<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:

    void func(int ind,vector<int>&dp,vector<int>nums){
        if(ind==nums.size()){
            return;
        }

        for(int i=0; i<ind; i++){
            if(nums[ind]>nums[i]){
                dp[ind] = max(dp[ind],dp[i]+1);
            }
        }
        func(ind+1,dp,nums);
    }

    int lengthOfLIS(vector<int>& nums) {
        
        vector<int>dp(nums.size(),1);

        func(1,dp,nums);

        int x=1;
        for(int i=0; i<nums.size(); i++){
            if(x<dp[i]){
                x=dp[i];
            }
        }
        return x;
    }
};
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