Learning-competitive-coding/Maximum_Sum_Circular_Subarray.py at master · raniyer/Learning-competitive-coding · GitHub

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"""
Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C.
Here, a circular array means the end of the array connects to the beginning of the array.
(Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)
Also, a subarray may only include each element of the fixed buffer A at most once.
(Formally, for a subarray C[i], C[i+1], ..., C[j], there does not exist i <= k1, k2 <= j with k1 % A.length = k2 % A.length.)

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

Example 2:
Input: [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10

Example 3:
Input: [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4

Example 4:
Input: [3,-2,2,-3]
Output: 3
Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3

Example 5:
Input: [-2,-3,-1]
Output: -1
Explanation: Subarray [-1] has maximum sum -1

Note:
    -30000 <= A[i] <= 30000
    1 <= A.length <= 30000
"""
#used kadane's algorithm

class Solution:
    def maxSubarraySumCircular(self, A: List[int]) -> int:
        def kadane(gen):
            ans = cur = -float('inf')
            for i in gen:
                cur = i + max(cur, 0)
                ans = max(ans, cur)
            return ans
        if len(A) == 1:
            return (A[0])
        S = sum(A)
        ans1 = kadane(iter(A))
        ans2 = S + kadane(-A[i] for i in range(1, len(A)))
        ans3 = S + kadane(-A[i] for i in range(len(A)-1))
        return max(ans1, ans2, ans3)