<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:
        '''
        subproblem
        DP[i] = longestLIS length end at i
        # compare against every thing that's smaller in front of it
        can we compare with the last smaller? NO .98.99.1...100
        can we compare with closest smaller? NO 99.1.2.3.4.5..100
        # n^2
        '''
        DP = {} # memo

        def LIS(i):
            # base case
            if i not in DP:
                dp_i = 1
                for j in range(i):
                    if nums[i] > nums[j]:
                        dp_i = max(dp_i, LIS(j)+1)
                DP[i] = dp_i
            return DP[i]
        return_value = 0
        for i in range(len(nums)):
            return_value = max(return_value, LIS(i))
        return return_value
19100