<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 find_position(vector<int> arr,int num,int start,int end){
        // i want to find the index of first element greater than or equal to num

        if(start == end){
            return start;
        }
        int mid = (start+end)/2;
        if(arr[mid]==num || ((mid-1)>=0 && (arr[mid-1]<num) && arr[mid]>num))
            return mid;
        if(arr[mid]>num){
            return find_position(arr,num,start,mid-1);
        }
        return find_position(arr,num,mid+1,end);
    }

    int lengthOfLIS(vector<int>& arr) {
        int n = arr.size();
        // toh ulte se shuru kar 
        // vector<int> dp(n,1);
        // int ans = 1;
        // for(int i=n-1;i>=0;i--){
        //     int prev = arr[i],temp_ans = 1;
        //     for(int j=i+1;j<n;j++){
        //         if(arr[j]>arr[i])
        //             temp_ans = max(temp_ans,1+dp[j]);
        //     }
        //     dp[i] = temp_ans;
        //     ans= max(ans,temp_ans);
        // }
        // return ans;

        vector<int> increasing;
        increasing.push_back(arr[0]);
        for(int i=1;i<n;i++){
            // find the position of arr[i] in the increasing array
            if(arr[i]>increasing[increasing.size()-1])
                increasing.push_back(arr[i]);
            else if(arr[i]>=increasing[0]){
                int index = find_position(increasing,arr[i],0,increasing.size()-1);
                increasing[index] = arr[i];
            }
            else{
                increasing[0] = arr[i];
            }
        }
        return increasing.size();
    }
};
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