334. Increasing Triplet Subsequence

Medium

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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?

class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        first = second = float("inf")
        for num in nums:
            if num <= first:
                first = num
            elif num <= second:
                second = num
            else:
                return True
        return False