334. Increasing Triplet Subsequence

Tags: #array #greedy #medium

Problem Statement

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.

Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?

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.

My approach

class Solution:
    def increasingTriplet(self, nums):
        f = float('inf')
        s = float('inf')

        for n in nums:
            if n <= f:
                f = n
            elif n <= s:
                s = n
            else:
                return True

        return False

In this solution, first we intialise two variable f and s to be infinity. Then, we loop over the nums array and compare the current element if it's smaller then f. Certainly in the first iteration, n will be smaller than f so, f is now assigned to be the current element. In the next iteration, we again check the condition and keep the smaller of the element in f (this means the smallest element will be in f) otherwise we reassign to s (this means that the element just larger to f will be in s) . Once again, checking through the conditions, if there are no more elements greater than f and s then we return True since our triplets condition is satisfied otherwise we return False.

Personal Thoughts

I couldn't come up with a solution to this problem on my own and had to go through the discussions tab in order to do this solution. I will come back to this problem once again, and try to solve this myself on a later date.