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>

divide_and_conquer/max_subarray.py

@@ -0,0 +1,112 @@
+"""
+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()