# LeetCode : [334] Increasing Triplet Subsequence

Given an integer array

nums, return

trueif there exists a triple of indices

(i, j, k)such that

i < j < kand

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 <= 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?

Approach -1 : Brute Force

Idea is to check for all combinations which holds true condition that is nums[i] < nums[j] < nums[k] for i < j < k .

Time Complexity : O(n³)

Space Complexity : O(1)

Approach - 2 : Using Stack

We can observe that we just need to check whether an element has smaller value in left side and greater value in right side , so we can find left smaller and right greater in O(n) using stack .

Since we really need only one combination that holds this condition true , so we are storing all left smaller and then check for right greater if we found then return true .

Time Complexity : O(n) , for scanning two times and each time every element will be accessed at most two times

Space Complexity : O(n) , for stack and left array

Approach - 3 : Storing smaller and next smaller

If we have two element a and b where a < b and now if we found c as b < c then we will return true , now if we found a less than previous a then we will update a .

Time Complexity : O(n)

Space Complexity : O(1)

Thanks …