<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) 
    {
        if (nums.size() == 1)
        {
            return 1;
        }

        int** memo = new int*[nums.size() + 1];

        for (int i = 0; i < nums.size() + 1; ++i)
        {
            memo[i] = new int[nums.size() + 1];

            for (int j = 0; j < nums.size() + 1; ++j)
            {
                memo[i][j] = -1;
            }
        }

        return helper(nums, 0, -1, memo);
    }

    int helper(const vector<int>& nums, int index, int prev, int** memo)
    {
        if (index >= nums.size())
        {
            return 0;
        }

        if (prev != -1 && memo[prev][index] != -1)
        {
            return memo[prev][index];
        }

        int count = 0;

        if (prev == -1 || nums[prev] < nums[index])
        {
            count = helper(nums, index + 1, index, memo) + 1; 
        }

        int next = helper(nums, index + 1, prev, memo);

        if (prev != -1)
        {
            return memo[prev][index] = max(count, next);
        }

        return max(count, next);
    }
};

/*
i must be set to index of recursive or base call

For i .. n - 1 in array
    Check i + 1 .. n - 1 for any strictly greater than value
        If found greater than value
            Recursively call function and increment count by 1

0 .. n -1 in array
    Check 0 + 1 .. n - 1 for any strictly greater than value
        If found greater than value
            Recursively call function and increment count by 1

1 .. n - 1 in array
    Check if 1 + 1 .. n - 1
*/
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