# Increasing Triplet Subsequence | Arrays | LeetCode

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?

```
class Solution {
public boolean increasingTriplet(int[] nums) {
// start with two largest values, as soon as we find a number bigger than both, while both have been updated, return true.
int first = Integer.MAX_VALUE, second = Integer.MAX_VALUE;
for( int n : nums){
if(n<= first) first = n; // update firstif n is smaller than both
else if(n<=second) second = n; // update second only if n is greater than first
else return true; // return if you find a number bigger than both
}
return false;
}
}
```