<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 {
    multiset<int> s;
    vector<vector<int>> dp;
    int rec(vector<pair<int,int>>& nums,int i,bool tk){
        if(i>=nums.size()) return 0;
        if(dp[i][tk]!=-1) return dp[i][tk];
        int mx=0;
        int skip=0;
        skip=rec(nums,i+1,s.contains(i+1<nums.size()?nums[i+1].second:0));
        if(tk){
            mx=rec(nums,i+1,s.contains(i+1<nums.size()?nums[i+1].second:0));

        }
        else{
            s.insert(nums[i].second-1);
            s.insert(nums[i].second+1);
        mx=nums[i].first+rec(nums,i+1,s.contains((i+1)<nums.size()?nums[i+1].second:0));
        s.erase(s.find(nums[i].second-1));
        s.erase(s.find(nums[i].second+1));
        }
        // cout<<i<<" "<<tk<<" "<<max(mx,skip)<<endl;
        // return  max(mx,skip);
        return dp[i][tk]= max(mx,skip);

    }
public:
    int deleteAndEarn(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<pair<int,int>> h;
        int pro=nums[0];
        for(int i=1;i<nums.size();i++){
            if(nums[i]!=nums[i-1]){
                h.push_back({pro,nums[i-1]});
                pro=0;

            }
            pro+=nums[i];

        }
                h.push_back({pro,nums[nums.size()-1]});
        dp.resize(h.size(),vector<int>(2,-1));
        return rec(h,0,0);
    }
};
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