<p>Given an integer array <code>nums</code>, return <em>the length of the longest <strong>strictly increasing </strong></em><span data-keyword="subsequence-array"><em><strong>subsequence</strong></em></span>.</p>

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

<pre>
<strong>Input:</strong> nums = [10,9,2,5,3,7,101,18]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,0,3,2,3]
<strong>Output:</strong> 4
</pre>

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

<pre>
<strong>Input:</strong> nums = [7,7,7,7,7,7,7]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2500</code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><b>Follow up:</b>&nbsp;Can you come up with an algorithm that runs in&nbsp;<code>O(n log(n))</code> time complexity?</p>

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        cache = {}

        def longest_increasing_sub(idx, last_element):
            if idx == len(nums):
                return 0
            
            if (idx, last_element) in cache:
                return cache[(idx, last_element)]

            len_incr_sub = 0

            for i in range(idx, len(nums)):
                if nums[i] > last_element:
                    len_incr_sub = max(len_incr_sub, 1 + longest_increasing_sub(i + 1, nums[i]))
            
            cache[(idx, last_element)] = len_incr_sub
            return len_incr_sub
        
        return longest_increasing_sub(0, float('-inf'))
18600