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.

Analyse:

First glance at the problem, I think of brute force, time complexity is O(n^3). Time limit exceeded. 

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        if(nums.size() < 3) return false;
        
        for(int i = 0; i < nums.size() - 2; i++){
            for(int j = i + 1; j < nums.size() - 1; j++){
                for(int k = j + 1; k < nums.size(); k++){
                    if(nums[i] < nums[j] && nums[j] < nums[k])
                        return true;
                }
            }
        }
        return false;
    }
};

Then I come up with a o(n^2) solution, it is fix the first number, for the second number, I scan the array and keep recording a number that is larger than the first number, when the third number is found, return true. For exmaple, I have 3, 8, 7, 2, 0, 1, 4. Firstly, fix 3, second number is 8, when it scans to 7, second number becomes 7; keep scaning, when it comes to 2, since 2 is smaller than 3, doesn't update, second number is still 7. After one pass scan, nothing returned... Then fix 8, and repeat the same process. 

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        if(nums.size() < 3) return false;
        
        for(int i = 0; i < nums.size() - 2; i++){
            int b = i + 1;
            while(nums[b] <= nums[i] && b < nums.size() - 1)
                b++;
            while(b + 1 < nums.size() && nums[b] >= nums[b + 1] && nums[b + 1] > nums[i])
                b++;
            if(b == nums.size() - 1) continue;
            int c = b + 1;
            while(c < nums.size() && nums[b] >= nums[c])
                c++;
            if(c == nums.size()) continue;
            return true;
        }
        return false;
    }
};

After all those tries, I gave a o(n) version. 

class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        if(nums.size() < 3) return false;
        
        int a = INT_MAX, b = INT_MAX;
        for(int i = 0; i < nums.size(); i++){
            a = min(nums[i], a);
            if(nums[i] > a) b = min(nums[i], b);
            if(nums[i] > b) return true;
        }
        return false;
    }
};