<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>

#define ll long long
ll a[2505];
ll dp[2505][2505];
int n;

ll func(int age, int akhon)
{
    if(akhon>n){
        return 0;
    }
    if(dp[age][akhon]!=-1) return dp[age][akhon];
    // nibo
    ll ans=0;
    if(a[akhon]>a[age])
    {
        ans=1+func(akhon, akhon+1);
    }
    // nibo na
    ans=max(ans, func(age, akhon+1));
    return dp[age][akhon]=ans;
}


class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        a[0]=-1e5;
        n=nums.size();
        for(int i=1;i<=n;i++)
        {
            a[i]=nums[i-1];
        }
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                dp[i][j]=-1;
            }
        }
        int now=func(0,1);
        return now;

        
    }
};
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