334. Increasing Triplet Subsequence

Problem:

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.

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

Examples:
Given [1, 2, 3, 4, 5],
return true.

Given [5, 4, 3, 2, 1],
return false.

Analysis:

这种题也是刷出来的，自己根本不可能想出来。

Use two pointers, min and secMin (second minmum). Initialize them to Integer.MAX_VALUE. Update their values on the go. If n is greater than secMin, we know that there is a triplet. 

Keep in mind that, min updates before second min. If secMin is not max value then we know that min is not max value. 

	4	3	6	2	8
min	4	3	3	2
2min	max	max	6	6

When we reach 6, we can make sure there is a increasing order of size 6. 2 starts the new sequence.

Solution:

class Solution {
    public boolean increasingTriplet(int[] nums) {
        int min = Integer.MAX_VALUE, secMin = Integer.MAX_VALUE;
        for (int n: nums) {
            if (n <= min) {
                min = n;
            } else if (n <= secMin) {
                secMin = n;
            } else if (n > secMin) {
                return true;
            }
        }
        return false;
    }
}