LeetCode #3186 — MEDIUM

Maximum Total Damage With Spell Casting

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

A magician has various spells.

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

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

Each spell can be cast only once.

Return the maximum possible total damage that a magician can cast.

Example 1:

Input: power = [1,1,3,4]

Output: 6

Explanation:

The maximum possible damage of 6 is produced by casting spells 0, 1, 3 with damage 1, 1, 4.

Example 2:

Input: power = [7,1,6,6]

Output: 13

Explanation:

The maximum possible damage of 13 is produced by casting spells 1, 2, 3 with damage 1, 6, 6.

Constraints:

1 <= power.length <= 10⁵
1 <= power[i] <= 10⁹

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: A magician has various spells. You are given an array power, where each element represents the damage of a spell. Multiple spells can have the same damage value. It is a known fact that if a magician decides to cast a spell with a damage of power[i], they cannot cast any spell with a damage of power[i] - 2, power[i] - 1, power[i] + 1, or power[i] + 2. Each spell can be cast only once. Return the maximum possible total damage that a magician can cast.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Two Pointers · Binary Search · Dynamic Programming

Example 1

[1,1,3,4]

Example 2

[7,1,6,6]

Step 02

Core Insight

What unlocks the optimal approach

If we ever decide to use some spell with power <code>x</code>, then we will use all spells with power <code>x</code>.
Think of dynamic programming.
<code>dp[i][j]</code> represents the maximum damage considering up to the <code>i</code>-th unique spell and <code>j</code> represents the number of spells skipped (up to 3 as per constraints).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3186: Maximum Total Damage With Spell Casting
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);
    }
}

// Accepted solution for LeetCode #3186: Maximum Total Damage With Spell Casting
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)
}

# Accepted solution for LeetCode #3186: Maximum Total Damage With Spell Casting
class Solution:
    def maximumTotalDamage(self, power: List[int]) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= n:
                return 0
            a = dfs(i + cnt[power[i]])
            b = power[i] * cnt[power[i]] + dfs(nxt[i])
            return max(a, b)

        n = len(power)
        cnt = Counter(power)
        power.sort()
        nxt = [bisect_right(power, x + 2, lo=i + 1) for i, x in enumerate(power)]
        return dfs(0)

// Accepted solution for LeetCode #3186: Maximum Total Damage With Spell Casting
use std::collections::HashMap;

impl Solution {
    pub fn maximum_total_damage(mut power: Vec<i32>) -> i64 {
        power.sort();
        let n = power.len();
        let mut cnt = HashMap::new();
        let mut nxt = vec![0; n];
        let mut f = vec![-1_i64; n];

        for i in 0..n {
            *cnt.entry(power[i]).or_insert(0) += 1;
            let j = match power[i + 1..].binary_search_by(|&x| x.cmp(&(power[i] + 2 + 1))) {
                Ok(pos) | Err(pos) => i + 1 + pos,
            };
            nxt[i] = j;
        }

        fn dfs(
            i: usize,
            n: usize,
            power: &Vec<i32>,
            nxt: &Vec<usize>,
            f: &mut Vec<i64>,
            cnt: &HashMap<i32, i32>,
        ) -> i64 {
            if i >= n {
                return 0;
            }
            if f[i] != -1 {
                return f[i];
            }
            let c = *cnt.get(&power[i]).unwrap();
            let a = dfs(i + c as usize, n, power, nxt, f, cnt);
            let b = power[i] as i64 * c as i64 + dfs(nxt[i], n, power, nxt, f, cnt);
            f[i] = a.max(b);
            f[i]
        }

        dfs(0, n, &power, &nxt, &mut f, &cnt)
    }
}

// Accepted solution for LeetCode #3186: Maximum Total Damage With Spell Casting
function maximumTotalDamage(power: number[]): number {
    const n = power.length;
    power.sort((a, b) => a - b);
    const f: number[] = Array(n).fill(0);
    const cnt: Record<number, number> = {};
    const nxt: number[] = Array(n).fill(0);
    for (let i = 0; i < n; ++i) {
        cnt[power[i]] = (cnt[power[i]] || 0) + 1;
        let [l, r] = [i + 1, n];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (power[mid] > power[i] + 2) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        nxt[i] = l;
    }
    const dfs = (i: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i]) {
            return f[i];
        }
        const a = dfs(i + cnt[power[i]]);
        const b = power[i] * cnt[power[i]] + dfs(nxt[i]);
        return (f[i] = Math.max(a, b));
    };
    return dfs(0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.