Problem
An inversion is a pair of positions i < j with numbers[i] > numbers[j]. Count all inversions. Lists can be long, so a merge-sort based O(n log n) count is needed.
Implement count_inversions(numbers).
Input
numbers: The list of integers.
Output
Return the number of inversions.
Constraints
0 ≤ length ≤ 200,000
Examples
Example 1
Input:
numbers = [2, 4, 1, 3, 5]
Expected output:
expected = 3Input:
std::vector<long long> numbers = {2, 4, 1, 3, 5};
Expected output:
long long expected = 3;Input:
long[] numbers = new long[] {2, 4, 1, 3, 5};
Expected output:
long expected = 3;Explanation
Out-of-order pairs: (2,1), (4,1), (4,3): 3 inversions.
Example 2
Input:
numbers = [1, 2, 3]
Expected output:
expected = 0Input:
std::vector<long long> numbers = {1, 2, 3};
Expected output:
long long expected = 0;Input:
long[] numbers = new long[] {1, 2, 3};
Expected output:
long expected = 0;Explanation
Already sorted: 0 inversions.
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.