Here are the two Java solutions for the subset sum problem.
First using Recursive Approach.
Second using Dynamic Programming Approach.
/*
Question: Given a set 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: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True //There is a subset (4, 5) with sum 9.
Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with
sum equal to sum. n is the number of elements in set[].
*/
package SubsetSumProblem;
import java.util.Scanner;
public class UsingResursiveAndDPApproach {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
try{
System.out.println("Enter the number of elements in the array");
int n =in.nextInt();
System.out.println("Enter the elements of the array");
int[] a=new int[n];
for(int i=0;i<n;i++)
a[i]=in.nextInt();
System.out.println("Enter the sum, which you need to find");
int sum = in.nextInt();
System.out.println("Using recursion, the result is: "+usingRecursion(a,a.length,sum));
System.out.println("Using Dynamic Programming, the result is: "+usingDP(a,sum));
}
finally{
in.close();
}
}
private static boolean usingRecursion(int[] a,int length, int sum) {
// 1. Base Cases
if(sum==0)
return true;
if(length==0 && sum!=0)
return false;
// 2. To avoid unnecessary steps, we will optimize the recursion method by avoiding
// recursive calls to areas where we are definite that we can SAFELY ignore the case since
// the SOLUTION does not exist there.
// If last element is greater than sum, then ignore it
if(a[a.length-1]>sum)
return usingRecursion(a,length-1,sum);
// 3. This is the recursion step where we will call the method again and again
/* else, check if sum can be obtained by any of the following
(a) including the last element
(b) excluding the last element */
return (usingRecursion(a, length-1, sum-a[length-1])|| usingRecursion(a, length-1, sum));
}
/*
Analysis:
Time Complexity = O(2^n)
Space Complexity = // Don't know
*/
private static boolean usingDP(int[] a, int sum) {
// using boolean matrix for DP
boolean dp[][] = new boolean[a.length+1][sum+1]; // +1 in row and column
// if the length of the array is variable (and sum is 0) then fill TRUE, since the SUM=0
for(int row=0;row<dp.length;row++){
dp[row][0] = true; // NOTE: dp[length=VARIABLE][sum=0], thus we satisfy the condition where length is VARIABLE
// and the SUM=0
}
// if the SUM is variable and length is 0 then FALSE, since (sum=variable && length=0)
for(int column=1;column<dp[0].length;column++){
dp[0][column] = false; // NOTE: dp[length=0][sum=VARIABLE], thus we satisfy the condition where
// (length=0 && sum=variable)
}
for(int i=1;i<dp.length;i++){
for(int j=1;j<dp[0].length;j++){
/* Check if sum can be obtained by any of the following
(a) including the last element
(b) excluding the last element */
// VERY VERY IMP: This is same as "excluding the last element" which is represented in DP
dp[i][j] = dp[i-1][j]; // the current position[i][j] would be same as previous position.
// the previous position means that SUM is ACHIEVED OR NOT-ACHIEVED
// int the previous position then it will ofcourse be ACHIEVED or NOT-ACHIEVED
// in the current position.
// VERY VERY IMP: This is same as "including the last element" which is represented in DP
// if the column[ sum is represented in column of the matrix i.e this sum exist] > = sum-a[last_index]
// then decrease the sum
if(j>=a[i-1]) // i.e sum >= array[last index element]. If it is true then include this last element by
// deducting it from the total sum
dp[i][j] = dp[i][j] || dp[i-1][j-a[i-1]]; // VERY VERY IMP NOTE: Here dp[i][j] on R.H.S represent
// dp[i-1][j] which we have assigned in the previous step
}
}
return dp[a.length][sum];
}
/*
Analysis:
Time Complexity = O(a.length*sum)
Space Complexity = O(a.length*sum)
*/
}
