mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-10-06 21:59:30 +00:00
cecf43d648
* Pyupgrade to Python 3.9
* updating DIRECTORY.md
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
54 lines
1.2 KiB
Python
54 lines
1.2 KiB
Python
"""
|
|
Finding the peak of a unimodal list using divide and conquer.
|
|
A unimodal array is defined as follows: array is increasing up to index p,
|
|
then decreasing afterwards. (for p >= 1)
|
|
An obvious solution can be performed in O(n),
|
|
to find the maximum of the array.
|
|
(From Kleinberg and Tardos. Algorithm Design.
|
|
Addison Wesley 2006: Chapter 5 Solved Exercise 1)
|
|
"""
|
|
from __future__ import annotations
|
|
|
|
|
|
def peak(lst: list[int]) -> int:
|
|
"""
|
|
Return the peak value of `lst`.
|
|
>>> peak([1, 2, 3, 4, 5, 4, 3, 2, 1])
|
|
5
|
|
>>> peak([1, 10, 9, 8, 7, 6, 5, 4])
|
|
10
|
|
>>> peak([1, 9, 8, 7])
|
|
9
|
|
>>> peak([1, 2, 3, 4, 5, 6, 7, 0])
|
|
7
|
|
>>> peak([1, 2, 3, 4, 3, 2, 1, 0, -1, -2])
|
|
4
|
|
"""
|
|
# middle index
|
|
m = len(lst) // 2
|
|
|
|
# choose the middle 3 elements
|
|
three = lst[m - 1 : m + 2]
|
|
|
|
# if middle element is peak
|
|
if three[1] > three[0] and three[1] > three[2]:
|
|
return three[1]
|
|
|
|
# if increasing, recurse on right
|
|
elif three[0] < three[2]:
|
|
if len(lst[:m]) == 2:
|
|
m -= 1
|
|
return peak(lst[m:])
|
|
|
|
# decreasing
|
|
else:
|
|
if len(lst[:m]) == 2:
|
|
m += 1
|
|
return peak(lst[:m])
|
|
|
|
|
|
if __name__ == "__main__":
|
|
import doctest
|
|
|
|
doctest.testmod()
|