## 334. 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 <= 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?

## Rust Solution

``````struct Solution;

impl Solution {
fn increasing_triplet(nums: Vec<i32>) -> bool {
let mut dp: Vec<i32> = vec![];
for x in nums {
if let Err(i) = dp.binary_search(&x) {
if i == dp.len() {
dp.push(x)
} else {
dp[i] = x;
}
}
if dp.len() == 3 {
return true;
}
}
false
}
}

#[test]
fn test() {
let nums = vec![1, 2, 3, 4, 5];
let res = true;
assert_eq!(Solution::increasing_triplet(nums), res);
let nums = vec![5, 4, 3, 2, 1];
let res = false;
assert_eq!(Solution::increasing_triplet(nums), res);
}
``````

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