leetcode_gym/_problems/increasingTripletSubsequence.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Increasing Triplet Subsequence	2020-07-05	dp	https://leetcode.com/problems/increasing-triplet-subsequence	Solved

Problem

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

Initially tried this, but it would fail for [2,5,3,4,5]

{% highlight python %}

def increasingTriplet(self, nums: List[int]) -> bool:
    
    if not nums: return false
    
    first = nums[0]
    second = None
    third = None
    
    for i in range(1, len(nums)):
        
        if first and not second and not third and nums[i]>first:
            second = nums[i]
        elif first and second and not third and nums[i]>second:
            third = nums[i]
    
    
    print(first, second, third)
    return True if third else False

{% endhighlight %}

---

A simpler solution involving thresholds

{% highlight python %}

def increasingTriplet(self, nums: List[int]) -> bool:

first = float('inf')
second = float('inf')
third = float('inf')

for n in nums:

    if n<first:
        first = n
    elif n>first and n<second:
        second = n
    elif n>second and n<third:
        third = n

print(first, second, third)

return third!=float('inf')

{% endhighlight %}

---

Looks like the way to approach this is similiar to the LIS (Longest increasing subsequence) problem.