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
func maximumTotalDamage(power []int) int64 {
	n := len(power)
	sort.Ints(power)
	cnt := map[int]int{}
	nxt := make([]int, n)
	f := make([]int64, n)
	for i, x := range power {
		cnt[x]++
		nxt[i] = sort.SearchInts(power, x+3)
	}
	var dfs func(int) int64
	dfs = func(i int) int64 {
		if i >= n {
			return 0
		}
		if f[i] != 0 {
			return f[i]
		}
		a := dfs(i + cnt[power[i]])
		b := int64(power[i]*cnt[power[i]]) + dfs(nxt[i])
		f[i] = max(a, b)
		return f[i]
	}
	return dfs(0)
}