LeetCode in Kotlin
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
class Solution {
private fun kadane(nums: IntArray, sign: Int): Int {
var currSum = Int.MIN_VALUE
var maxSum = Int.MIN_VALUE
for (i in nums) {
currSum = sign * i + currSum.coerceAtLeast(0)
maxSum = maxSum.coerceAtLeast(currSum)
}
return maxSum
}
fun maxSubarraySumCircular(nums: IntArray): Int {
if (nums.size == 1) {
return nums[0]
}
var sumOfArray = 0
for (i in nums) {
sumOfArray += i
}
val maxSumSubarray = kadane(nums, 1)
val minSumSubarray = kadane(nums, -1) * -1
return if (sumOfArray == minSumSubarray) {
maxSumSubarray
} else {
maxSumSubarray.coerceAtLeast(sumOfArray - minSumSubarray)
}
}
}