[UCSD白板题] Number of Inversions

Problem Introduction

An inversion of a sequence \(a_0,a_1,\cdots,a_{n-1}\) is a pair of indices \(0 \leq i < j < n\) such that \(a_i>a_j\). The number of inversions of a sequence in some sense measures how close the sequence is to being sorted. For example, a sorted(in non-descending order) sequence contains no inversions at all, while in a sequence sorted in descending order any two elements constitute an inversion (for a total of \(n(n-1)/2\) inversions).

Problem Description

Task.The goal in this problem is to count the number of inversions of a given sequence.

Input Format.The first line contains an integer \(n\), the next one contains a sequence of integers \(a_0,a_1,\cdots,a_{n-1}\)

Constraints.\(1 \leq n \leq 10^5, 1 \leq a_i \leq 10^9\) for all \(0 \leq i < n\).

Output Format.Output the number of inversions in the sequence.

Sample 1.
Input:

5
2 3 9 2 9

Output:

2

Solution

# Uses python3
import sys

def merge_and_count(a, b):
    c = []
    number_of_inversions = 0
    i = j = 0
    while i < len(a) and j < len(b):
        if a[i] <= b[j]:
            c.append(a[i]); i += 1
        else:
            c.append(b[j]); j += 1
            number_of_inversions += len(a)-i
    while i < len(a):
        c.append(a[i]); i += 1
    while j < len(b):
        c.append(b[j]); j += 1
    return c, number_of_inversions

def get_number_of_inversions(a, b, left, right):
    number_of_inversions = 0
    if right - left <= 1:
        return number_of_inversions
    ave = (left + right) // 2
    number_of_inversions += get_number_of_inversions(a, b, left, ave)
    number_of_inversions += get_number_of_inversions(a, b, ave, right)
    b, count = merge_and_count(a[left:ave], a[ave:right])
    a[left:right] = b
    return number_of_inversions + count

if __name__ == '__main__':
    input = sys.stdin.read()
    n, *a = list(map(int, input.split()))
    b = n * [0]
    print(get_number_of_inversions(a, b, 0, len(a)))