<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 LIStabulation(vector<int>& nums, set<int>& st) {
        int n = nums.size();
        int m = st.size();
        vector<int> v2(st.begin(), st.end());

        vector<vector<int>> dp(n+1, vector<int> (m+1, 0));

        for(int i=1; i<n+1; i++) {
            for(int j=1; j<m+1; j++) {
                if(nums[i-1] == v2[j-1]) {
                    dp[i][j] = 1 + dp[i-1][j-1];
                }
                else {
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        return dp[n][m];
    }
    int lengthOfLIS(vector<int>& nums) {
        set<int> st;
        for(int i=0; i<nums.size(); i++) {
            st.insert(nums[i]);
        }

        return LIStabulation(nums, st);
    }
};
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