<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 maxi = 0;
    int solve(vector<int>& nums,int index,int prev,vector<vector<int>>& dp){
        if(index >= nums.size()){
            return 0;
        }

        if(prev != -1 && dp[index][prev] != -1) return dp[index][prev];

        // exclude
        int ex = solve(nums,index + 1,prev,dp);

        // include
        int in = 0;
        if(prev == -1 || nums[index] > nums[prev]){
            in = 1 + solve(nums,index+1,index,dp);
        }
        if(prev != -1)
            dp[index][prev] = max(in,ex);
        return max(in,ex);
    }
    int lengthOfLIS(vector<int>& nums) {
        vector<vector<int>> dp(nums.size(),vector<int>(nums.size(),-1));
        return solve(nums,0,-1,dp);
    }
};
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