<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:
        n = len(nums)

        # memo = [ [-1] * (n+1) for _ in range(n)]
        def memoization(idx, prev):
            if idx == n:
                return 0
            if memo[idx][prev+1] != -1:
                return memo[idx][prev+1]
                
            not_take = 0 + memoization(idx+1, prev)
            take = 1 + memoization(idx+1, idx) if (nums[idx] > nums[prev] or prev == -1) else float('-inf')

            memo[idx][prev+1] = max(take, not_take)
            return memo[idx][prev+1]
        
        # trying memoization
        # return memoization(0, -1)

        def tabulation():
            dp = [ [0] * (n+1) for _ in range(n+1) ]

            for idx in range(n-1, -1, -1):
                for prev in range(idx-1, -2, -1):
                    not_take = dp[idx+1][prev+1]
                    take = 1 + dp[idx+1][idx+1] if (prev == -1 or nums[idx] > nums[prev]) else float('-inf')
                    dp[idx][prev+1] = max(take, not_take)

            return dp[0][-1+1]
        
        # trying tabulation
        # return tabulation()

        def optimized():
            if not nums:
                return 0
            
            n = len(nums)
            dp = [1] * n

            for idx in range(1, n):
                for prev in range(idx):
                    if nums[idx] > nums[prev]:
                        dp[idx] = max(dp[idx], 1 + dp[prev])

            return max(dp)
        
        return optimized()
17700