<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 max(int a, int b){
        return a > b ? a : b;
    }
    int maxinVec(vector<int> vec){
        int maxi = vec[0];
        for(int i : vec){
            if(i > maxi){
                maxi = i;
            }

        }
        return maxi;
    }
    int lengthOfLIS(vector<int>& nums) {
        vector<int> dp(nums.size(), 0);  //
        int length = 0;
        for(int i = nums.size() - 1;i >= 0; i--){  //go backwards
            vector<int> larger = {0};
            for(int j = i; j < nums.size(); j++){  //for a number look for a higher next number
                //vector<int> larger = {0}; //check how many larger values u got
                if(nums[j] > nums[i]){
                    larger.push_back(dp[j]);
                      //use previously computed value for the higher number and add 1 to it
                }
                 //find the next larger element with the highest increaing sequence length
                //if(dp[i] == 1){ dp[i]++;}
            }
            dp[i] = maxinVec(larger) + 1 ;
            length = max(dp[i], length);
        }

        for(int x : dp){
            cout<<x<<" ";
        }
        return length;        
        
    }
};
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