<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:

    int find(vector<int> &nums,int i,int prev,vector<vector<short>>&dp)
    {
        if(i==nums.size())
        return 0;
        if(dp[i][prev+1]!=-1)
        return dp[i][prev+1];

        int take=INT_MIN;
        if(prev==-1||nums[prev]<nums[i])
        take=1+find(nums,i+1,i,dp);
        int not_take=0+find(nums,i+1,prev,dp);
        return dp[i][prev+1]=max(not_take,take);

    }
    int lengthOfLIS(vector<int>& nums) {
        vector<vector<short> > dp(nums.size(),vector<short>(nums.size()+1,-1));
        return find(nums,0,-1,dp);
    }
};
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