# Increasing Triplet Subsequence

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```

History is always recurrent.

At the very first, I thought it is a typical dp question, and then I wrote a classical dp solution, and then I got a TLE…

This AC solution I actually checked it out from discuss board, that is really smart.

```class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
N = len(nums)
first = float("inf")
second = float("inf")

for i in range(N):
if nums[i] < first:
first = nums[i]

elif first < nums[i] < second:
second = nums[i]

if nums[i] > second:
return True

return False```

# Increasing Triplet Subsequence

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`

Since this problem is literally similar another one (longest-increasing-subsequence), I have tried the traditional dynamic programming at very first. However, this question is not as easy as it looks be, I got a Time Limit Exceeded.

After that, I noticed that we actually can iterate the entire array by one pass with maintaining two pointers.

```class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
#         # TLE
#         if not nums:
#             return False

#         dp =  * len(nums)

#         for i in range(1, len(nums)):
#             for j in range(i):
#                 if nums[i] > nums[j]:
#                     dp[i] = max(dp[i], dp[j] + 1)
#                 if dp[i] == 3:
#                     return True

#         return False

first = second = float('inf')

for n in nums:
if n <= first:
first = n

elif n <= second:
second = n
else:
return True

return False```