# 918. Maximum Sum Circular Subarray

## Description

Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums.

A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].

A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.

Example 1:

Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3.


Example 2:

Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.


Example 3:

Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.


Constraints:

• n == nums.length
• 1 <= n <= 3 * 104
• -3 * 104 <= nums[i] <= 3 * 104

## Solutions

• class Solution {
public int maxSubarraySumCircular(int[] nums) {
int s1 = nums[0], s2 = nums[0], f1 = nums[0], f2 = nums[0], total = nums[0];
for (int i = 1; i < nums.length; ++i) {
total += nums[i];
f1 = nums[i] + Math.max(f1, 0);
f2 = nums[i] + Math.min(f2, 0);
s1 = Math.max(s1, f1);
s2 = Math.min(s2, f2);
}
return s1 > 0 ? Math.max(s1, total - s2) : s1;
}
}

• class Solution {
public:
int maxSubarraySumCircular(vector<int>& nums) {
int s1 = nums[0], s2 = nums[0], f1 = nums[0], f2 = nums[0], total = nums[0];
for (int i = 1; i < nums.size(); ++i) {
total += nums[i];
f1 = nums[i] + max(f1, 0);
f2 = nums[i] + min(f2, 0);
s1 = max(s1, f1);
s2 = min(s2, f2);
}
return s1 > 0 ? max(s1, total - s2) : s1;
}
};

• class Solution:
def maxSubarraySumCircular(self, nums: List[int]) -> int:
s1 = s2 = f1 = f2 = nums[0]
for num in nums[1:]:
f1 = num + max(f1, 0)
f2 = num + min(f2, 0)
s1 = max(s1, f1)
s2 = min(s2, f2)
return s1 if s1 <= 0 else max(s1, sum(nums) - s2)


• func maxSubarraySumCircular(nums []int) int {
s1, s2, f1, f2, total := nums[0], nums[0], nums[0], nums[0], nums[0]
for i := 1; i < len(nums); i++ {
total += nums[i]
f1 = nums[i] + max(f1, 0)
f2 = nums[i] + min(f2, 0)
s1 = max(s1, f1)
s2 = min(s2, f2)
}
if s1 <= 0 {
return s1
}
return max(s1, total-s2)
}

• function maxSubarraySumCircular(nums: number[]): number {
let pre1 = nums[0],
pre2 = nums[0];
let ans1 = nums[0],
ans2 = nums[0];
let sum = nums[0];

for (let i = 1; i < nums.length; ++i) {
let cur = nums[i];
sum += cur;
pre1 = Math.max(pre1 + cur, cur);
ans1 = Math.max(pre1, ans1);

pre2 = Math.min(pre2 + cur, cur);
ans2 = Math.min(pre2, ans2);
}
return ans1 > 0 ? Math.max(ans1, sum - ans2) : ans1;
}