mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-18 09:10:16 +00:00
287bf26bc8
* A divide and conquer method in finding the maximum difference pair * fix formatting issues * fix formatting issues * add doctest runner
48 lines
1.3 KiB
Python
48 lines
1.3 KiB
Python
from typing import List
|
|
|
|
|
|
def max_difference(a: List[int]) -> (int, int):
|
|
"""
|
|
We are given an array A[1..n] of integers, n >= 1. We want to
|
|
find a pair of indices (i, j) such that
|
|
1 <= i <= j <= n and A[j] - A[i] is as large as possible.
|
|
|
|
Explanation:
|
|
https://www.geeksforgeeks.org/maximum-difference-between-two-elements/
|
|
|
|
>>> max_difference([5, 11, 2, 1, 7, 9, 0, 7])
|
|
(1, 9)
|
|
"""
|
|
# base case
|
|
if len(a) == 1:
|
|
return a[0], a[0]
|
|
else:
|
|
# split A into half.
|
|
first = a[: len(a) // 2]
|
|
second = a[len(a) // 2 :]
|
|
|
|
# 2 sub problems, 1/2 of original size.
|
|
small1, big1 = max_difference(first)
|
|
small2, big2 = max_difference(second)
|
|
|
|
# get min of first and max of second
|
|
# linear time
|
|
min_first = min(first)
|
|
max_second = max(second)
|
|
|
|
# 3 cases, either (small1, big1),
|
|
# (min_first, max_second), (small2, big2)
|
|
# constant comparisons
|
|
if big2 - small2 > max_second - min_first and big2 - small2 > big1 - small1:
|
|
return small2, big2
|
|
elif big1 - small1 > max_second - min_first:
|
|
return small1, big1
|
|
else:
|
|
return min_first, max_second
|
|
|
|
|
|
if __name__ == "__main__":
|
|
import doctest
|
|
|
|
doctest.testmod()
|