Can the merging step in “Counting Inversions” problem with n integers, be done in time? Why?

Question

Can the merging step in “Counting Inversions” problem with n integers, be done in  time? Why?

BlackTom AI · Accepted Answer

First, a quick note: the provided data has no answer options listed, only one answer attempt. I will evaluate the given answer statement itself and explain the merging step in counting inversions.

Option analysis:
Option statement: 'Yes. When merging, we compare the smallest value of first half of elements with the smallest value of the second half of elements and, add the minimum of these two to the output and erase if from its list. This takes time O(n).'
- This description resembles the standard merge operation in a merge sort, where two sorted halves are merged by repeatedly selecting the smaller of the two current front elements. Indeed, in that process each element from both halves is looked at exactly once, so the merge step runs in linear time with respect to the total number of elements being merged. Therefore, the claim that the merging step can be done in O(n) time is correct in terms of time complexity for merging two sorted lists.
- However, there is a subtlety: in the context of counting inversions, the merge step not only outputs the merged sequence but also contributes to counting inversions. The actual counting during merge requires additional accounting (e.g., when the current element from the right half is placed before elements remaining in the left half, you add the number of remaining left elements to the inversion count). This still executes in linear time overall, i.e., O(n) for the merge itself, though the counting adds a little more logic rather than changing the time bound.

Why this reasoning is correct: merging two sorted lists by two pointers ensures every element is visited once, so the total work is proportional to the total number of elements being merged, which is linear in n. Since each comparison moves one of the pointers forward, you perform exactly as many steps as there are elements in the combined lists, yielding O(n).

Why the option as stated is potentially confusing or incomplete:
- It describes selecting the minimum of two current elements and appending it, which is a valid description of the merge operation, but a precise count of inversions during merge is not explicitly described. The core idea that the merge step runs in linear time remains correct, but the statement should more clearly connect how elements are moved and counted.
- If one expects a detailed explanation of counting inversions specifically, the answer would also include how many inversions are accounted for when taking from the right half (i.e., how many elements remain in the left half at that moment).

Putting it together: the assertion that the merging step can be done in O(n) time is correct, because a standard merge of two sorted halves processes each element exactly once. The described mechanism aligns with the two-pointer merge, though a complete treatment for counting inversions would also describe the inversion-count updates during certain merge decisions.

In summary, the provided answer correctly identifies that merging two sorted halves operates in linear time, O(n), by iterating through elements with two pointers and appending the smaller current element at each step.

CPSC_V 320 201/202/203 2024W2 Reading Quiz #4 (Divide and Conquer 1)

Can the merging step in “Counting Inversions” problem with n integers, be done in time? Why?

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