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
class Solution {
    public boolean increasingTriplet(int[] nums) {
        int n = nums.length;
        int[] lmi = new int[n];
        int[] rmx = new int[n];
        lmi[0] = Integer.MAX_VALUE;
        rmx[n - 1] = Integer.MIN_VALUE;
        for (int i = 1; i < n; ++i) {
            lmi[i] = Math.min(lmi[i - 1], nums[i - 1]);
        }
        for (int i = n - 2; i >= 0; --i) {
            rmx[i] = Math.max(rmx[i + 1], nums[i + 1]);
        }
        for (int i = 0; i < n; ++i) {
            if (lmi[i] < nums[i] && nums[i] < rmx[i]) {
                return true;
            }
        }
        return false;
    }
}