## DEV Community

Abhishek Chaudhary

Posted on

# Increasing Triplet Subsequence

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?SOLUTION:

``````class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
n = len(nums)
minval, maxval = min(nums), max(nums)
for i in range(1, n - 1):
if nums[i] > minval and nums[i] < maxval:
if min(nums[:i]) < nums[i] and nums[i] < max(nums[i+1:]):
return True
return False
``````