<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) {

        if (nums.size() == 0) {
            return 0;
        }
        
        int ans = 1;
        map<int, int> hashmap;
        for(int i =0;i < nums.size();i++) {
            ans = max(ans, helper(nums, i, hashmap));
        }
        return ans;
    }

    int helper(vector<int>& nums, int end, map<int, int>& hashmap) {

        if (end == 0) {
            return 1;
        } else if (hashmap.count(end) == 1) {
            return hashmap[end];
        } 
        int sub_ans = 1;
        for(int i = 0; i < end; i++) {
            if (nums[i] < nums[end]) {
                sub_ans = max(sub_ans, helper(nums, i, hashmap) + 1);
            }
        }
        hashmap[end] = sub_ans;
        return sub_ans;
    }
};
17995