Challenge lab in Sorting and Divide-and-Conquer

Mission

Count i < j pairs where a[i] is greater than a[j].

Learning outcome: Count inversions during merge

Correctness contract

Invariant: A selection prefix is final; a merge emits the smallest remaining head.

Required technique: Count split inversions during merge by adding the number of unmerged left values.

Complexity target: time O(n log n); space O(n).

Input and output

Input: n followed by exactly n integers. Whitespace may be spaces or line breaks.

Output: Print the single exact numeric result with no label. Return it as a String; Main.java prints it without adding other text.

Assumptions:

• n is positive and is followed by exactly n signed 32-bit integers.

Before you code

1. Restate the input and output contract, then predict the visible example without running code.
2. Implement the core state transition: Count split inversions during merge by adding the number of unmerged left values.
3. Trace the smallest boundary case, verify exact formatting, and justify the authored time and auxiliary-space bounds.

Implement Practice.solve(Scanner sc). Keep every provided filename and public class name unchanged.

Sample input

5 2 4 1 3 5

Sample output

3

Why the sample works: Visible walkthrough for the ordinary non-trivial path. When a right value is emitted before remaining left values, all of those remaining values form split inversions. Input `5 2 4 1 3 5` therefore produces `3`.

Progressive hints

Try the trace and first milestone before opening a hint. Open them in order.

Hint 1 — Contract: identify what each parsed variable represents and write the invariant beside the loop or recursive method.

Hint 2 — Next step: When right[j] is smaller, all remaining values in the sorted left half are also larger.

Hint 3 — Verification: compare the structure state before and after one operation, then test the smallest valid input and a duplicate or unreachable case when allowed.