<p>You are given an integer array <code>nums</code>. You want to maximize the number of points you get by performing the following operation any number of times:</p>

<ul>
	<li>Pick any <code>nums[i]</code> and delete it to earn <code>nums[i]</code> points. Afterwards, you must delete <b>every</b> element equal to <code>nums[i] - 1</code> and <strong>every</strong> element equal to <code>nums[i] + 1</code>.</li>
</ul>

<p>Return <em>the <strong>maximum number of points</strong> you can earn by applying the above operation some number of times</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,4,2]
<strong>Output:</strong> 6
<strong>Explanation:</strong> You can perform the following operations:
- Delete 4 to earn 4 points. Consequently, 3 is also deleted. nums = [2].
- Delete 2 to earn 2 points. nums = [].
You earn a total of 6 points.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,3,3,3,4]
<strong>Output:</strong> 9
<strong>Explanation:</strong> You can perform the following operations:
- Delete a 3 to earn 3 points. All 2&#39;s and 4&#39;s are also deleted. nums = [3,3].
- Delete a 3 again to earn 3 points. nums = [3].
- Delete a 3 once more to earn 3 points. nums = [].
You earn a total of 9 points.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int maxi;
    int c;
    int DP[10000];
    int solve(int i, vector<int>& unique, unordered_map<int,int>& elements){
        // c++;
        // maxi=max(maxi,c);
        if(DP[i]!=-1) return DP[i];

        // if(maxi>30000)cout<<(maxi>50000)<<endl;
        // if(maxi>50000) return 0;
        if(i==unique.size()){
            return 0;
        }

        int take=0;
        int nottake=0;

        int cur_freq = elements[unique[i]];
        int pno_freq = elements[unique[i]-1];
        int nno_freq = elements[unique[i]+1];
        // elements.erase(unique[i]);
        // elements.erase(unique[i]+1);
        // elements.erase(unique[i]-1);
        take = unique[i]*cur_freq;
        if(i+1<unique.size() && unique[i+1]==unique[i]+1) take+=solve(i+2,unique,elements);
        else take+=solve(i+1,unique,elements);
        // elements[unique[i]]=cur_freq;
        // elements[unique[i]+1]= nno_freq;
        // elements[unique[i]-1]= pno_freq;
        nottake = solve(i+1, unique, elements);

        // c--;
        return  DP[i]=max(take,nottake);

    }
    int deleteAndEarn(vector<int>& nums) {
        maxi=-1;
        c=0;
        unordered_map<int,int> elements;
        vector<int> unique;
        for(auto it: nums){
            elements[it]++;
        }
        memset(DP,-1,sizeof(DP));
    
        for(auto it:elements){
            unique.push_back(it.first);
        }

        sort(unique.begin(),unique.end());
        int count = solve(0,unique, elements);

        return count;
    }
};
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