mirror of
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a0eec90466
* 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>
113 lines
3.4 KiB
Python
113 lines
3.4 KiB
Python
"""
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The maximum subarray problem is the task of finding the continuous subarray that has the
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maximum sum within a given array of numbers. For example, given the array
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[-2, 1, -3, 4, -1, 2, 1, -5, 4], the contiguous subarray with the maximum sum is
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[4, -1, 2, 1], which has a sum of 6.
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This divide-and-conquer algorithm finds the maximum subarray in O(n log n) time.
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"""
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from __future__ import annotations
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import time
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from collections.abc import Sequence
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from random import randint
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from matplotlib import pyplot as plt
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def max_subarray(
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arr: Sequence[float], low: int, high: int
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) -> tuple[int | None, int | None, float]:
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"""
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Solves the maximum subarray problem using divide and conquer.
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:param arr: the given array of numbers
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:param low: the start index
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:param high: the end index
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:return: the start index of the maximum subarray, the end index of the
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maximum subarray, and the maximum subarray sum
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>>> nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
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>>> max_subarray(nums, 0, len(nums) - 1)
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(3, 6, 6)
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>>> nums = [2, 8, 9]
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>>> max_subarray(nums, 0, len(nums) - 1)
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(0, 2, 19)
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>>> nums = [0, 0]
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>>> max_subarray(nums, 0, len(nums) - 1)
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(0, 0, 0)
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>>> nums = [-1.0, 0.0, 1.0]
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>>> max_subarray(nums, 0, len(nums) - 1)
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(2, 2, 1.0)
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>>> nums = [-2, -3, -1, -4, -6]
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>>> max_subarray(nums, 0, len(nums) - 1)
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(2, 2, -1)
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>>> max_subarray([], 0, 0)
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(None, None, 0)
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"""
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if not arr:
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return None, None, 0
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if low == high:
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return low, high, arr[low]
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mid = (low + high) // 2
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left_low, left_high, left_sum = max_subarray(arr, low, mid)
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right_low, right_high, right_sum = max_subarray(arr, mid + 1, high)
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cross_left, cross_right, cross_sum = max_cross_sum(arr, low, mid, high)
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if left_sum >= right_sum and left_sum >= cross_sum:
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return left_low, left_high, left_sum
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elif right_sum >= left_sum and right_sum >= cross_sum:
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return right_low, right_high, right_sum
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return cross_left, cross_right, cross_sum
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def max_cross_sum(
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arr: Sequence[float], low: int, mid: int, high: int
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) -> tuple[int, int, float]:
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left_sum, max_left = float("-inf"), -1
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right_sum, max_right = float("-inf"), -1
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summ: int | float = 0
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for i in range(mid, low - 1, -1):
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summ += arr[i]
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if summ > left_sum:
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left_sum = summ
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max_left = i
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summ = 0
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for i in range(mid + 1, high + 1):
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summ += arr[i]
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if summ > right_sum:
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right_sum = summ
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max_right = i
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return max_left, max_right, (left_sum + right_sum)
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def time_max_subarray(input_size: int) -> float:
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arr = [randint(1, input_size) for _ in range(input_size)]
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start = time.time()
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max_subarray(arr, 0, input_size - 1)
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end = time.time()
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return end - start
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def plot_runtimes() -> None:
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input_sizes = [10, 100, 1000, 10000, 50000, 100000, 200000, 300000, 400000, 500000]
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runtimes = [time_max_subarray(input_size) for input_size in input_sizes]
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print("No of Inputs\t\tTime Taken")
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for input_size, runtime in zip(input_sizes, runtimes):
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print(input_size, "\t\t", runtime)
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plt.plot(input_sizes, runtimes)
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plt.xlabel("Number of Inputs")
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plt.ylabel("Time taken in seconds")
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plt.show()
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if __name__ == "__main__":
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"""
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A random simulation of this algorithm.
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"""
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from doctest import testmod
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testmod()
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