<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 recFun(vector<int>nums,vector<int>&dp,int ind){
        if(ind == 0)
        return dp[ind]=1;
        if(dp[ind]!=-1)
        return dp[ind];
        int res=1;
        for(int i=0;i<ind;i++){
            if(dp[i]==-1)
            recFun(nums,dp,i);
            if(nums[i] < nums[ind]){
                res = max(res,1+dp[i]);
            }
        }
        return dp[ind]=res;
    }
    int lengthOfLIS(vector<int>& nums) {
        vector<int>dp(nums.size(),-1);
        recFun(nums,dp,nums.size()-1);
        int res=1;
        for(int i=1;i<nums.size();i++){
            cout<<dp[i]<<" ";
           res = max(res,dp[i]); 
        }
        return res;
        // dp[0]=1;
        // int r,res=1;
        // for(int i=1;i<nums.size();i++){
        //     r = 1;
        //     for(int j=0;j<i;j++){
        //         if(nums[i]>nums[j])
        //         r = max(r,dp[j]+1);
        //     }
        //     dp[i]=r;
        //     res = max(res,dp[i]);
        // }
        // return res;
    }
};
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