Increasing Triplet Subsequence
<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?
Greedy; Array
function increasingTriplet(nums: number[]): boolean {
    let n = nums.length;
    if (n < 3) return false;
    let min = nums[0],
        mid = Number.MAX_SAFE_INTEGER;
    for (let num of nums) {
        if (num <= min) {
            min = num;
        } else if (num <= mid) {
            mid = num;
        } else if (num > mid) {
            return true;
        }
    }
    return false;
}
