<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[2501][2501];
    bool checkdp[2501][2501];
    
    int rec(int level, int prev_idx, vector<int>&nums) {
        int n = nums.size();
        // pruning case
        if (level >= n) return 0;
        // cache check
        if (checkdp[level][prev_idx + 1]) return dp[level][prev_idx + 1];
        // compute/transition
        int ans = rec(level + 1, prev_idx, nums);
        if (prev_idx == -1 || nums[level] > nums[prev_idx]) 
            ans = max(ans, 1 + rec(level + 1, level, nums));
        // save and return
        checkdp[level][prev_idx + 1] = true;
        return dp[level][prev_idx + 1] = ans;
    }
    
    int lengthOfLIS(vector<int>& nums) {
        memset(dp, -1, sizeof(dp));
        memset(checkdp, 0, sizeof(checkdp));
        return rec(0, -1, nums);
    }
};
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