<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 lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        // vector<vector<int>> dp(n, vector<int>(n,-1));
        // dp[n-1][n-1] = 1;
        vector<int> dp(n,-1);
        dp[n-1] = 1;
        int ret=1;
        for(int i= n-2; i>=0; i--){
            int maxx=1;
            vector<int> temp(n,-1);
            for(int j= n-1; j>i; j--){
                temp[j]=dp[j];
                if(nums[j] > nums[i]){
                    maxx = max(maxx,1+temp[j]);
                // }else{
                //     dp[i][j] = max(dp[i][j-1] , dp[i+1][j]) ;
                }
                // dp[i][i] = max(dp[i][i], dp[i][j]);
            }
            temp[i] = maxx;
            dp = temp;
            ret = max(ret, maxx);
        }
        // for(auto i:dp){
        //     for(auto j:i) cout<<j<<" ";
        //     cout<<endl;
        // }
        return ret;
    }
};
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