<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[5000000];
 int lis(int i, vector<int>&a){
    if (dp[i] != -1){
        return dp[i];
    }
    int ans =1;
    for (int j=0; j<i; j++){
        if (a[i]>a[j]){
            ans = max(ans,lis(j,a)+1);
        }
    }
    return dp[i] = ans;
 }

    int lengthOfLIS(vector<int>& nums) {
        memset(dp,-1,sizeof(dp));
        int ans =0;
        for (int i=0; i<nums.size(); i++){
            ans = max(ans,lis(i,nums));
        
        }
        return ans;
    }
};
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