<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>& arr) {
          vector<int>nums;
    nums.push_back(INT_MIN);
    int n=arr.size();
    int i=0;
    while(i<n){
        nums.push_back(arr[i]);
        i++;
    }
int n1=nums.size();
    vector<int>dp(n1,0);

    int index=n1-1;
    while(index>=1){
        int prev=0;
        vector<int>dp1(n1);
        while(prev<=index){
          int prev_num=nums[prev];
          int curr=nums[index];
        if(curr>prev_num){
    dp1[prev]=max(dp[prev],1+dp[index]);
        }
        else if(index!=prev){
            dp1[prev]=dp[prev];
        }
        else if(index==prev){
            dp1[prev]=1;
        }
        prev++;
        }
        dp=dp1;
        index--;
    }
    return dp[0];
    }
};
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