<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 solve(int ind, vector<int>& nums, vector<vector<int>>& dp, int prev) {
        if (ind == nums.size()) {
            return 0;
        }
        if (dp[ind][prev + 1] != -1)
            return dp[ind][prev + 1];
        int take = 0;
        int ntake = solve(ind + 1, nums, dp, prev);
        if (prev == -1 || nums[ind] > nums[prev])
            take = 1 + solve(ind + 1, nums, dp, ind);
        return dp[ind][prev + 1] = max(take, ntake);
    }
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        /* vector<vector<int>>dp(n, vector<int>(n + 1, -1));
         return solve(0, nums, dp, -1); */

        
       /* vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
        for (int i = n - 1; i >= 0; i--) {
            for (int prev = n - 1; prev >= -1; prev--) {
                int take = 0;
                int ntake = dp[i + 1][prev + 1];
                if (prev == -1 || nums[i] > nums[prev])
                    take = 1 + dp[i + 1][i + 1];
                dp[i][prev + 1] = max(take, ntake);
            }
        }
        return dp[0][0]; */
        vector<int>dp(n+1, 0);
        for (int i = n - 1; i >= 0; i--) {
            vector<int>temp(n + 1, 0);
            for (int prev = n - 1; prev >= -1; prev--) {
                int take = 0;
                int ntake = dp[prev + 1];
                if (prev == -1 || nums[i] > nums[prev])
                    take = 1 + dp[i + 1];
                temp[prev + 1] = max(take, ntake);
            }
            dp = temp;
        }
        return dp[0];
    }
};
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