because we
            // don't know which value is
            // full-fill our criteria
            // that's why we doing two calls
            if (subsetSum(a, n - 1, sum) != 0
                || subsetSum(a, n - 1, sum - a[n - 1])
                    != 0) {
                return tab[n - 1][sum] = 1;
            }
            else
                return tab[n - 1][sum] = 0;
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 5;
        int a[] = { 1, 5, 3, 7, 4 };
        int sum = 12;
 
        if (subsetSum(a, n, sum) != 0) {
            System.out.println("YES\n");
        }
        else
            System.out.println("NO\n");
    }
}
 
// This code is contributed by rajsanghavi9.


Output

YES



Time Complexity: O(sum*n)
Auxiliary space: O(n)

Java Program for Subset Sum Problem using Dynamic Programming:

We can solve the problem in Pseudo-polynomial time we can use the Dynamic programming approach.

So we will create a 2D array of size (n + 1) * (sum + 1) of type boolean. The state dp[i][j] will be true if there exists a subset of elements from set[0 . . . i] with sum value = ‘j’.

The dynamic programming relation is as follows:

if (A[i-1] > j)
dp[i][j] = dp[i-1][j]
else
dp[i][j] = dp[i-1][j] OR dp[i-1][j-set[i-1]]

Below is the implementation of the above approach:

Java




// A Dynamic Programming solution for subset
// sum problem
import java.io.*;
class GFG {
 
    // Returns true if there is a subset of
    // set[] with sum equal to given sum
    static boolean isSubsetSum(int set[], int n, int sum)
    {
        // The value of subset[i][j] will be
        // true if there is a subset of
        // set[0..j-1] with sum equal to i
        boolean subset[][] = new boolean[sum + 1][n + 1];
 
        // If sum is 0, then answer is true
        for (int i = 0; i <= n; i++)
            subset[0][i] = true;
 
        // If sum is not 0 and set is empty,
        // then answer is false
        for (int i = 1; i <= sum; i++)
            subset[i][0] = false;
 
        // Fill the subset table in bottom
        // up manner
        for (int i = 1; i <= sum; i++) {
            for (int j = 1; j <= n; j++) {
                subset[i][j] = subset[i][j - 1];
                if (i >= set[j - 1])
                    subset[i][j]
                        = subset[i][j]
                        || subset[i - set[j - 1]][j - 1];
            }
        }
 
        return subset[sum][n];
    }
 
    // Driver code
    public static void main(String args[])
    {
        int set[] = { 3, 34, 4, 12, 5, 2 };
        int sum = 9;
        int n = set.length;
        if (isSubsetSum(set, n, sum) == true)
            System.out.println("Found a subset"
                            + " with given sum");
        else
            System.out.println("No subset with"
                            + " given sum");
    }
}
 
/* This code is contributed by Rajat Mishra */


Output

Found a subset with given sum


Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum*n), as the size of the 2-D array is sum*n.

Java Program for Subset Sum Problem using Dynamic Programming with space optimization to linear:

In previous approach of dynamic programming we have derive the relation between states as given below:
if (A[i-1] > j)
dp[i][j] = dp[i-1][j]
else
dp[i][j] = dp[i-1][j] OR dp[i-1][j-set[i-1]]
If we observe that for calculating current dp[i][j] state we only need previous row dp[i-1][j] or dp[i-1][j-set[i-1]].
There is no need to store all the previous states just one previous state is used to compute result.

Step-by-step approach:

  • Define two arrays prev and curr of size Sum+1 to store the just previous row result and current row result respectively.
  • Once curr array is calculated then curr becomes our prev for the next row.
  • When all rows are processed the answer is stored in prev array.

Below is the implementation of the above approach:

Java




import java.util.Arrays;
 
public class SubsetSum {
    public static boolean isSubsetSum(int[] set, int n, int sum) {
        boolean[] prev = new boolean[sum + 1];
 
        Arrays.fill(prev, false);
 
        for (int i = 0; i <= n; i++) {
            prev[0] = true;
        }
 
        boolean[] curr = new boolean[sum + 1];
 
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= sum; j++) {
                if (j < set[i - 1]) {
                    curr[j] = prev[j];
                }
                if (j >= set[i - 1]) {
                    curr[j] = prev[j] || prev[j - set[i - 1]];
                }
            }
            // Now curr becomes prev for (i + 1)th element
            System.arraycopy(curr, 0, prev, 0, sum + 1);
        }
 
        return prev[sum];
    }
 
    public static void main(String[] args) {
        int[] set = {3, 34, 4, 12, 5, 2};
        int sum = 9;
        int n = set.length;
 
        if (isSubsetSum(set, n, sum)) {
            System.out.println("Found a subset with given sum");
        } else {
            System.out.println("No subset with given sum");
        }
    }
}


Output

Found a subset with given sum


Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum), as the size of the 1-D array is sum+1.

Please refer complete article on Subset Sum Problem | DP-25 for more details!



Last Updated : 10 Nov, 2023
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