# Maximum Sum Circular Subarray

#1

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`

#2

The below code considers the intuition on what would have been the max sum if there were no negative numbers in between.

``````class Solution {
public int maxSubarraySumCircular(int[] A) {
int total = 0, minSum = 0, min = Integer.MAX_VALUE, maxSum =0, max = Integer.MIN_VALUE;
for(int num : A){
minSum = Math.min(num, num+minSum);
min = Math.min(min, minSum);
maxSum = Math.max(num, num+maxSum);
max = Math.max(max, maxSum);
total += num;
}

return max>0? Math.max(total-min, max):max;
}
}
``````