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.
Restricciones
0 ≤ length ≤ 200,000
Ejemplos
Ejemplo 1
Entrada:
numbers = [2, 4, 1, 3, 5]
Salida esperada:
expected = 3Entrada:
std::vector<long long> numbers = {2, 4, 1, 3, 5};
Salida esperada:
long long expected = 3;Entrada:
long[] numbers = new long[] {2, 4, 1, 3, 5};
Salida esperada:
long expected = 3;Explicación
Out-of-order pairs: (2,1), (4,1), (4,3): 3 inversions.
Ejemplo 2
Entrada:
numbers = [1, 2, 3]
Salida esperada:
expected = 0Entrada:
std::vector<long long> numbers = {1, 2, 3};
Salida esperada:
long long expected = 0;Entrada:
long[] numbers = new long[] {1, 2, 3};
Salida esperada:
long expected = 0;Explicación
Already sorted: 0 inversions.
Referencias
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