<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:
    long long helper(int i, vector<int> &nums, vector<long long> &dp){
        if(i>=nums.size() || i<0){
            return 0;
        }
        if(dp[i]!=-1){
            return dp[i];
        }
        long long cv = nums[i];
        int idx = i+1;
        while(idx<nums.size() &&   nums[idx]==nums[i]){
            cv += nums[i];
            idx++;
        }
        int ni = idx;
        while(idx<nums.size() &&  (nums[idx]==nums[i]+1)){
            idx++;
        }
        long long take=0, non_take=0;
        take = cv + helper(idx, nums, dp);
        non_take = helper(ni, nums, dp);

        return dp[i] = max(take, non_take);
    }
    int deleteAndEarn(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<long long> dp(n+1,-1);
        return helper(0, nums, dp);
    }
};
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