<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[3000][3000];
    int  func(vector<int>& nums, int prev,int idx,int len){
        if(idx == nums.size()){
            return 0;
        }
        if(dp[prev][idx] != -1) return dp[prev][idx];
        int ans = 0;
        if(nums[idx]>nums[prev]){
            ans += (1+func(nums,idx,idx+1,len+1));
        }
        ans = max(ans,func(nums,prev,idx+1,len));
        return dp[prev][idx]=ans;
    }
    int lengthOfLIS(vector<int>& nums) {
        int ans = 0;
        memset(dp,-1,sizeof(dp));
        for(int i=0; i<nums.size(); i++){
            ans = max(ans,func(nums,i,i+1,1));

        }
        return ans+1;
    }
};
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