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918. Maximum Sum Circular Subarray

Medium

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

Solution

using System;
using System.Linq;

public class Solution {
    private int Kadane(int[] nums, int sign) {
        int currSum = int.MinValue;
        int maxSum = int.MinValue;
        foreach (int i in 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;
        foreach (int i in nums) {
            sumOfArray += i;
        }
        int maxSumSubarray = Kadane(nums, 1);
        int minSumSubarray = Kadane(nums, -1) * -1;
        // If all numbers are negative, minSumSubarray will be sumOfArray.
        // In this case, the circular sum (sumOfArray - minSumSubarray) would be 0,
        // which is incorrect if the actual answer is the largest negative number.
        // The largest negative number would be captured by maxSumSubarray.
        if (sumOfArray == minSumSubarray) {
            return maxSumSubarray;
        } else {
            return Math.Max(maxSumSubarray, sumOfArray - minSumSubarray);
        }
    }
}

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