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.
Java
import java.io.*;
class GFG {
static boolean isSubsetSum( int set[], int n, int sum)
{
if (sum == 0 )
return true ;
if (n == 0 )
return false ;
if (set[n - 1 ] > sum)
return isSubsetSum(set, n - 1 , sum);
return isSubsetSum(set, n - 1 , sum)
|| isSubsetSum(set, n - 1 , sum - set[n - 1 ]);
}
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" );
}
}
|
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
import java.io.*;
class GFG {
static int subsetSum( int a[], int n, int sum)
{
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 (sum == 0 )
return 1 ;
if (n <= 0 )
return 0 ;
if (tab[n - 1 ][sum] != - 1 )
return tab[n - 1 ][sum];
if (a[n - 1 ] > sum)
return tab[n - 1 ][sum]
= subsetSum(a, n - 1 , sum);
else {
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 ;
}
}
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" );
}
}
|
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
import java.io.*;
class GFG {
static boolean isSubsetSum( int set[], int n, int sum)
{
boolean subset[][] = new boolean [sum + 1 ][n + 1 ];
for ( int i = 0 ; i <= n; i++)
subset[ 0 ][i] = true ;
for ( int i = 1 ; i <= sum; i++)
subset[i][ 0 ] = false ;
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];
}
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" );
}
}
|
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 ]];
}
}
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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