Java Program for Subset Sum Problem | DP-25
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.
// 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 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:
// 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:
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!
