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

Solution:

class Solution {
    public int maxSubarraySumCircular(int[] A) {
        int maxSubSum = A[0], minSubSum = A[0], currMax = A[0], currMin = A[0], total = A[0];
        for (int i = 1; i < A.length; i ++) {
            currMax = Math.max(currMax + A[i], A[i]);
            maxSubSum = Math.max(currMax, maxSubSum);
            currMin = Math.min(currMin + A[i], A[i]);
            minSubSum = Math.min(currMin, minSubSum);
            total += A[i];
        }
        return maxSubSum > 0 ? Math.max(maxSubSum, total - minSubSum) : maxSubSum;
    }
}