Algorithm Optimized

divide_and_conquer/inversions.py

@@ -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])