Tags: "leetcode", "python", "two-pointers", access_time 1-min read

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Increasing Triplet Subsequence

Created: November 8, 2018 by [lek-tin]

Last updated: March 24, 2020

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:

Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Note Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example 1

Input: [1,2,3,4,5]
Output: true

Example 2

Input: [5,4,3,2,1]
Output: false

Solution

import math

class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        if not nums or len(nums) < 3:
            return False
        # min_1 < min_2
        min_1, min_2 = math.inf-1, math.inf
        for n in nums:
            # use <= to skip duplicate value, for example, [1,1,1,1,1,1,1,1,1,1]
            if n <= min_1:
                min_1 = n
            elif n <= min_2:
                min_2 = n
            else:
                return True

        return False