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# 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`

![](/files/-MCM7_QzAUhLZFQBni3K)

```java
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;
    }
}
```


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