LeetCode in Kotlin
3105. Longest Strictly Increasing or Strictly Decreasing Subarray
Easy
You are given an array of integers nums. Return the length of the longest subarray of nums which is either strictly increasing or strictly decreasing.
Example 1:
Input: nums = [1,4,3,3,2]
Output: 2
Explanation:
The strictly increasing subarrays of nums are [1], [2], [3], [3], [4], and [1,4].
The strictly decreasing subarrays of nums are [1], [2], [3], [3], [4], [3,2], and [4,3].
Hence, we return 2.
Example 2:
Input: nums = [3,3,3,3]
Output: 1
Explanation:
The strictly increasing subarrays of nums are [3], [3], [3], and [3].
The strictly decreasing subarrays of nums are [3], [3], [3], and [3].
Hence, we return 1.
Example 3:
Input: nums = [3,2,1]
Output: 3
Explanation:
The strictly increasing subarrays of nums are [3], [2], and [1].
The strictly decreasing subarrays of nums are [3], [2], [1], [3,2], [2,1], and [3,2,1].
Hence, we return 3.
Constraints:
1 <= nums.length <= 501 <= nums[i] <= 50
Solution
import kotlin.math.max
class Solution {
fun longestMonotonicSubarray(nums: IntArray): Int {
var inc = 1
var dec = 1
var res = 1
for (i in 1 until nums.size) {
if (nums[i] > nums[i - 1]) {
inc += 1
dec = 1
} else if (nums[i] < nums[i - 1]) {
dec += 1
inc = 1
} else {
inc = 1
dec = 1
}
res = max(res, max(inc, dec))
}
return res
}
}