<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 count2(int n, int m, vector<int>& str1, vector<int>& str2, vector<vector<int>>& dp) {
        if(dp[n][m]!=-1)return dp[n][m];
        if(n==0||m==0)return 0;
        if(str1[n-1]==str2[m-1]){
            dp[n][m] = 1+count2(n-1,m-1,str1,str2,dp);
        }else{
            dp[n][m] = max(count2(n-1,m,str1,str2,dp),count2(n,m-1,str1,str2,dp));
        }
        return dp[n][m];
    }
    int lengthOfLIS(vector<int>& nums) {
        int m = nums.size();
        set<int> newSet(nums.begin(),nums.end());
        vector<int> newNums(newSet.begin(),newSet.end());
        sort(newNums.begin(),newNums.end());
        int n = newNums.size();

        vector<vector<int>> dp(m+1, vector<int> (n+1, -1));
        return this->count2(m,n,nums,newNums,dp);
    }
};
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