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Simplify code by dropping support for legacy Python (#1143)

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* Simplify code by dropping support for legacy Python

* sort() --> sorted()
This commit is contained in:
Christian Clauss
2019-08-19 15:37:49 +02:00
committed by GitHub
parent 32aa7ff081
commit 47a9ea2b0b
145 changed files with 367 additions and 976 deletions
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12
divide_and_conquer/convex_hull.py
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@@ -1,18 +1,16 @@
from __future__ import print_function, absolute_import, division
from numbers import Number
"""
The convex hull problem is problem of finding all the vertices of convex polygon, P of
The convex hull problem is problem of finding all the vertices of convex polygon, P of
a set of points in a plane such that all the points are either on the vertices of P or
inside P. TH convex hull problem has several applications in geometrical problems,
computer graphics and game development.
inside P. TH convex hull problem has several applications in geometrical problems,
computer graphics and game development.
Two algorithms have been implemented for the convex hull problem here.
Two algorithms have been implemented for the convex hull problem here.
1. A brute-force algorithm which runs in O(n^3)
2. A divide-and-conquer algorithm which runs in O(n^3)
There are other several other algorithms for the convex hull problem
which have not been implemented here, yet.
which have not been implemented here, yet.
"""

14
divide_and_conquer/inversions.py
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@@ -1,15 +1,13 @@
from __future__ import print_function, absolute_import, division
"""
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
(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.
algorithm which runs in nlogn and the brute-force n^2 algorithm.
"""