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[10/11] 334. Increasing Triplet Subsequence

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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?
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Runtime: 605 ms, faster than 90.03% of Python3 online submissions for Increasing Triplet Subsequence.
Memory Usage: 24.6 MB, less than 80.42% of Python3 online submissions for Increasing Triplet Subsequence.
class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        first_small = second_small = float('inf')
        
        for num in nums:
            if num <= first_small:
                first_small = num
            elif num <= second_small:
                second_small = num
            else:
                return True
        
        return False
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