<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>

vector<int> numList;
const int MAX = 2500 + 1;
int memory[MAX][MAX];

// called with LIS(0, m)
int LIS(int i, int prev) {
    if (i == numList.size())
        return 0;

    auto& ret = memory[i][prev];

    if (ret != -1)
        return ret;

    int leave = LIS(i + 1, prev);
    int take = 0;

    if (prev >= (int)numList.size() || numList[prev] < numList[i])
        take = 1 + LIS(i + 1, i);

    return ret = max(leave, take);
}

void printLIS(int i, int prev) {
    if (i == numList.size())
        return;
        
    int leave = LIS(i + 1, prev); // Leave
    int take = 0;

    if (prev >= (int)numList.size() || numList[prev] < numList[i])
        take = 1 + LIS(i + 1, i); // Take
    int optimal = max(leave, take);
    if (optimal == leave)
        printLIS(i + 1, prev);
    else {
        cout << numList[i] << " ";
        printLIS(i + 1, i);
    }
}
class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        numList = nums;
        memset(memory, -1, sizeof(memory));
        int ans = LIS(0, numList.size());
        printLIS(0, numList.size());
        return ans;
    }
};

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