Leetcode: Increasing Triplet Subsequence

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.

Naive Solution: use DP, Time O(N^2), Space O(N)

　　dp[i] represents the length of longest increasing subsequence till i including element i in nums array. dp[i] is initialized to be 1.

　　dp[i] = max(dp[i], dp[j]+1), where j is an index before i

public class Solution {
    public boolean increasingTriplet(int[] nums) {
        int[] dp = new int[nums.length];
        for (int i=0; i<nums.length; i++) {
            dp[i] = 1;
            for (int j=0; j<i; j++) {
                if (nums[j] < nums[i]) {
                    dp[i] = Math.max(dp[i], dp[j]+1);
                }
                if (dp[i] == 3) return true;
            }
        }
        return false;
    }
}

Better Solution: keep two values. Once find a number bigger than both, while both values have been updated, return true.

small: is the minimum value ever seen untill now

big: the smallest value that has something before it that is even smaller. That 'something before it that is even smaller' does not have to be the current min value.

Example:
3,2,1,4,0,5

When you see 5, min value is 0, and the smallest second value is 4, which is not after the current min value.

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