**Objective: **Given a set of positive integers, and a value *sum S*, find out if there exist a subset in array whose sum is equal to given *sum S.*

Example:

int[] A = { 3, 2, 7, 1}, S = 6 Output: True, subset is (3, 2, 1}

We will first discuss the recursive approach and then we will improve it using Dynamic Programming.

**Recursive Approach:**

For every element in the array has two options, either we will include that element in subset or we don’t include it.

- So if we take example as int[] A = { 3, 2, 7, 1}, S = 6
- If we consider another int array with the same size as A.
- If we include the element in subset we will put 1 in that particular index else put 0.
- So we need to make every possible subsets and check if any of the subset makes the sum as S.
*If we think carefully this problem is quite similar to “**Generate All Strings of n bits“**See the code for better explanation.*

**Complete Code:**

**Time Complexity: O(2 ^{n}).**

**Approach: Dynamic Programming (Bottom-Up)**

**Base Cases:**

- If no elements in the set then we can’t make any subset except for 0.
- If sum needed is 0 then by returning the empty subset we can make the subset with sum 0.

**Given** – Set = **arrA[]**, Size = **n**, sum = **S**

- Now for every element in he set we have 2 options, either we include it or exclude it.
- for any i
^{th}element- - If include it => S = S-arrA[i], n=n-1
- If exclude it => S, n=n-1.

Recursive Equation:Base Cases:SubsetSum(arrA, n, S)= false, if sum > 0 and n == 0 SubsetSum(arrA, n, S)= true, if sum == 0 (return empty set)Rest CasesSubsetSum(arrA, n, S) = SubsetSum(arrA, n-1, S)|| SubsetSum(arrA, n-1, S-arrA[n-1])

**How to track the elements.**

- Start from the bottom-right corner and backtrack and check from the True is coming.
- If value in the cell above if false that means current cell has become true after including the current element. So include the current element and check for the sum = sum – current element.

**Complete Code:**

Output: From DP: true