Subset Sum Problem: Dynamic Programming & Recursion Based Solution

Most of the time, dynamic programming is just a faster way to do things with recursion. You use dynamic programming to optimize any recursive solution that makes use of the same inputs repeatedly. Subproblem results will be stored, so they do not have to be recomputed in the future as needed. This straightforward optimization decreases the time complexity from exponential to polynomial levels.

By the end of this tutorial, you will better understand the recursion and dynamic programming approach to the subset sum problem with all the necessary details and practical implementations.

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What Is the Problem Statement for the Subset Sum Problem?

You will be given a set of non-negative integers and a value of variable sum, and you must determine if there is a subset of the given set with a sum equal to a given sum.

Now, look at the recursive solution to solve the subset sum problem.

What Is the Recursion Based Solution to the Subset Sum Problem?

For the recursion based approach, you will look at two scenarios:

1. Start with the last element and work your way up to the goal sum by subtracting the value of the 'last' element from the total number of elements.
2. 'Final' element will be removed, and the required sum will equal the target sum, with the number of elements equal to the total elements minus 1.

And this is how you solve the subset sum problem using recursion. Now, implement this solution through a simple C++ code.

How Do You Implement the Recursion-Based Solution of the Subset Sum Problem?

You will be provided with a set with elements {10, 7, 8, 9, 1, 5}. You are to find a subset whose sum must be equal to 16, which is set {7, 9}.

Code:

// A c++ program to illustrate recursion based solution

#include <bits/stdc++.h>

using namespace std;

//It will return true if subset of set[] 

//has sum equal to given sum; false otherwise

bool is_subset_sum(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 we will ignore it

if (set[n - 1] > sum)

return is_subset_sum(set, n - 1, sum);

/* otherwise,we will check if the sum can be obtained by any

of the following:

(a) including the last element

(b) excluding the last element */

return is_subset_sum(set, n - 1, sum)

|| is_subset_sum(set, n - 1, sum - set[n - 1]);

}

// Driver code

int main()

{

int set[] = { 10, 7, 8, 9, 1, 5 };

int sum = 16;

int n = sizeof(set) / sizeof(set[0]);

if (is_subset_sum(set, n, sum) == true)

cout <<"Found a subset with given sum";

else

cout <<"No subset with given sum was found";

return 0;

}

You have now explored the recursive approach to the subset sum problem with a code. Now, look at the dynamic programming-based solution to this problem.

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What Is the Dynamic Programming Based Solution to the Subset Sum Problem?

You use dynamic programming to solve the problem in pseudo-polynomial time,

• Make a boolean-type 2D array of size (arr.size() + 1) * (target + 1).
• If there is a subset of elements from A[0....i] with sum value = 'j,' the state DP[i][j] is true.
• To avoid duplicating previous cases' answers, you will have to ensure that the current element has a value greater than the "current sum value."
• So, check if any prior states have already encountered the sum='j' OR have previously experienced the value 'j – A[i], which will help you achieve your purpose if it is bigger than the current total value.

And this is how you solve this problem using dynamic programming. Now implement this solution through a simple C++ code.

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How Do You Implement the Dynamic Programming Based Solution of the Subset Sum Problem?

You will be given a set with elements as {10, 7, 8, 4, 1, 6}. You have to find a subset whose sum must be equal to 16, which is set {10, 6}.

Code: 

// A C++ program to demonstrate Dynamic Programming 

//approach to solve this problem

#include <bits/stdc++.h>

using namespace std;

//It will return true if subset of set[] 

//has sum equal to given sum; false otherwise

bool is_subset_sum(int set[], int n, int target_sum)

{

//If set[0..j-1] has a subset with 

//target_sum equal to i then the 

//value of sub_set[i][j] will be true.

bool sub_set[n + 1][target_sum + 1];

// If target_sum is 0, then answer is true

for (int i = 0; i <= n; i++)

sub_set[i][0] = true;

// If target_sum is not 0 and set is empty,

// then answer is false

for (int i = 1; i <= target_sum; i++)

sub_set[0][i] = false;

// we will now fill the subset table in bottom up manner

for (int i = 1; i <= n; i++) {

for (int j = 1; j <= target_sum; j++) {

if (j < set[i - 1])

sub_set[i][j] = sub_set[i - 1][j];

if (j >= set[i - 1])

sub_set[i][j] = sub_set[i-1][j]

|| sub_set[i - 1][j - set[i - 1]];

}

}

/* // we can uncomment this code to print table

for (int i = 0; i <= n; i++)

{

for (int j = 0; j <= target_sum; j++)

printf ("%4d", sub_set[i][j]);

cout <<"\n";

}*/

return sub_set[n][target_sum];

}

int main()

{

int set[] = { 10, 7, 8, 4, 1, 6 };

int target_sum = 16;

int n = sizeof(set) / sizeof(set[0]);

if (is_subset_sum(set, n, target_sum) == true)

cout <<"Found a subset with given sum = "<<target_sum;

else

cout <<"No subset found with given sum = "<<target_sum;

return 0;

}

With this, you have come to an end of this tutorial. You will now look at what could be your next steps to conquer dynamic programming problems.

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Next Steps

Your next stop in mastering dynamic programming problems should be matrix chain multiplication. Given a list of matrices, you must find the most efficient way to multiply a set of matrices together. The issue isn't whether or not to multiply, but rather which multiplications to perform first.

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