334. Increasing Triplet Subsequence

https://leetcode.com/problems/increasing-triplet-subsequence/description/

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:

Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.

Note: Your algorithm should run in O(n) time complexity and O(1) space complexity.

Example 1:

Input: [1,2,3,4,5]
Output: true

Example 2:

Input: [5,4,3,2,1]
Output: false

Thoughts

给定数组中是否存在递增的三元组，要求O(N)时间O(1)空间。LIS的简化版，由于只要求找长度为3的子序列，tails数组长度为2，元素分别记录以长度为1和2的最小子序列尾，O(1)空间，且遍历tails只需要O(1)时间。

Code

class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        tails = [math.inf] * 2
        for num in nums:
            if num <= tails[0]:
                tails[0] = num
            elif num <= tails[1]:
                tails[1] = num
            else: return True
        return False
        

class Solution {
    public boolean increasingTriplet(int[] nums) {
        int[] tails = new int[2];
        tails[0] = Integer.MAX_VALUE; tails[1] = Integer.MAX_VALUE;
        for (int num : nums) {
            if (num <= tails[0]) {
                tails[0] = num;
            } else if (num <= tails[1]) {
                tails[1] = num;
            } else {
                return true;
            }
        }

        return false;
    }
}

Analysis

时间复杂度O(N), 空间O(1).