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# 0334. Increasing Triplet Subsequence

<https://leetcode.com/problems/increasing-triplet-subsequence>

## Description

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?

## ac

no topic, trick

```java
class Solution {
    public boolean increasingTriplet(int[] nums) {
        int small = Integer.MAX_VALUE, big = Integer.MAX_VALUE;

        for (int n : nums) {
            if (n <= small) {
                small = n;
            } else if (n <= big) {
                big = n;
            } else {
                return true;
            }
        }

        return false;
    }
}
```