<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>

#define LIMIT 1e4+1
class Solution {
    int util(int i, int prev, vector<int>& nums, vector<vector<int>> & dp){
        // cout << i << ", " << prev << "\n";
        if(i == nums.size()){
            return 0;
        }
        if(dp[i][prev + LIMIT] !=-1 ){
            return dp[i][prev + LIMIT];
        }
        if(nums[i] > prev){
            return dp[i][prev + LIMIT] = max(util(i+1, nums[i], nums, dp)+1, util(i+1, prev, nums, dp));
        }

        return dp[i][prev + LIMIT] = util(i+1, prev, nums, dp);
    }
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        // vector<vector<int>> dp(n+1, vector<int>(2*LIMIT, 0));
        vector<int> curr(2*LIMIT, 0), ahead(2*LIMIT, 0);
        for(int i = n-1; i >= 0; i--){
            for(int j = 0; j < 2*LIMIT; j++){
                if(nums[i]+LIMIT > j){
                    curr[j] = max(ahead[nums[i] + LIMIT]+1, ahead[j]);
                }
                else
                    curr[j] = ahead[j];
            }
            ahead = curr;
        }

        return ahead[0];
    }
};
25055