LeetCode Challenge Day 7 — 3005. Count Elements With Maximum Frequency

Hey folks

This is Day 7 of my LeetCode streak.
Even though it’s Monday and things are already busy, I promised myself I’d keep this streak alive for 199 days straight — so here’s today’s grind 💪.

Today’s problem is 3005. Count Elements With Maximum Frequency — a neat little frequency-counting puzzle. It’s not about system design like yesterday’s beast, but rather about practicing clean hash-map logic and keeping things simple.

📌 Problem Statement

You’re given an array nums consisting of positive integers.

• The frequency of an element is the number of times it appears.

• First, find the maximum frequency among all numbers.

• Then return the total number of elements in the array that have this maximum frequency.

Examples:

• Input: [1,2,2,3,1,4]
  Output: 4
  Explanation: 1 and 2 appear twice each → total = 2 + 2 = 4.

• Input: [1,2,3,4,5]
  Output: 5
  Explanation: All numbers appear once (max frequency = 1) → total = 5.

Constraints:

• 1 <= nums.length <= 100

• 1 <= nums[i] <= 100

💡 Intuition

This is a frequency map problem.
We need to:

1. Count how many times each number appears.

2. Find the largest count.

3. Add up the counts of all numbers that match this largest count.

It’s just Count → Find Max → Sum.

🔑 Approach

1. Use a Map (hash table) to store the frequency of each number.

2. Loop through the map to find the maximum frequency.

3. Loop again and sum all counts equal to that maximum.

4. Return the sum.

That’s it — no tricks needed here.

⏱️ Complexity Analysis

• Time complexity: O(n) (one pass to count + small passes over the map).

• Space complexity: O(k) (k = number of unique elements, at most 100).

🧑‍💻 Code (JavaScript)

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxFrequencyElements = function(nums) {
    // Step 1: Count frequencies
    const freq = new Map();
    for (const x of nums) {
        freq.set(x, (freq.get(x) || 0) + 1);
    }

    // Step 2: Find the maximum frequency
    let maxCount = 0;
    for (const count of freq.values()) {
        if (count > maxCount) maxCount = count;
    }

    // Step 3: Add up all counts that match the max frequency
    let total = 0;
    for (const count of freq.values()) {
        if (count === maxCount) total += count;
    }

    return total;
};

// ✅ Quick tests
console.log(maxFrequencyElements([1,2,2,3,1,4])); // 4
console.log(maxFrequencyElements([1,2,3,4,5]));   // 5
console.log(maxFrequencyElements([7,7,7]));       // 3
console.log(maxFrequencyElements([1]));           // 1

🧪 Edge Cases

• Array length = 1 → just return 1.

• All numbers unique → max frequency = 1 → return nums.length.

• All numbers same → max frequency = n → return n.

🎥 Video walkthrough

I recorded a short walkthrough explaining the idea and a step-by-step run of the example. Watch it here (or the embedded player below):

YouTube: [https://youtu.be/qyptT7sp_gg]

🎥 Reflections

After yesterday’s heavy system design style problem, today felt like a quick breather. It reminded me why simple hash map questions are so important — they train your fundamentals and force you to keep solutions clean and efficient.

That’s it for Day 7 of my LeetCode journey!
See you tomorrow for Day 8 🚀.

Happy Coding 👨‍💻