<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 size = nums.size();
        int** memo = new int*[size];

        for (int i = 0 ; i < size; ++i)
        {
            memo[i] = new int[size];
            memset(memo[i], -1, sizeof(int) * size);
        }

        int result = lengthOfLISHelper(nums, -1, 0, memo);

        for (int i = 0; i < size; ++i)
        {
            delete[] memo[i];
        }

        delete[] memo;

        return result;
    }

    int lengthOfLISHelper(const vector<int>& nums, int prevIndex, int currIndex, int** memo)
    {
        if (currIndex >= nums.size())
        {
            return 0;
        }

        if (prevIndex != -1 && memo[prevIndex][currIndex] != -1)
        {
            return memo[prevIndex][currIndex];
        }

        int continuePath = 0;

        if (prevIndex == -1 || nums[prevIndex] < nums[currIndex])
        {
            continuePath = lengthOfLISHelper(nums, currIndex, currIndex + 1, memo) + 1;
        }

        int tryNextStartingPath = lengthOfLISHelper(nums, prevIndex, currIndex + 1, memo);

        if (prevIndex != -1)
        {
            memo[prevIndex][currIndex] = max(continuePath, tryNextStartingPath);
            return memo[prevIndex][currIndex];
        }

        return max(continuePath, tryNextStartingPath);
    }
};


/*
    Use recursion with 2 indices:
    Assume CurrIndex, PrevIndex.
    CurrIndex = 0
    PrevIndex = -1

    Base case:

        If CurrIndex == nums.size()
            Return 0
    
    Assume ContinuePath = 0

    Other cases:

        If PrevIndex != -1 || nums[PrevIndex] < nums[currIndex]
            Case1 = recurse(CurrIndex = CurrIndex + 1, PrevIndex = CurrIndex) + 1
        
        GoNextPath = recurse(CurrIndex = CurrIndex + 1, PrevIndex)

    We need to return max(ContinuePath, goNextPath)
    
*/
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