C Program for Subset Sum Problem | DP-25
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.
C Program for Subset Sum Problem using Recursion:
For the recursive approach, there will be two cases.
- Consider the âlastâ element to be a part of the subset. Now the new required sum = required sum â value of âlastâ element.
- Donât include the âlastâ 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.
C
// 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; } |
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 â 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:
C
#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; } |
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 = âjâ.
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:
C
// 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. |
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:
C
#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; } |
Found a subset with given sum
Time Complexity: O(sum * n), where n is the size of the array.
Auxiliary Space: O(sum), as the size of the 1-D array is sum+1.
Please refer complete article on Subset Sum Problem | DP-25 for more details!