334. Increasing Triplet Subsequence

334. Increasing Triplet Subsequence (Medium)

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?

Companies:
Google, Facebook, Amazon

Related Topics:
Array, Greedy

Similar Questions:

• Longest Increasing Subsequence (Medium)
• Count Special Quadruplets (Easy)
• Count Good Triplets in an Array (Hard)

Solution 1. Longest Increasing Subsequence

This is a simplified version of 300. Longest Increasing Subsequence (Medium).

It's a regret greedy algorithm. We maintain a mono-increasing subsequence. For each A[i], we use it to replace the first number in the subsequence that is >= A[i].

It's simplified compared to LIS because we only need to keep a subsequence of length 3.

// OJ: https://leetcode.com/problems/increasing-triplet-subsequence/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool increasingTriplet(vector<int>& A) {
        int first = INT_MAX, second = INT_MAX;
        for (int n : A) {
            if (n <= first) first = n;
            else if (n <= second) second = n;
            else return true;
        }
        return false;
    }
};