Problem statement:

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 Aat 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

Solution

class Solution {
    public int maxSubarraySumCircular(int[] a) {
        int currMax = a[0];
        int max = a[0];
        int currMin = a[0], min = currMin, total = a[0];

        for(int i = 1; i < a.length; i++) {
            currMax = Math.max(currMax + a[i], a[i]);
            max = Math.max(currMax, max);
            currMin = Math.min(currMin + a[i], a[i]);
            min = Math.min(currMin, min);
            total += a[i];
        }
        
        return max > 0 ? Math.max(max, total - min) : max;
    }
}

Time complexity:

O(N) One pass

Space complexity:

O(1) no extra space