max subarray sum Algorithm
This problem can be solved use several different algorithmic techniques, including brute force, divide and conquer, dynamic programming, and reduction to shortest paths. If the array contains all non-positive numbers, then a solution is any subarray of size 1 containing the maximal value of the array (or the empty subarray, if it is allowed). The maximal subarray problem was proposed by Ulf Grenander in 1977 as a simplified model for maximal likelihood estimate of shapes in digitized pictures.
There is some evidence that no significantly faster algorithm exists; an algorithm that solves the two-dimensional maximal subarray problem in O(n3−ε) time, for any ε>0, would imply a similarly fast algorithm for the all-pairs shortest paths problem. Grenander derived an algorithm that solves the one-dimensional problem in O(n2) time,
better the brute force working time of O(n3).
"""
Given a array of length n, max_subarray_sum() finds
the maximum of sum of contiguous sub-array using divide and conquer method.
Time complexity : O(n log n)
Ref : INTRODUCTION TO ALGORITHMS THIRD EDITION
(section : 4, sub-section : 4.1, page : 70)
"""
def max_sum_from_start(array):
""" This function finds the maximum contiguous sum of array from 0 index
Parameters :
array (list[int]) : given array
Returns :
max_sum (int) : maximum contiguous sum of array from 0 index
"""
array_sum = 0
max_sum = float("-inf")
for num in array:
array_sum += num
if array_sum > max_sum:
max_sum = array_sum
return max_sum
def max_cross_array_sum(array, left, mid, right):
""" This function finds the maximum contiguous sum of left and right arrays
Parameters :
array, left, mid, right (list[int], int, int, int)
Returns :
(int) : maximum of sum of contiguous sum of left and right arrays
"""
max_sum_of_left = max_sum_from_start(array[left : mid + 1][::-1])
max_sum_of_right = max_sum_from_start(array[mid + 1 : right + 1])
return max_sum_of_left + max_sum_of_right
def max_subarray_sum(array, left, right):
""" Maximum contiguous sub-array sum, using divide and conquer method
Parameters :
array, left, right (list[int], int, int) :
given array, current left index and current right index
Returns :
int : maximum of sum of contiguous sub-array
"""
# base case: array has only one element
if left == right:
return array[right]
# Recursion
mid = (left + right) // 2
left_half_sum = max_subarray_sum(array, left, mid)
right_half_sum = max_subarray_sum(array, mid + 1, right)
cross_sum = max_cross_array_sum(array, left, mid, right)
return max(left_half_sum, right_half_sum, cross_sum)
array = [-2, -5, 6, -2, -3, 1, 5, -6]
array_length = len(array)
print(
"Maximum sum of contiguous subarray:", max_subarray_sum(array, 0, array_length - 1)
)
