/**
 * 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 * 10^5
 * -2^31 <= nums[i] <= 2^31 - 1
 *
 *
 * Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?
 * @see <a href="https://leetcode.com/problems/increasing-triplet-subsequence/">LeetCode</a>
 */
fun increasingTriplet(nums: IntArray): Boolean {
    var first = Int.MAX_VALUE
    var second = Int.MAX_VALUE

    for (num in nums) {
        when {
            num <= first -> first = num
            num <= second -> second = num
            else -> return true
        }
    }

    return false
}
