<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 dp[2502][2502];
    int func(int idx, int prev, vector<int>& nums){
        int n=nums.size();
        if(idx>=n) return 0;
        if(dp[idx][prev]!=-1) return dp[idx][prev];
        int x=0;
        for(int i=idx; i<n; i++){
            if(nums[prev] < nums[i]){
                x = max(1+func(i+1, i, nums), x);
            } 
        }
        // cout<<x<<" ";
        return dp[idx][prev] = x;
    }
    int lengthOfLIS(vector<int>& nums) {
        int n=nums.size();
        vector<int> vec;
        vec.push_back(-10001);
        for(auto i: nums) vec.push_back(i);
        memset(dp, -1, sizeof(dp));
        return func(0, 0, vec);
    }
};
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