# 918. Maximum Sum Circular Subarray | LeetCode Solution

## 918. Maximum Sum Circular Subarray | LeetCode Solution

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

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]`.

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  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`
class Solution {
public:
int maxSubarraySumCircular(vector<int>& nums) {
int pre=nums;
int post=nums;
int total=accumulate(nums.begin(),nums.end(),0);
int currmax=0,currmin=0;

for(int i=0; i<nums.size(); i++){
currmax=max(currmax+nums[i],nums[i]);
currmin=min(currmin+nums[i],nums[i]);
pre=max(pre,currmax);
post=min(post,currmin);
}
if(total==post)return pre;
return max(pre,total-post);
}
};