918. Maximum Sum Circular Subarray

            package
             LeetCode_918


            /**
            
 * 918. Maximum Sum Circular Subarray
 * 
            https://leetcode.com/problems/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
 * 
            */
            class
             Solution {
    
            /*
            
    * solution: base in get maximum in sub-array,
    * maximum in circular: max(max sub-array sum, total sum - min sub-array sum).
    * Prove this case:
      max(prefix+suffix)
      = max(total sum - subarray)
      = total sum + max(-subarray)
      = total sum - min(subarray)
    * Time:O(n), Space:O(1)
    * 
            */
            
    fun maxSubarraySumCircular(A: IntArray): Int {
        var currentMax = 0
        var currentMin = 0
        var maxSum = Int.MIN_VALUE
        var minSum = Int.MAX_VALUE
        var total = 0
        for (item in A) {
            currentMax = Math.max(item, currentMax + item)
            maxSum = Math.max(maxSum, currentMax)
            currentMin = Math.min(item, currentMin + item)
            minSum = Math.min(minSum, currentMin)
            total += item
        }
        //if maxSum<=0 need return maxSum, case: [-2,-3,-1]returnif (maxSum > 0) Math.max(maxSum, total - minSum) else maxSum
    }
}