MaximumSumCircularSubarray.py

"""
Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums.

A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].

A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.



Example 1:

Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3.
Example 2:

Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.
Example 3:

Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.


Constraints:

n == nums.length
1 <= n <= 3 * 104
-3 * 104 <= nums[i] <= 3 * 104
"""

class Solution:
    def maxSubarraySumCircular(self, nums: List[int]) -> int:
        def kadane(nums):
            max_sum = current_sum = nums[0]
            for num in nums[1:]:
                current_sum = max(num, current_sum + num)
                max_sum = max(max_sum, current_sum)
            return max_sum

        max_straight_sum = kadane(nums)
        total_sum = sum(nums)
        inverted_nums = [-num for num in nums]
        max_inverse_sum = kadane(inverted_nums) + total_sum

        if max_inverse_sum == 0:
            return max_straight_sum

        return max(max_straight_sum, max_inverse_sum)

# Example usage:
nums = [1, -2, 3, -2]
solution = Solution()
result = solution.maxSubarraySumCircular(nums)
print(result)  # Output: 3