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 ‘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.
C
#include <stdio.h>
#include <stdbool.h>
bool isSubsetSum( int set[], int n, int sum)
{
if (sum == 0)
return true ;
if (n == 0)
return false ;
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
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:
C
#include <stdio.h>
#include <string.h>
int tab[2000][2000];
int subsetSum( int a[], int n, int sum) {
if (sum == 0)
return 1;
if (n <= 0)
return 0;
if (tab[n - 1][sum] != -1)
return tab[n - 1][sum];
if (a[n - 1] > sum)
return tab[n - 1][sum] = subsetSum(a, n - 1, sum);
else {
return tab[n - 1][sum] = subsetSum(a, n - 1, sum) ||
subsetSum(a, n - 1, sum - a[n - 1]);
}
}
int main() {
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;
}
|
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:
C
#include <stdio.h>
#include <stdbool.h>
bool isSubsetSum( int set[], int n, int sum)
{
bool subset[n + 1][sum + 1];
for ( int i = 0; i <= n; i++)
subset[i][0] = true ;
for ( int i = 1; i <= sum; i++)
subset[0][i] = false ;
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];
}
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(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>
bool isSubsetSum( int set[], int n, int sum) {
bool prev[sum + 1];
for ( int i = 0; i <= n; i++)
prev[0] = true ;
for ( int i = 1; i <= sum; i++)
prev[i] = false ;
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]];
}
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;
}
|
Output
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!
Last Updated :
09 Nov, 2023
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...