Hà Phan（河）

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:

• `-30000 <= A[i] <= 30000`
• `1 <= A.length <= 30000`

## Solution

https://leetcode.com/problems/maximum-sum-circular-subarray/discuss/178422/One-Pass

``````class Solution {
func maxSubarraySumCircular(_ A: [Int]) -> Int {
var total = 0
var maxSum = -30000, curMax = 0
var minSum = 30000, curMin = 0

for a in A {
curMax = max(curMax + a, a)
maxSum = max(maxSum, curMax)
curMin = min(curMin + a, a)
minSum = min(minSum, curMin)
total += a
}

return maxSum > 0 ? max(maxSum, total - minSum) : maxSum
}
}
``````