<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 {
    map<int, int> mp;
    int solve(int i, int taked, vector<pair<int, int>>& vec, vector<vector<int>>& dp) {
        if (i >= vec.size()) return 0;
        if (dp[i][taked] != -1) return dp[i][taked];
        if (!taked) {
            int op1 = vec[i].first * vec[i].second + solve(i + 1, 1 - taked, vec, dp);
            int op2 = solve(i + 1, taked, vec, dp);
            return dp[i][taked] = max(op1, op2);
        } else {
            int op1 = solve(i + 1, 1 - taked, vec, dp);
            int op2 = 0;
            if (mp.find(vec[i].first - 1) == mp.end()) {
                op2 = vec[i].first * vec[i].second + solve(i + 1, taked, vec, dp);
            }
            return dp[i][taked] = max(op1, op2);
        }
    }
public:
    int deleteAndEarn(vector<int>& nums) {
        for (int num : nums) mp[num]++;
        vector<pair<int, int>> vec;
        for (auto m : mp) vec.push_back({m.first, m.second});
        int n = vec.size();
        vector<vector<int>> dp(n + 1, vector<int> (2, -1));
        return solve(0, 0, vec, dp);
    }
};
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