# Dynamic Programming – Subset Sum Problem

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:

 public class SubSetSumRecursion { public static void find(int[] A, int currSum, int index, int sum, int[] solution) { if (currSum == sum) { System.out.println("\nSum found"); for (int i = 0; i < solution.length; i++) { if (solution[i] == 1) { System.out.print(" " + A[i]); } } } else if (index == A.length) { return; } else { solution[index] = 1;// select the element currSum += A[index]; find(A, currSum, index + 1, sum, solution); currSum -= A[index]; solution[index] = 0;// do not select the element find(A, currSum, index + 1, sum, solution); } return; } public static void main(String[] args) { int[] A = { 3, 2, 7, 1}; int[] solution = new int[A.length]; find(A, 0, 0, 6, solution); } }

Time Complexity: O(2n).

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 ith 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 Cases
SubsetSum(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:

 public class SubSetSum { public static boolean subSetDP(int[] A, int sum) { boolean[][] solution = new boolean[A.length + 1][sum + 1]; // if sum is not zero and subset is 0, we can't make it for(int i=1;i<=sum;i++){ solution[i]=false; } // if sum is 0 the we can make the empty subset to make sum 0 for(int i=0;i<=A.length;i++){ solution[i]=true; } // for(int i=1;i<=A.length;i++){ for(int j=1;j<=sum;j++){ //first copy the data from the top solution[i][j] = solution[i–1][j]; //If solution[i][j]==false check if can be made if(solution[i][j]==false && j>=A[i–1]) solution[i][j] = solution[i][j] || solution[i–1][j–A[i–1]]; } } return solution[A.length][sum]; } public static void main(String[] args) { int[] A = { 3, 2, 7, 1}; System.out.println("\nFrom DP: " + subSetDP(A, 6) ); } }

view raw
SubSetSum.java
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```Output: From DP: true
```