Maximum Total Damage With Spell Casting
<p>A magician has various spells.</p>

<p>You are given an array <code>power</code>, where each element represents the damage of a spell. Multiple spells can have the same damage value.</p>

<p>It is a known fact that if a magician decides to cast a spell with a damage of <code>power[i]</code>, they <strong>cannot</strong> cast any spell with a damage of <code>power[i] - 2</code>, <code>power[i] - 1</code>, <code>power[i] + 1</code>, or <code>power[i] + 2</code>.</p>

<p>Each spell can be cast <strong>only once</strong>.</p>

<p>Return the <strong>maximum</strong> possible <em>total damage</em> that a magician can cast.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">power = [1,1,3,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">6</span></p>

<p><strong>Explanation:</strong></p>

<p>The maximum possible damage of 6 is produced by casting spells 0, 1, 3 with damage 1, 1, 4.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">power = [7,1,6,6]</span></p>

<p><strong>Output:</strong> <span class="example-io">13</span></p>

<p><strong>Explanation:</strong></p>

<p>The maximum possible damage of 13 is produced by casting spells 1, 2, 3 with damage 1, 6, 6.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= power.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= power[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Hash Table; Two Pointers; Binary Search; Dynamic Programming; Counting; Sorting
class Solution {
public:
    long long maximumTotalDamage(vector<int>& power) {
        sort(power.begin(), power.end());
        this->power = power;
        n = power.size();
        f.resize(n);
        nxt.resize(n);
        for (int i = 0; i < n; ++i) {
            cnt[power[i]]++;
            nxt[i] = upper_bound(power.begin() + i + 1, power.end(), power[i] + 2) - power.begin();
        }
        return dfs(0);
    }

private:
    unordered_map<int, int> cnt;
    vector<long long> f;
    vector<int> power;
    vector<int> nxt;
    int n;

    long long dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i]) {
            return f[i];
        }
        long long a = dfs(i + cnt[power[i]]);
        long long b = 1LL * power[i] * cnt[power[i]] + dfs(nxt[i]);
        return f[i] = max(a, b);
    }
};