p

```
"""
Given an array-like data structure A[1..n], how many pairs
(i, j) for all 1 <= i < j <= n such that A[i] > A[j]? These pairs are
called inversions. Counting the number of such inversions in an array-like
object is the important. Among other things, counting inversions can help
us determine how close a given array is to being sorted.
In this implementation, I provide two algorithms, a divide-and-conquer
algorithm which runs in nlogn and the brute-force n^2 algorithm.
"""
def count_inversions_bf(arr):
"""
Counts the number of inversions using a naive brute-force algorithm
Parameters
----------
arr: arr: array-like, the list containing the items for which the number
of inversions is desired. The elements of `arr` must be comparable.
Returns
-------
num_inversions: The total number of inversions in `arr`
Examples
---------
>>> count_inversions_bf([1, 4, 2, 4, 1])
4
>>> count_inversions_bf([1, 1, 2, 4, 4])
0
>>> count_inversions_bf([])
0
"""
num_inversions = 0
n = len(arr)
for i in range(n - 1):
for j in range(i + 1, n):
if arr[i] > arr[j]:
num_inversions += 1
return num_inversions
def count_inversions_recursive(arr):
"""
Counts the number of inversions using a divide-and-conquer algorithm
Parameters
-----------
arr: array-like, the list containing the items for which the number
of inversions is desired. The elements of `arr` must be comparable.
Returns
-------
C: a sorted copy of `arr`.
num_inversions: int, the total number of inversions in 'arr'
Examples
--------
>>> count_inversions_recursive([1, 4, 2, 4, 1])
([1, 1, 2, 4, 4], 4)
>>> count_inversions_recursive([1, 1, 2, 4, 4])
([1, 1, 2, 4, 4], 0)
>>> count_inversions_recursive([])
([], 0)
"""
if len(arr) <= 1:
return arr, 0
mid = len(arr) // 2
p = arr[0:mid]
q = arr[mid:]
a, inversion_p = count_inversions_recursive(p)
b, inversions_q = count_inversions_recursive(q)
c, cross_inversions = _count_cross_inversions(a, b)
num_inversions = inversion_p + inversions_q + cross_inversions
return c, num_inversions
def _count_cross_inversions(p, q):
"""
Counts the inversions across two sorted arrays.
And combine the two arrays into one sorted array
For all 1<= i<=len(P) and for all 1 <= j <= len(Q),
if P[i] > Q[j], then (i, j) is a cross inversion
Parameters
----------
P: array-like, sorted in non-decreasing order
Q: array-like, sorted in non-decreasing order
Returns
------
R: array-like, a sorted array of the elements of `P` and `Q`
num_inversion: int, the number of inversions across `P` and `Q`
Examples
--------
>>> _count_cross_inversions([1, 2, 3], [0, 2, 5])
([0, 1, 2, 2, 3, 5], 4)
>>> _count_cross_inversions([1, 2, 3], [3, 4, 5])
([1, 2, 3, 3, 4, 5], 0)
"""
r = []
i = j = num_inversion = 0
while i < len(p) and j < len(q):
if p[i] > q[j]:
# if P[1] > Q[j], then P[k] > Q[k] for all i < k <= len(P)
# These are all inversions. The claim emerges from the
# property that P is sorted.
num_inversion += len(p) - i
r.append(q[j])
j += 1
else:
r.append(p[i])
i += 1
if i < len(p):
r.extend(p[i:])
else:
r.extend(q[j:])
return r, num_inversion
def main():
arr_1 = [10, 2, 1, 5, 5, 2, 11]
# this arr has 8 inversions:
# (10, 2), (10, 1), (10, 5), (10, 5), (10, 2), (2, 1), (5, 2), (5, 2)
num_inversions_bf = count_inversions_bf(arr_1)
_, num_inversions_recursive = count_inversions_recursive(arr_1)
assert num_inversions_bf == num_inversions_recursive == 8
print("number of inversions = ", num_inversions_bf)
# testing an array with zero inversion (a sorted arr_1)
arr_1.sort()
num_inversions_bf = count_inversions_bf(arr_1)
_, num_inversions_recursive = count_inversions_recursive(arr_1)
assert num_inversions_bf == num_inversions_recursive == 0
print("number of inversions = ", num_inversions_bf)
# an empty list should also have zero inversions
arr_1 = []
num_inversions_bf = count_inversions_bf(arr_1)
_, num_inversions_recursive = count_inversions_recursive(arr_1)
assert num_inversions_bf == num_inversions_recursive == 0
print("number of inversions = ", num_inversions_bf)
if __name__ == "__main__":
main()
```