<p>Given an integer array <code>nums</code>, return <code>true</code><em> if there exists a triple of indices </em><code>(i, j, k)</code><em> such that </em><code>i &lt; j &lt; k</code><em> and </em><code>nums[i] &lt; nums[j] &lt; nums[k]</code>. If no such indices exists, return <code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> Any triplet where i &lt; j &lt; k is valid.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,4,3,2,1]
<strong>Output:</strong> false
<strong>Explanation:</strong> No triplet exists.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,1,5,0,4,6]
<strong>Output:</strong> true
<strong>Explanation:</strong> The triplet (3, 4, 5) is valid because nums[3] == 0 &lt; nums[4] == 4 &lt; nums[5] == 6.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>5</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you implement a solution that runs in <code>O(n)</code> time complexity and <code>O(1)</code> space complexity?
class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        // O(n*log(n)) to insert all values to BST. O(n*log(n)) to check if each value has values higher and smaller than him. 
        if (nums.size() < 3) {
            return false;
        }
        // map of all occurences of nums
        unordered_map<int, int> nums_occurences;
        set<int> nums_set;  // BST - O(log(n)) insertion and access.
        for (int num: nums) {
            if (!nums_occurences.count(num)) {
                nums_occurences[num] = 1;
                nums_set.insert(num);
            } else {
                nums_occurences[num]++;
            }
        }
        int last_min = nums[0];
        remove_occurences(nums_occurences, nums_set, nums[0]);
        int cur_num;
        for (int num_index = 1; num_index < nums.size(); num_index++) {
            cur_num = nums[num_index];
            if (cur_num < last_min) {
                last_min = cur_num;
            }
            else if (cur_num > last_min) {
                if (cur_num < *nums_set.rbegin()) {  // biggest key in nums_set
                    return true;
                }
            }
            remove_occurences(nums_occurences, nums_set, cur_num);
        }
        return false;
    }

    void remove_occurences(unordered_map<int,int>& nums_occurences, set<int>& nums_set, int cur_num) {
        nums_occurences[cur_num]--;
        if (nums_occurences[cur_num] == 0) {
            nums_occurences.erase(cur_num);
            nums_set.erase(cur_num);
        }
    }
};
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