918. Maximum Sum Circular Subarray

#### QUESTION:

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  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  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
``````

#### SOLUTION:

``````class Solution {
public int maxSubarraySumCircular(int[] A) {
int total = 0, maxSum = -30000, curMax = 0, minSum = 30000, curMin = 0;
for (int a : A) {
curMax = Math.max(curMax + a, a);
maxSum = Math.max(maxSum, curMax);
curMin = Math.min(curMin + a, a);
minSum = Math.min(minSum, curMin);
total += a;
}
return maxSum > 0 ? Math.max(maxSum, total - minSum) : maxSum;
}
}
``````
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