<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 deleteAndEarn(vector<int>& nums) {
        vector<int> sums, arr;         
        sort(nums.begin(), nums.end()); 
        int sum = nums[0];
        
        for (int i = 1; i < nums.size(); i++){
            if (nums[i] != nums[i - 1]){
                sums.push_back(sum);
                arr.push_back(nums[i - 1]);
                sum = nums[i];
            } else sum += nums[i];
        }
        
        sums.push_back(sum);
        arr.push_back(nums[nums.size() - 1]);
        vector<int> memo(sums.size(), 0);
        
        return dp(0, sums, arr, memo);
    }
    
    int dp(int i, vector<int> & sums, vector<int> & arr, vector<int> & memo){
        if (i == sums.size() - 1) return sums[i];
        if (sums.size() - 2 == i) {
            if (arr[i + 1] - 1 == arr[i]) return max(sums[i], sums[i + 1]);
            else return sums[i] + sums[i + 1];
        }
        
        if (memo[i] != 0) return memo[i];
        
        return memo[i] = (arr[i] + 1 == arr[i + 1]) ? max(sums[i] + dp(i + 2, sums,arr, memo), dp(i + 1, sums, arr, memo)): sums[i] + dp(i + 1, sums, arr, memo);
    }
};
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