<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 func(vector<int>&nums,int id,int prev,vector<vector<int>>&dp){
    //     if(id>=nums.size())return 0;
    //     if(dp[id][prev+1]!=-1)return dp[id][prev+1];
    //     int len = func(nums,id+1,prev,dp);
    //     if(prev==-1 || nums[id]>nums[prev])
    //     len = max(len,1+func(nums,id+1,id,dp));
    //     return dp[id][prev+1]  = len;
    // }
    
    int lengthOfLIS(vector<int>& nums) {
        int n =nums.size();
        // int maxi = *max_element(nums.begin(),nums.end());
        vector<int>nxt(n+1,0);
        // return func(nums,0,-1,dp);
        for(int id=n-1;id>=0;id--){
            vector<int>curr(n,0);
            for(int prev=id-1;prev>=-1;prev--){
                int len = nxt[prev+1];
                if(prev==-1 || nums[id]>nums[prev])
                len = max(len,1+nxt[id+1]);
                curr[prev+1]  = len;
            }
            nxt = curr;
            
        }
        return nxt[-1+1];

        // for(int i =1; i<=n;i++){
        //     for(int j =1;j<=maxi;j++){
        //         if(nums[i-1]==j){
        //             dp[i][j] =1;
        //         }
        //         else{
        //             dp[i][j] = 2;
        //         }
        //     }
        // }
    }
};
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