Question

Given an array with n>3 elements, you can apply one swap operation on arbitrary two of the elements. What is the maximum number of inversions that can be reduced by one swap operation?

Answer 1

To maximize the number of inversions reduced by swapping two elements in an array, you should select the two elements that are furthest apart. The maximum number of inversions reduced can be calculated using the formula k + (n-j) + (n-i-1), where k is the number of inversions before the swap and i and j are the indices of the elements.

Step-by-step explanation:

When swapping two elements in an array, the number of inversions that can be reduced by one operation depends on the current arrangement of the elements. An inversion occurs when two elements are out of order. To maximize the number of inversions reduced, we should swap the two elements that are furthest apart in the array. Let's say the two elements are at indices i and j, where i < j. If both i and j have k inversions before the swap, then the maximum number of inversions reduced is k + (n - j) + (n - i - 1).

For example, consider an array with n = 6 elements: [4, 2, 5, 1, 6, 3]. The two elements that are furthest apart are 1 and 6, which are at indices 3 and 4. Before the swap, there are 3 inversions involving 1: (4, 2), (5, 1), and (6, 1). There are also 2 inversions involving 6: (6, 4) and (6, 5). After the swap, the array becomes [4, 2, 5, 6, 1, 3], and the inversions involving 1 and 6 are reduced. This results in a total reduction of 3 + (6 - 4) + (6 - 3 - 1) = 10 inversions.