<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) {
        vector<int> t1 = nums;
        vector<int> t;
        int n = nums.size();
        sort(t1.begin(),t1.end());
        int a = t1[0];
        t.push_back(a);
        for(int i=1;i<nums.size();i++){
            if(t1[i]==a){
                continue;
            }else{
                t.push_back(t1[i]);
                a=t1[i];
            }
        }
        int m = t.size();
        int maxLen=0;
        vector<vector<int>> v(m+1,vector<int>(n+1,0));
        for(int i=1;i<=m;i++){
            for(int j=1;j<=n;j++){
                if(nums[j-1] == t[i-1]) {
                v[i][j] = v[i-1][j-1] + 1;
                // maxLen = max(maxLen, v[i][j]);
            } else {
                v[i][j] = max(v[i-1][j],v[i][j-1]);
            }
            }
        }
        return v[m][n];
    }
};
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