<p>Given a <strong>circular integer array</strong> <code>nums</code> of length <code>n</code>, return <em>the maximum possible sum of a non-empty <strong>subarray</strong> of </em><code>nums</code>.</p>

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

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,-2,3,-2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Subarray [3] has maximum sum 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,-3,5]
<strong>Output:</strong> 10
<strong>Explanation:</strong> Subarray [5,5] has maximum sum 5 + 5 = 10.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [-3,-2,-3]
<strong>Output:</strong> -2
<strong>Explanation:</strong> Subarray [-2] has maximum sum -2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>-3 * 10<sup>4</sup> &lt;= nums[i] &lt;= 3 * 10<sup>4</sup></code></li>
</ul>

class Solution {
    int maxSub(vector<int>& nums, int n)
    {
        int s = 0;
        int res = -1e9;
        for (auto num : nums)
        {
            s += num;
            res = max(res, s);
            if (s < 0) s = 0;
        }
        return res;
    }

    int minSub(vector<int>& nums, int n)
    {
        int s = 0;
        int res = 1e9;
        for (auto num : nums)
        {
            s += num;
            res = min(res, s);
            if (s > 0) s = 0;
        }
        return res;
    }

public:
    int maxSubarraySumCircular(vector<int>& nums) {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);

        int sum = 0;
        for (auto num : nums) sum += num;

        int op1 = maxSub(nums, nums.size());
        int op2 = sum - minSub(nums, nums.size());

        if (op1 > op2 || op2 == 0)
            return op1;
        else
            return op2;
    }
};
42900