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

maximumTotalDamage.js

/**
 * @param {number[]} power
 * @return {number}
 */
var maximumTotalDamage = function(power) {
    // Step 1: Count how many times each damage value appears
    const count = new Map();
    for (const dmg of power) {
        count.set(dmg, (count.get(dmg) || 0) + 1);
    }

    // Step 2: Sort unique damage values in ascending order
    const unique = Array.from(count.keys()).sort((a, b) => a - b);
    const n = unique.length;

    // Edge case: only one unique damage value
    if (n === 1) {
        return unique[0] * count.get(unique[0]);
    }

    // Step 3: Initialize DP array
    // dp[i] stores the max damage using the first i unique damage values
    const dp = new Array(n + 1).fill(0);
    dp[0] = 0;
    dp[1] = unique[0] * count.get(unique[0]);

    // Step 4: Handle second damage value with conflict check
    if (unique[1] - unique[0] > 2) {
        // No conflict: safe to add both
        dp[2] = dp[1] + unique[1] * count.get(unique[1]);
    } else {
        // Conflict: choose max of taking current or skipping it
        dp[2] = Math.max(dp[1], unique[1] * count.get(unique[1]));
    }

    // Step 5: Fill DP table for remaining values
    for (let i = 3; i <= n; i++) {
        const curr = unique[i - 1];
        const prev1 = unique[i - 2];
        const prev2 = unique[i - 3];
        const currTotal = curr * count.get(curr);

        if ((curr - prev1 > 2) && (curr - prev2 > 2)) {
            // No conflict with previous two: safe to add
            dp[i] = dp[i - 1] + currTotal;
        } else if ((curr - prev2 > 2) && (curr - prev1 <= 2)) {
            // Conflict with prev1 only: choose max of skipping or taking with dp[i-2]
            dp[i] = Math.max(dp[i - 2] + currTotal, dp[i - 1]);
        } else {
            // Conflict with both prev1 and prev2: choose max of skipping or taking with dp[i-3]
            dp[i] = Math.max(dp[i - 3] + currTotal, dp[i - 1]);
        }
    }

    // Final result: max damage using all unique values
    return dp[n];
};