README.md

918. Maximum Sum Circular Subarray

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.

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.

Constraints:

• n == nums.length
• 1 <= n <= 3 * 104
• -3 * 104 <= nums[i] <= 3 * 104

Solutions (Rust)

1. Solution

impl Solution {
    pub fn max_subarray_sum_circular(nums: Vec<i32>) -> i32 {
        let sum = nums.iter().sum();
        let mut max_sum = nums[0];
        let mut min_sum = nums[0];
        let mut prefix_sum = nums[0];
        let mut ret = nums[0].max(sum);

        for i in 1..nums.len() {
            prefix_sum += nums[i];
            ret = ret
                .max(prefix_sum - min_sum)
                .max(sum - prefix_sum + max_sum);
            max_sum = max_sum.max(prefix_sum);
            min_sum = min_sum.min(prefix_sum);
        }

        ret
    }
}