<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) {
        unordered_map<int, int> mp;
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            mp[nums[i]]++;
        }
        vector<int> unq;
        for (auto it : mp) {
            unq.push_back(it.first);
        }
        sort(unq.begin(), unq.end());
        vector<int> dp(unq.size() + 1);
        dp[0] = 0;
        dp[1] = mp[unq[0]] * unq[0];
        for (int i = 2; i <= unq.size(); i++) {
            if (unq[i - 2] < unq[i - 1] - 1) {
                dp[i] = dp[i - 1] + unq[i - 1] * mp[unq[i - 1]];
            } else {
                dp[i] = max(dp[i - 1], dp[i - 2] + unq[i - 1] * mp[unq[i - 1]]);
            }
        }
        return dp[unq.size()];
    }
};
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