918. Maximum Sum Circular Subarray

1. 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.

2. Example

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.

3. Constraints

• n == nums.length
• 1 <= n <= 3 * $10^4$
• -3 * $10^4$ <= nums[i] <= 3 * $10^4$

4. Solutions

Kadane

n = nums.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int maxSubarraySumCircular(const vector<int> &nums) {
        const int n = nums.size();
        int sum = nums[0];
        int max_current_sum = nums[0], min_current_sum = nums[0];
        int max_subarray_sum = nums[0], min_subarray_sum = nums[0];
        for (int i = 1; i < n; ++i) {
            sum += nums[i];

            max_current_sum = nums[i] + max(0, max_current_sum);
            max_subarray_sum = max(max_current_sum, max_subarray_sum);

            min_current_sum = nums[i] + min(0, min_current_sum);
            min_subarray_sum = min(min_current_sum, min_subarray_sum);
        }

        return max_subarray_sum < 0 ? max_subarray_sum
                                    : max(max_subarray_sum, sum - min_subarray_sum);
    }
};