increasing-triplet-subsequence.py

# 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.
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# Example 1:
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# Input: nums = [1,2,3,4,5]
# Output: true
# Explanation: Any triplet where i < j < k is valid.
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# Example 2:
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# Input: nums = [5,4,3,2,1]
# Output: false
# Explanation: No triplet exists.
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# Example 3:
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# 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.
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# Constraints:
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# 	1 <= nums.length <= 5 * 105
# 	-231 <= nums[i] <= 231 - 1
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# 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:
        # [1,2,3,4,5] [2,1,5,0,4,6]
        smallest = secondSmallest = float("inf")
        # They don't need to be adjacent, so we can just use min approach
        for num in nums:
            if num <= smallest:
                smallest = num
            elif num <= secondSmallest:
                secondSmallest = num
            else:
                return True
        return False