<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 helper(vector<int>&nums,int index,int prev, vector<vector<int>>&dp){

if(index==nums.size()){
    return 0;
}
if(dp[index][prev+1]!=-1){
    return dp[index][prev+1];
}
    //not pick
    int notpick=helper(nums,index+1,prev,dp);
    int pick=INT_MIN;
    if(prev==-1 || nums[index]>nums[prev]){
        pick=1+helper(nums,index+1,index,dp);
    }
    return dp[index][prev+1]=max(pick,notpick);
}
    int lengthOfLIS(vector<int>& nums) {
         vector<vector<int>>dp(nums.size(),vector<int>(nums.size(),-1));
        return helper(nums,0,-1,dp);
    }
};
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