Problema
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.
Vincoli
0 ≤ length ≤ 200,000
Esempi
Esempio 1
Input:
numbers = [2, 4, 1, 3, 5]
Output atteso:
expected = 3Input:
std::vector<long long> numbers = {2, 4, 1, 3, 5};
Output atteso:
long long expected = 3;Input:
long[] numbers = new long[] {2, 4, 1, 3, 5};
Output atteso:
long expected = 3;Spiegazione
Out-of-order pairs: (2,1), (4,1), (4,3): 3 inversions.
Esempio 2
Input:
numbers = [1, 2, 3]
Output atteso:
expected = 0Input:
std::vector<long long> numbers = {1, 2, 3};
Output atteso:
long long expected = 0;Input:
long[] numbers = new long[] {1, 2, 3};
Output atteso:
long expected = 0;Spiegazione
Already sorted: 0 inversions.
Riferimenti
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