<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 findanswer(vector<int>&nums,int prev,int curr,vector<vector<int>>&dp)
{
    if(curr>=nums.size())
    {
        return 0;
    }
    if(dp[prev+1][curr+1]!=-1)
    {
        return dp[prev+1][curr+1];
    }
    int ans1=findanswer(nums,prev,curr+1,dp);
    int ans2=INT_MIN;
    if(prev==-1 || nums[curr]>nums[prev])
    {
        
        ans2=1+findanswer(nums,curr,curr+1,dp);
    }
    return dp[prev+1][curr+1]=max(ans1,ans2);
    
}
int tabulation(vector<int>& nums,vector<vector<int>>&dp)
{
    // base case ke haisaab se cheezo ko rkhlo
//    prev=-1 to nums.size()-1;
//  curr=0 to nums.size();
    for(int i=0;i<=nums.size();i++)
    {
        dp[i][nums.size()+1]=0;
    }
    for(int prev=nums.size()-1;prev>=-1;prev--)
    {
        for(int curr=nums.size()-1;curr>prev;curr--)
        {
            int ans1=dp[prev+1][curr+2];
            int ans2=INT_MIN;
            if(prev==-1 ||nums[curr]>nums[prev])
            {
                ans2=1+dp[curr+1][curr+2];

            }
            dp[prev+1][curr+1]=max(ans1,ans2);
        }
    }
    return dp[0][1];
}
int optimisation(vector<int>&nums)
{
    vector<int>next(nums.size()+2);
    //i current ko bta rha h
    int i=nums.size();
    while(i>0)
    {
        vector<int>curr(i);
        for(int prev=i-1;prev>=0;prev--)
        {
            
            int ans1=next[prev];
            int ans2=INT_MIN;
            if(prev==0||nums[i-1]>nums[prev-1])
            {
                ans2=1+next[i];

            }
            curr[prev]=max(ans1,ans2);

        }
        next=curr;
        i--;

    }
    return next[0];
}
    int lengthOfLIS(vector<int>& nums) {
    //   int prev=-1;
    //   int curr=0;
    //   vector<vector<int>>dp(nums.size()+2,vector<int>(nums.size()+2,0));
     
    //     // return findanswer(nums,prev,curr,dp);
    //     return tabulation(nums,dp);
    return optimisation(nums);



        
    }
};
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