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Remove unnecessary branch (#4824)

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* Algorithm Optimized

* Update divide_and_conquer/inversions.py

Co-authored-by: John Law <johnlaw.po@gmail.com>

* Update divide_and_conquer/inversions.py

Co-authored-by: John Law <johnlaw.po@gmail.com>

* Update divide_and_conquer/inversions.py

Co-authored-by: John Law <johnlaw.po@gmail.com>

Co-authored-by: John Law <johnlaw.po@gmail.com>
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Muhammad Hammad Sani 2021-10-11 21:33:06 +05:00 committed by GitHub
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divide_and_conquer/inversions.py
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@ -2,31 +2,25 @@
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
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 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])
@ -49,20 +43,16 @@ def count_inversions_bf(arr):
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])
@ -72,40 +62,34 @@ def count_inversions_recursive(arr):
"""
if len(arr) <= 1:
return arr, 0
else:
mid = len(arr) // 2
P = arr[0:mid]
Q = arr[mid:]
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)
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
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])