918. Maximum Sum Circular Subarray

Problem

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 * 10^4

• -3 * 10^4 <= nums[i] <= 3 * 10^4

Solution (Java)

class Solution {
    private int kadane(int[] nums, int sign) {
        int currSum = Integer.MIN_VALUE;
        int maxSum = Integer.MIN_VALUE;
        for (int i : nums) {
            currSum = sign * i + Math.max(currSum, 0);
            maxSum = Math.max(maxSum, currSum);
        }
        return maxSum;
    }

    public int maxSubarraySumCircular(int[] nums) {
        if (nums.length == 1) {
            return nums[0];
        }
        int sumOfArray = 0;
        for (int i : nums) {
            sumOfArray += i;
        }
        int maxSumSubarray = kadane(nums, 1);
        int minSumSubarray = kadane(nums, -1) * -1;
        if (sumOfArray == minSumSubarray) {
            return maxSumSubarray;
        } else {
            return Math.max(maxSumSubarray, sumOfArray - minSumSubarray);
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).