<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 lcs(vector<int> x, vector<int> y)
{
    int n =x.size();
    int m =y.size();
    vector<vector<int>> dp(n+1,vector<int>(m+1));
    for(int i =0;i<n+1;i++)
    {
        for(int j =0;j<m+1;j++)
        {
            if(i==0 || j==0)
            {
                dp[i][j]=0;
            }
        }
    }
    for(int i =1;i<n+1;i++)
    {
        for(int j =1;j<m+1;j++)
        {
            if(x[i-1]==y[j-1])
            {
                dp[i][j]=1+dp[i-1][j-1];
            }
            else
            {
                dp[i][j]=max(dp[i][j-1],dp[i-1][j]);
            }
        }
    }
    return dp[n][m];

}
    int lengthOfLIS(vector<int>& nums) {
        set<int>st;
        for(int i =0;i<nums.size();i++)
        {
            st.insert(nums[i]);
        }
        vector<int>nums2(st.begin(),st.end());
        return lcs(nums,nums2);
    }
};
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