Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.
Examples: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True //There is a subset (4, 5) with sum 9.
Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with sum equal to sum. n is the number of elements in set[].
The isSubsetSum problem can be divided into two subproblems
…a) Include the last element, recur for n = n-1, sum = sum – set[n-1]
…b) Exclude the last element, recur for n = n-1.
If any of the above the above subproblems return true, then return true.
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Following is the recursive formula for isSubsetSum() problem.
isSubsetSum(set, n, sum) = isSubsetSum(set, n-1, sum) ||
isSubsetSum(set, n-1, sum-set[n-1])
Base Cases:
isSubsetSum(set, n, sum) = false, if sum > 0 and n == 0
isSubsetSum(set, n, sum) = true, if sum == 0
Following is naive recursive implementation that simply follows the recursive structure mentioned above.
Java
// A recursive solution for subset sum problem
class subset_sum
{
// Returns true if there is a subset of set[] with sum
// equal to given sum
static boolean isSubsetSum(int set[], int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0 && sum != 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 program to test above function */
public static void main (String args[])
{
int set[] = {3, 34, 4, 12, 5, 2};
int sum = 9;
int n = set.length;
if (isSubsetSum(set, n, sum) == true)
System.out.println("Found a subset with given sum");
else
System.out.println("No subset with given sum");
}
}
Output :
Found a subset with given sum
The above solution may try all subsets of given set in worst case. Therefore time complexity of the above solution is exponential. The problem is in-fact NP-Complete (There is no known polynomial time solution for this problem).
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We can solve the problem in Pseudo-polynomial time using Dynamic programming. We create a boolean 2D table subset[][] and fill it in bottom up manner. The value of subset[i][j] will be true if there is a subset of set[0..j-1] with sum equal to i., otherwise false. Finally, we return subset[sum][n]
Java
// A Dynamic Programming solution for subset sum problem
class subset_sum
{
// Returns true if there is a subset of set[] with sun equal to given sum
static boolean 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
boolean subset[][] = new boolean[sum+1][n+1];
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++)
subset[0][i] = true;
// If sum is not 0 and set is empty, then answer is false
for (int i = 1; i <= sum; i++)
subset[i][0] = false;
// Fill the subset table in botton up manner
for (int i = 1; i <= sum; i++)
{
for (int j = 1; j <= n; j++)
{
subset[i][j] = subset[i][j-1];
if (i >= set[j-1])
subset[i][j] = subset[i][j] ||
subset[i - set[j-1]][j-1];
}
}
/* // uncomment this code to print table
for (int i = 0; i <= sum; i++)
{
for (int j = 0; j <= n; j++)
printf ("%4d", subset[i][j]);
printf("\n");
} */
return subset[sum][n];
}
/* Driver program to test above function */
public static void main (String args[])
{
int set[] = {3, 34, 4, 12, 5, 2};
int sum = 9;
int n = set.length;
if (isSubsetSum(set, n, sum) == true)
System.out.println("Found a subset with given sum");
else
System.out.println("No subset with given sum");
}
}
Output:
Found a subset with given sum
Time complexity of the above solution is O(sum*n).
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