C Program for Subset Sum Problem | DP-25

Last Updated : 09 Nov, 2023

Write a C 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.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C 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.

// A recursive solution for subset sum problem #include <stdio.h> #include <stdbool.h> // Returns true if there is a subset // of set[] with sum equal to given sum bool isSubsetSum(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 ignore it  if (set[n - 1] > sum)  return isSubsetSum(set, n - 1, sum);  // Else, check if sum can be obtained by any  // of the following:  // (a) including the last element  // (b) excluding the last element  return isSubsetSum(set, n - 1, sum)  || isSubsetSum(set, n - 1, sum - set[n - 1]); } // Driver code int main() {  int set[] = { 3, 34, 4, 12, 5, 2 };  int sum = 9;  int n = sizeof(set) / sizeof(set[0]);  if (isSubsetSum(set, n, sum) == true)  printf("Found a subset with given sum");  else  printf("No subset with given sum");  return 0; }

Output

Found a subset with given sum

Time Complexity: O(2ⁿ)
Auxiliary space: O(n)

C 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:

#include <stdio.h> #include <string.h> // Taking the matrix as globally int tab[2000][2000]; // Check if a possible subset with  // the given sum is possible or not int subsetSum(int a[], int n, int sum) {  // If the sum is zero, it means   // we got our expected sum  if (sum == 0)  return 1;  if (n <= 0)  return 0;  // If the value is not -1, it means it   // already called the function   // with the same value.  // It will save us from repetition.  if (tab[n - 1][sum] != -1)  return tab[n - 1][sum];  // If the value of a[n-1] is  // greater than the sum,  // we call for the next value  if (a[n - 1] > sum)  return tab[n - 1][sum] = subsetSum(a, n - 1, sum);  else {  // Here we do two calls because we   // don't know which value fulfills our criteria.  // That's why we're doing two calls  return tab[n - 1][sum] = subsetSum(a, n - 1, sum) ||   subsetSum(a, n - 1, sum - a[n - 1]);  } } int main() {  // Storing the value -1 to the matrix  memset(tab, -1, sizeof(tab));  int n = 5;  int a[] = {1, 5, 3, 7, 4};  int sum = 12;  if (subsetSum(a, n, sum)) {  printf("YES\n");  } else {  printf("NO\n");  }  return 0; }

Output

YES

Time Complexity: O(sum*n)
Auxiliary space: O(n)

C 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:

// A Dynamic Programming solution // for subset sum problem #include <stdio.h> #include <stdbool.h> // Returns true if there is a subset of set[] // with sum equal to given sum bool isSubsetSum(int set[], int n, int sum) {  // The value of subset[i][j] will be true if  // there is a subset of set[0..j-1] with sum  // equal to i  bool subset[n + 1][sum + 1];  // If sum is 0, then answer is true  for (int i = 0; i <= n; i++)  subset[i][0] = true;  // If sum is not 0 and set is empty,  // then answer is false  for (int i = 1; i <= sum; i++)  subset[0][i] = false;  // Fill the subset table in bottom up manner  for (int i = 1; i <= n; i++) {  for (int j = 1; j <= sum; j++) {  if (j < set[i - 1])  subset[i][j] = subset[i - 1][j];  if (j >= set[i - 1])  subset[i][j]  = subset[i - 1][j]  || subset[i - 1][j - set[i - 1]];  }  }  return subset[n][sum]; } // Driver code int main() {  int set[] = { 3, 34, 4, 12, 5, 2 };  int sum = 9;  int n = sizeof(set) / sizeof(set[0]);  if (isSubsetSum(set, n, sum) == true)  printf("Found a subset with given sum");  else  printf("No subset with given sum");  return 0; } // This code is contributed by Arjun Tyagi.

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.

C 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:

#include <stdio.h> #include <stdbool.h> // Returns true if there is a subset of set[] // with a sum equal to the given sum bool isSubsetSum(int set[], int n, int sum) {  bool prev[sum + 1];  // If sum is 0, then the answer is true  for (int i = 0; i <= n; i++)  prev[0] = true;  // If sum is not 0 and set is empty,  // then the answer is false  for (int i = 1; i <= sum; i++)  prev[i] = false;  // curr array to store the current row result generated  // with the help of the prev array  bool curr[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]];  }  // now curr becomes prev for the (i + 1)-th element  for (int j = 0; j <= sum; j++)  prev[j] = curr[j];  }  return prev[sum]; } int main() {  int set[] = {3, 34, 4, 12, 5, 2};  int sum = 9;  int n = sizeof(set) / sizeof(set[0]);  if (isSubsetSum(set, n, sum)) {  printf("Found a subset with given sum\n");  } else {  printf("No subset with given sum\n");  }  return 0; }