Maximum Sum Circular Subarray

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:

1. -30000 <= A[i] <= 30000

2. 1 <= A.length <= 30000

class Solution {
    // Max ( Non circular max sum + circular max sum )
    public int maxSubarraySumCircular(int[] A) {
        int nonCircularSum = kadaneMaxSum(A);

        int totalSum = 0;
        for (int i = 0; i < A.length; i++) {
            totalSum += A[i];
            A[i] = -A[i];
        }
        // When we invert the array the min subarray becomes the max subarray
        // and kadane will return this max subarray sum
        // therefore max circular sum will be total sum -(-kadane) => sum+kadane
        int circularSum = totalSum + kadaneMaxSum(A);
        if (circularSum == 0)
            return nonCircularSum;
        return Math.max(circularSum, nonCircularSum);
    }

    int kadaneMaxSum(int[] A) {
        int currentSum = A[0];
        int maxSum = A[0];
        for (int i = 1; i < A.length; i++) {
            if (currentSum < 0)
                currentSum = 0;
            currentSum = A[i] + currentSum;
            maxSum = Math.max(maxSum, currentSum);
        }
        return maxSum;
    }
}