<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>

#include<bits/stdc++.h>
class Solution {
public:
    int solve(vector<int>& nums, vector<vector<int>> &dp, int i, int maxm) {
        if(i == nums.size()) {
            dp[i][maxm + 10001] = 0;
            return 0;
        }
        
        if(dp[i][maxm + 10001] != -1) {
            return dp[i][maxm + 10001];
        }
        
        if(nums[i] > maxm) {
            dp[i][maxm + 10001] = max(1 + solve(nums, dp, i + 1, nums[i]), solve(nums, dp, i + 1, maxm));
        } else {
            dp[i][maxm + 10001] = solve(nums, dp, i + 1, maxm);
        }
        return dp[i][maxm + 10001];
    }
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> prev(20002, -1);
        vector<int> curr(20002, -1);
        
        for(int i = n; i >= 0; i--) {
            for(int maxm = -10001; maxm <= 10000; maxm++) {
                if(i == n) {
                    curr[maxm + 10001] = 0;
                } else {
                    if(nums[i] > maxm) {
                        curr[maxm + 10001] = max(1 + prev[nums[i] + 10001], prev[maxm + 10001]);
                    } else {
                        curr[maxm + 10001] = prev[maxm + 10001];
                    }
                }
                
            }
             prev = curr;
        }
        
        return curr[0];
        // return solve(nums, dp, 0, -10001);
    }
};
25055