Question

Let S be a set of n positive integers, where n is even. a) Give an efficient o(n 2 ) algorithm to partition S into two subsets S 1 and S2

of 2n elements each with the property that the difference between the sum of the elements in S 1 and the sum of the elements in S 2
is maximum. b) Analyze the time complexity of your algorithm?

Answer 1

To partition the set S into two subsets S1 and S2 with the maximum difference in their sums, follow this algorithm: sort the set in non-decreasing order, initialize two empty sets, and then iterate over the elements in reverse order and place them in S1 or S2 alternatingly. The time complexity is O(n^2), where n is the number of elements in the set.

Step-by-step explanation:

To partition the set S into two subsets S1 and S2 with the property that the difference between the sum of the elements in S1 and the sum of the elements in S2 is maximum, we can use the following algorithm:

1. Sort the set S in non-decreasing order.
2. Initialize two empty sets, S1 and S2.
3. Iterate over the elements in S in reverse order and place each element in either S1 or S2 by alternating between the two sets.

This algorithm ensures that the elements in S are divided as evenly as possible between S1 and S2, maximizing the difference between their sums.

The time complexity of this algorithm is O(n^2), where n is the number of elements in S. This is because sorting the set S takes O(n log n) time, and the iteration over the elements in S takes O(n) time. Therefore, the overall time complexity is dominated by the sorting step.