<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:
    long long dp[2502][2502];
    long long solve(vector<int> &nums, int pos, int prev_ind) {
        int n = nums.size();
        if(pos >= n) return 0;
        //to handle prev_ind = -1 we are using 2501 for prev_ind
        if(prev_ind == -1) {
            if(dp[pos][2501] != -1) return dp[pos][2501];
        }
        else {
            if(dp[pos][prev_ind] != -1) return dp[pos][prev_ind];

        }
        //when can we pick?
        //if it is first element or it is greater than
        //previously picked element
        int pick = 0,not_pick = 0;
        if(prev_ind == -1 || nums[prev_ind] < nums[pos]) {
            pick = 1 + solve(nums,pos+1,pos);
        }
        //we can also choose to not pick
        not_pick = solve(nums,pos+1,prev_ind);
        //the one which generates max is result
        if(prev_ind == -1) return dp[pos][2501] = max(pick,not_pick);
        return dp[pos][prev_ind] = max(pick,not_pick);
    }
    int lengthOfLIS(vector<int>& nums) {
        memset(dp,-1,sizeof(dp));
        return solve(nums,0,-1);
    }
};
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