<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>

#include <vector>
#include <numeric>

using namespace std;


class Solution {
public:
    int maxSubarraySumCircular(vector<int>& nums) {
        int max_kadane = kadane(nums);
        int total_sum = accumulate(nums.begin(), nums.end(), 0);

        for (int i = 0; i < nums.size(); i++) {
            nums[i] = -nums[i];
        }

        int max_inverted_kadane = kadane(nums);
        int min_subarray_sum = -max_inverted_kadane;
        int max_circular = total_sum - min_subarray_sum;

        // Edge case: If all elements are negative, max_circular will be 0 or negative.#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("O3", "unroll-loops", "-ffloat-store")
#pragma GCC target("avx,mmx,sse2,sse3,sse4")

// "_": A lambda function to enable faster I/O operations
auto _=[]()noexcept{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);return 0;}();

#include <vector>
#include <numeric>

using namespace std;


class Solution {
public:
    int maxSubarraySumCircular(vector<int>& nums) {
        int max_kadane = kadane(nums);
        int total_sum = accumulate(nums.begin(), nums.end(), 0);

        for (int i = 0; i < nums.size(); i++) {
            nums[i] = -nums[i];
        }

        int max_inverted_kadane = kadane(nums);
        int min_subarray_sum = -max_inverted_kadane;
        int max_circular = total_sum - min_subarray_sum;

        // Edge case: If all elements are negative, max_circular will be 0 or negative.
        // In this case, return max_kadane (non-circular max sum).
        if (max_circular == 0) return max_kadane;

        return max(max_kadane, max_circular);        
    }

private:
    int kadane(vector<int>& nums) {
        int best_sum = INT_MIN;
        int current_sum = 0;

        for (int x : nums) {
            current_sum = max(x, current_sum + x);
            best_sum = max(best_sum, current_sum);
        }

        return best_sum;
    }
};
        // In this case, return max_kadane (non-circular max sum).
        if (max_circular == 0) return max_kadane;

        return max(max_kadane, max_circular);        
    }

private:
    int kadane(vector<int>& nums) {
        int best_sum = INT_MIN;
        int current_sum = 0;

        for (int x : nums) {
            current_sum = max(x, current_sum + x);
            best_sum = max(best_sum, current_sum);
        }

        return best_sum;
    }
};
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