LeetCode-in-Go
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.length1 <= n <= 3 * 104-3 * 104 <= nums[i] <= 3 * 104
Solution
func maxSubarraySumCircular(nums []int) int {
total := 0
maxSum := nums[0]
curMax := 0
minSum := nums[0]
curMin := 0
for _, num := range nums {
total += num
curMax = max(curMax+num, num)
maxSum = max(maxSum, curMax)
curMin = min(curMin+num, num)
minSum = min(minSum, curMin)
}
if maxSum > 0 {
return max(maxSum, total-minSum)
}
return maxSum
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
func min(a, b int) int {
if a < b {
return a
}
return b
}