<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 solution(vector<int>& nums, int n, vector<int>& incNums, int m, vector<vector<int>>& t )
    {
       if(n==0 || m==0)
       {
        t[n][m] = 0;
        return t[n][m];
       }
       if(t[n][m] != -1)
       {
        return t[n][m];
       }

       if(nums[n-1] == incNums[m-1])
       {
        t[n][m] = 1 + solution(nums, n-1, incNums, m-1, t); 
        return t[n][m];
       }

       t[n][m] = max(solution(nums, n-1, incNums, m, t),solution(nums, n, incNums, m-1, t) );
       return t[n][m];
    }
    int lengthOfLIS(vector<int>& nums) {
        vector<int> incNums;
        set<int> st;
        for(auto a: nums)
        {
            st.insert(a);
        }
        for(auto a:st)
        {
            incNums.push_back(a);
        }

        int n = nums.size();
        int m = incNums.size();
        vector<vector<int>> t(n+1, vector<int>(m+1, -1));

        
        return solution(nums,n, incNums,m ,t);
    }
};
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