<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) {
        set<int> tmp(nums.begin(), nums.end());
        vector<int> mySet(tmp.begin(), tmp.end());
        int n = nums.size(), m = mySet.size();

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

        // initial
        dp[0][0] = nums[0] == mySet[0] ? 1 : 0;
        for (int i = 1; i < n; i++) {
            dp[i][0] = nums[i] == mySet[0] ? 1 : dp[i-1][0];
        }
        for (int j = 1; j < m; j++) {
            dp[0][j] = mySet[j] == nums[0] ? 1 : dp[0][j-1];
        }

        // build
        for (int i = 1; i < n; i++) {
            for (int j = 1; j < m; j++) {
                if (nums[i] == mySet[j])
                    dp[i][j] = dp[i-1][j-1] + 1;
                else
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
            }
        }
        return dp[n-1][m-1];
    }
};
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