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;
    }
}