918. Maximum Sum Circular Subarray

Description

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

Solution

maximum-sum-circular-subarray.py

class Solution:
    def maxSubarraySumCircular(self, nums: List[int]) -> int:
        N = len(nums)
        currMax = mmax = currMin = mmin = total = nums[0]

        for i in range(1, N):
            currMax = max(nums[i], nums[i] + currMax)
            mmax = max(mmax, currMax)
            currMin = min(nums[i], nums[i] + currMin)
            mmin = min(mmin, currMin)
            total += nums[i]

        return max(mmax, total - mmin) if mmax > 0 else mmax