<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 solve(vector<pair<int, int>>& v, int index){
        if(index >= v.size()) return 0;

        int val=v[index].first * v[index].second;
        // take the curr val
        int take=val;
        
        // check if next number is consecutive
        if(index+1 < v.size() && v[index+1].first == v[index].first+1){
            take += solve(v, index+2);
        }else{
            take += solve(v, index+1);
        }

        // skip the current val
        int notTake = solve(v, index+1);

        return max(take, notTake);
    }
    int memo(vector<pair<int, int>>& v, int index, vector<int>& dp){
        if(index >= v.size()) return 0;

        if(dp[index] != -1) return dp[index];

        int val=v[index].first * v[index].second;
        // take the curr val
        int take=val;
        
        // check if next number is consecutive
        if(index+1 < v.size() && v[index+1].first == v[index].first+1){
            take += memo(v, index+2, dp);
        }else{
            take += memo(v, index+1, dp);
        }

        // skip the current val
        int notTake = memo(v, index+1, dp);

        return dp[index] = max(take, notTake);
    }

    int deleteAndEarn(vector<int>& nums) {
        // count freq of each number
        unordered_map<int, int> freq;
        for(auto i:nums){
            freq[i]++;
        }

        // transfer to a vector for sorting
        vector<pair<int, int>> v;
        for(auto pair:freq){
            v.push_back({pair.first, pair.second});
        }

        // sorting
        sort(v.begin(), v.end());

        // recursion
        // return solve(v, 0);

        // memoization
        int n=v.size();
        vector<int> dp(n+1, -1);
        return memo(v, 0, dp);
    }
};
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