LeetCode 334: Increasing Triplet Subsequence

Explanation:

To solve this problem in O(n) time complexity and O(1) space complexity, we can use a two-pointer approach. We maintain two variables, first and second, to keep track of the smallest and second smallest elements found so far. We iterate through the array and update these variables accordingly. If we encounter a number greater than both first and second, we have found a triplet and return true. If not, we keep updating first and second. If we finish iterating through the array without finding a triplet, we return false.

• Time complexity: O(n) where n is the number of elements in the array.
• Space complexity: O(1)

:

class Solution {
    public boolean increasingTriplet(int[] nums) {
        if(nums == null || nums.length < 3) {
            return false;
        }
        
        int first = Integer.MAX_VALUE;
        int second = Integer.MAX_VALUE;
        
        for(int num : nums) {
            if(num <= first) {
                first = num;
            } else if(num <= second) {
                second = num;
            } else {
                return true;
            }
        }
        
        return false;
    }
}