<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 {
public:
  int maxSubarraySumCircular(vector<int> &nums) {
    int prefixSum = 0, minPrefixSum = 0, maxPrefixSum = 0, res = INT_MIN;
    prefixSums.resize(1);
    prefixSums[0] = make_pair(0, 0);

    for (size_t i = 0; i < nums.size(); ++i) {
      prefixSum += nums[i];
      int mySum = prefixSum - minPrefixSum;
      res = max(res, mySum);

      minPrefixSum = min(minPrefixSum, prefixSum);
      if (prefixSum > maxPrefixSum) {
        prefixSums.push_back(make_pair(static_cast<uint32_t>(i) + 1, prefixSum));
        maxPrefixSum = prefixSum;
      }
    }

    if (prefixSums.back().first == nums.size())
      prefixSums.pop_back();

    int suffixSum = 0;
    for (size_t i = nums.size(); i != 1; --i) {
      suffixSum += nums[i - 1];
      if (prefixSums.back().first == i - 1)
        prefixSums.pop_back();

      int mySum = suffixSum + prefixSums.back().second;
      res = max(res, mySum);
    }

    return res;
  }

private:
  static vector<pair<uint32_t, int>> prefixSums;
};

vector<pair<uint32_t, int>> Solution::prefixSums;


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