<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 {
public:
    long dp[2501][2501];
    int n; 
    
    int solve(vector<int>& nums, int i, int p){
        if(i >= n)
            return 0;
        
        if(p!=-1 && dp[i][p] != -1)
            return dp[i][p];
        
        int take = 0;
        if(p == -1  || nums[p] < nums[i]){
            take = 1 + solve(nums, i+1, i);  
        }
        
        int skip = solve(nums, i+1, p);
        
        if(p != -1)
            dp[i][p] = max(take, skip);
        
        return max(take, skip);
    }
    
    int lengthOfLIS(vector<int>& nums) {
         n = nums.size();
        
        memset(dp, -1, sizeof(dp));
        
        return solve(nums, 0, -1);
    }
};
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