<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 f(int i, int prev, vector<int>& nums) {
        if (i < 0) {
            return 0;
        }
        if (dp[i][prev+1]!=-1) {
            return dp[i][prev+1];
        }
        int take = -1e8;
        int not_take = f(i - 1, prev, nums);
        if (prev == -1 || nums[i] < nums[prev]) {
            take = 1 + f(i - 1, i, nums);
        }
        return dp[i][prev+1] = max(take, not_take);
    }
    int lengthOfLIS(vector<int>& nums) {
        memset(dp,0,sizeof(dp));
        int n = nums.size();
        for(int i=n-1;i>=0;i--){
            for(int prev = i-1;prev>=-1;prev--){
                int take = -1e8;
                int not_take = dp[i + 1][prev+1];
                if (prev == -1 || nums[i] > nums[prev]) {
                    take = 1 + dp[i + 1][i+1];
                }
                dp[i][prev+1] = max(take, not_take);
            }
        }
        return dp[0][0];
    }
};
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