Python/divide_and_conquer/max_subarray.py
Tianyi Zheng a0eec90466
Consolidate duplicate implementations of max subarray (#8849)
* Remove max subarray sum duplicate implementations

* updating DIRECTORY.md

* Rename max_sum_contiguous_subsequence.py

* Fix typo in dynamic_programming/max_subarray_sum.py

* Remove duplicate divide and conquer max subarray

* updating DIRECTORY.md

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Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
2023-07-11 11:44:12 +02:00

113 lines
3.4 KiB
Python

"""
The maximum subarray problem is the task of finding the continuous subarray that has the
maximum sum within a given array of numbers. For example, given the array
[-2, 1, -3, 4, -1, 2, 1, -5, 4], the contiguous subarray with the maximum sum is
[4, -1, 2, 1], which has a sum of 6.
This divide-and-conquer algorithm finds the maximum subarray in O(n log n) time.
"""
from __future__ import annotations
import time
from collections.abc import Sequence
from random import randint
from matplotlib import pyplot as plt
def max_subarray(
arr: Sequence[float], low: int, high: int
) -> tuple[int | None, int | None, float]:
"""
Solves the maximum subarray problem using divide and conquer.
:param arr: the given array of numbers
:param low: the start index
:param high: the end index
:return: the start index of the maximum subarray, the end index of the
maximum subarray, and the maximum subarray sum
>>> nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
>>> max_subarray(nums, 0, len(nums) - 1)
(3, 6, 6)
>>> nums = [2, 8, 9]
>>> max_subarray(nums, 0, len(nums) - 1)
(0, 2, 19)
>>> nums = [0, 0]
>>> max_subarray(nums, 0, len(nums) - 1)
(0, 0, 0)
>>> nums = [-1.0, 0.0, 1.0]
>>> max_subarray(nums, 0, len(nums) - 1)
(2, 2, 1.0)
>>> nums = [-2, -3, -1, -4, -6]
>>> max_subarray(nums, 0, len(nums) - 1)
(2, 2, -1)
>>> max_subarray([], 0, 0)
(None, None, 0)
"""
if not arr:
return None, None, 0
if low == high:
return low, high, arr[low]
mid = (low + high) // 2
left_low, left_high, left_sum = max_subarray(arr, low, mid)
right_low, right_high, right_sum = max_subarray(arr, mid + 1, high)
cross_left, cross_right, cross_sum = max_cross_sum(arr, low, mid, high)
if left_sum >= right_sum and left_sum >= cross_sum:
return left_low, left_high, left_sum
elif right_sum >= left_sum and right_sum >= cross_sum:
return right_low, right_high, right_sum
return cross_left, cross_right, cross_sum
def max_cross_sum(
arr: Sequence[float], low: int, mid: int, high: int
) -> tuple[int, int, float]:
left_sum, max_left = float("-inf"), -1
right_sum, max_right = float("-inf"), -1
summ: int | float = 0
for i in range(mid, low - 1, -1):
summ += arr[i]
if summ > left_sum:
left_sum = summ
max_left = i
summ = 0
for i in range(mid + 1, high + 1):
summ += arr[i]
if summ > right_sum:
right_sum = summ
max_right = i
return max_left, max_right, (left_sum + right_sum)
def time_max_subarray(input_size: int) -> float:
arr = [randint(1, input_size) for _ in range(input_size)]
start = time.time()
max_subarray(arr, 0, input_size - 1)
end = time.time()
return end - start
def plot_runtimes() -> None:
input_sizes = [10, 100, 1000, 10000, 50000, 100000, 200000, 300000, 400000, 500000]
runtimes = [time_max_subarray(input_size) for input_size in input_sizes]
print("No of Inputs\t\tTime Taken")
for input_size, runtime in zip(input_sizes, runtimes):
print(input_size, "\t\t", runtime)
plt.plot(input_sizes, runtimes)
plt.xlabel("Number of Inputs")
plt.ylabel("Time taken in seconds")
plt.show()
if __name__ == "__main__":
"""
A random simulation of this algorithm.
"""
from doctest import testmod
testmod()