<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:




        def dp(nums, last):

            if not nums:
                return 0

            best = 0
            for ind, num in enumerate(nums):

                if num > last:
                    best = max(
                        best,
                        dp(nums[ind+1:],num) + 1)
            
            return best




        def dp_two(nums):

            if not nums:
                return 0

            num = nums[0]
            best = 0
            for ind in range(1, len(nums)):

                if nums[ind] > num:
                    best = max(
                        best,
                        dp_two(nums[ind:]) + 1)
            
            return best



        @lru_cache(None)
        def dp3(i):

            # if i >= len(nums):
            #     return 0

            best = 0
            for ind, num in enumerate(nums[:i]):

                if nums[i] > num:
                    best = max(best, dp3(ind) + 1)
            
            return best




        b = 0
        # for ind, num in enumerate(nums):
        #     b = max(b, dp_two(nums[ind:]) + 1)
    

        for ind in range(len(nums)):
            b = max(b, dp3(ind))


        return b  + 1
18800