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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.

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 :

Include the last element, recur for n = n-1, sum = sum – set[n-1]
Exclude the last element, recur for n = n-1

If any of the above the above subproblems return true, then return true.

Recursive formula for isSubsetSum() problem

isSubsetSum(set, n, sum) = isSubsetSum(set, n-1, sum) || isSubsetSum(arr, n-1, sum-set[n-1])

Base Cases
1. isSubsetSum(set, n, sum) = false, if sum > 0 and n == 0
2. isSubsetSum(set, n, sum) = true, if sum == 0

Naive recursive implementation that simply follows the recursive structure mentioned above

// Returns true if there is a subset of set[] with sun equal to given sum
function isSubsetSum($set, $n, $sum)
   // Base Cases
   if ($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
function main(){
  $set = array(3, 34, 4, 12, 5, 2);
  $sum = 9;
  $n = sizeof($set)/sizeof($set[0]);
  if (isSubsetSum($set, $n, $sum) == true)
     print("Found a subset with given sum");
     print("No subset with given sum");



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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