<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 function(vector<int>&a1,vector<int>&a2,int n,int m,vector<vector<int>>&dp){
        if(n==0 || m==0){
            return 0;
        }
        if(dp[n][m]!=-1){
            return dp[n][m];
        }
        if(a1[n-1]==a2[m-1]){
            return dp[n][m]=1+function(a1,a2,n-1,m-1,dp);
        }
        else{
            return dp[n][m]=max(function(a1,a2,n-1,m,dp),function(a1,a2,n,m-1,dp));
        }
    }
    int lengthOfLIS(vector<int>& a1) {
        vector<int>a2=a1;
        sort(a2.begin(),a2.end());
        vector<int>a3;
        for(int i=0;i<a2.size()-1;i++){
            if(a2[i]!=a2[i+1]){
                // i++;
                 a3.push_back(a2[i]);
            }
            
        }
        a3.push_back(a2[a2.size()-1]);
        int n=a1.size();
        int m=a3.size();
        vector<vector<int>>dp(n+1,vector<int>(m+1,-1));
        return function(a1,a3,n,m,dp);

    }
    // 001233
};
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