<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:
    
    vector<int> nums;
    int dp[2509][2509];
    int go(int idx, int preIdx) {

        if (idx == nums.size()) return 0;

        if (dp[idx][preIdx + 1] != -1) 
            return dp[idx][preIdx + 1];
        
        int take = 0, notTake = 0;

        if (preIdx == -1 || nums[preIdx] < nums[idx]) {
            take = go(idx + 1, idx) + 1;
        }
        notTake = go(idx + 1, preIdx);

        return dp[idx][preIdx + 1] = max(take, notTake);
    }

    int lengthOfLIS(vector<int>& _nums) {
        nums = _nums;
        memset(dp, -1, sizeof(dp));
        return go(0, -1);
    }
};
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