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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