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 {
    private Long[] f;
    private int[] power;
    private Map<Integer, Integer> cnt;
    private int[] nxt;
    private int n;

    public long maximumTotalDamage(int[] power) {
        Arrays.sort(power);
        this.power = power;
        n = power.length;
        f = new Long[n];
        cnt = new HashMap<>(n);
        nxt = new int[n];
        for (int i = 0; i < n; ++i) {
            cnt.merge(power[i], 1, Integer::sum);
            int l = Arrays.binarySearch(power, power[i] + 3);
            l = l < 0 ? -l - 1 : l;
            nxt[i] = l;
        }
        return dfs(0);
    }

    private long dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        long a = dfs(i + cnt.get(power[i]));
        long b = 1L * power[i] * cnt.get(power[i]) + dfs(nxt[i]);
        return f[i] = Math.max(a, b);
    }
}