LeetCode in Kotlin
334. Increasing Triplet Subsequence
Medium
Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.
Example 1:
Input: nums = [1,2,3,4,5]
Output: true
Explanation: Any triplet where i < j < k is valid.
Example 2:
Input: nums = [5,4,3,2,1]
Output: false
Explanation: No triplet exists.
Example 3:
Input: nums = [2,1,5,0,4,6]
Output: true
Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.
Constraints:
1 <= nums.length <= 5 * 105-231 <= nums[i] <= 231 - 1
Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?
Solution
class Solution {
fun increasingTriplet(nums: IntArray): Boolean {
if (nums.size == 1) {
return false
}
var big: Int = nums.lastIndex
var medium: Int? = null
for (i in nums.lastIndex - 1 downTo 0) {
if (nums[i] > nums[big]) {
big = i
continue
} else if ((medium != null && nums[i] > nums[medium] && nums[i] < nums[big]) ||
(medium == null && nums[i] < nums[big])
) {
medium = i
continue
} else if (medium != null && nums[i] < nums[medium]) {
return true
}
}
return false
}
}