Problem Description

You are given an array nums consisting of positive integers.

Your task is to find all elements that appear with the maximum frequency in the array, and return the sum of their frequencies.

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

For example:

• If nums = [1, 2, 2, 3, 3], the frequencies are: 1 appears once, 2 appears twice, and 3 appears twice. The maximum frequency is 2. Both 2 and 3 have this maximum frequency, so the total frequencies would be 2 + 2 = 4.
• If nums = [1, 2, 3, 4, 5], each element appears once (frequency of 1). Since all elements have the maximum frequency of 1, the total would be 1 + 1 + 1 + 1 + 1 = 5.

The solution uses a Counter to track the frequency of each element, finds the maximum frequency value mx, and then sums up all frequencies that equal mx.

Intuition

To solve this problem, we need to identify which elements appear most frequently and then count their total occurrences.

The key insight is that we're looking for a two-step process:

1. First, determine what the maximum frequency is across all elements
2. Then, sum up the frequencies of all elements that have this maximum frequency

Think of it this way: if we have nums = [1, 2, 2, 3, 3, 3], we first count how many times each number appears:

• 1 appears 1 time
• 2 appears 2 times
• 3 appears 3 times

The maximum frequency here is 3. Only one element (3) has this maximum frequency, so our answer is simply 3.

But if we had nums = [1, 2, 2, 3, 3], both 2 and 3 appear 2 times (the maximum frequency). So we need to add up their frequencies: 2 + 2 = 4.

This naturally leads us to use a counting structure (like Python's Counter) to track frequencies, find the maximum value among all frequencies, and then sum up all frequency values that equal this maximum. The elegance of this approach is that it handles all cases uniformly - whether one element or multiple elements share the maximum frequency.

Solution Approach

The solution uses a counting approach with a hash table to efficiently track element frequencies.

Step 1: Count frequencies of all elements

We use Python's Counter from the collections module to create a frequency map:

cnt = Counter(nums)

This creates a dictionary-like object where keys are the elements from nums and values are their frequencies. For example, if nums = [1, 2, 2, 3, 3, 3], then cnt would be {1: 1, 2: 2, 3: 3}.

Step 2: Find the maximum frequency

We extract all frequency values and find the maximum:

mx = max(cnt.values())

The cnt.values() gives us all the frequency counts, and max() finds the highest frequency among them.

Step 3: Sum frequencies equal to maximum

We iterate through all frequency values and sum up those that equal the maximum frequency:

return sum(x for x in cnt.values() if x == mx)

This generator expression filters for frequencies equal to mx and sums them up.

Time Complexity: O(n) where n is the length of the input array, as we traverse the array once to count frequencies and then traverse the frequency values (at most n unique elements).

Space Complexity: O(n) in the worst case when all elements are unique, requiring space to store each element's count in the hash table.

Example Walkthrough

Let's walk through the solution with nums = [4, 7, 7, 1, 1, 7].

Step 1: Count frequencies We create a frequency map using Counter(nums):

• 4 appears 1 time
• 7 appears 3 times
• 1 appears 2 times

So cnt = {4: 1, 7: 3, 1: 2}

Step 2: Find maximum frequency We get all frequency values: [1, 3, 2] The maximum frequency is mx = 3

Step 3: Sum frequencies equal to maximum We check each frequency value:

• Is 1 == 3? No, skip
• Is 3 == 3? Yes, include in sum
• Is 2 == 3? No, skip

Only the frequency 3 (from element 7) equals the maximum, so our sum is 3.

Result: The function returns 3.

Let's trace through one more example where multiple elements share the maximum frequency: nums = [5, 5, 8, 8, 9]

• Count: {5: 2, 8: 2, 9: 1}
• Maximum frequency: mx = 2
• Sum frequencies equal to max: 2 (from 5) + 2 (from 8) = 4

The function returns 4 because both 5 and 8 appear with the maximum frequency of 2.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because:

• Creating the Counter object requires iterating through all n elements once: O(n)
• Finding the maximum value among the frequencies using max(cnt.values()) takes O(k) where k is the number of unique elements, and k ≤ n, so this is O(n) in the worst case
• The final sum operation iterates through all unique elements' frequencies, which is again O(k) where k ≤ n, giving us O(n)
• Overall: O(n) + O(n) + O(n) = O(n)

The space complexity is O(n), where n is the length of the array nums. This is because:

• The Counter object cnt stores at most n key-value pairs (in the worst case where all elements are unique), requiring O(n) space
• The generator expression in the sum creates values on-the-fly without storing them, so it uses O(1) additional space
• Overall: O(n) + O(1) = O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Empty Array Handling

Pitfall: The code will raise a ValueError when nums is empty because max() cannot operate on an empty sequence.

# This will fail if nums = []
max_frequency = max(frequency_counter.values())  # ValueError: max() arg is an empty sequence

Solution: Add a check for empty input:

def maxFrequencyElements(self, nums: List[int]) -> int:
    if not nums:  # Handle empty array
        return 0
  
    frequency_counter = Counter(nums)
    max_frequency = max(frequency_counter.values())
    total_count = sum(freq for freq in frequency_counter.values() if freq == max_frequency)
    return total_count

2. Inefficient Double Iteration Over Values

Pitfall: The current solution iterates through frequency_counter.values() twice - once to find the maximum and once to sum matching frequencies. This could be optimized to a single pass.

Solution: Use a single iteration approach:

def maxFrequencyElements(self, nums: List[int]) -> int:
    if not nums:
        return 0
  
    frequency_counter = Counter(nums)
    frequencies = list(frequency_counter.values())
  
    max_frequency = frequencies[0]
    total_count = frequencies[0]
  
    for freq in frequencies[1:]:
        if freq > max_frequency:
            max_frequency = freq
            total_count = freq  # Reset count with new maximum
        elif freq == max_frequency:
            total_count += freq  # Add to count if equal to maximum
  
    return total_count

3. Misunderstanding the Problem Requirements

Pitfall: Some might mistakenly return just the count of distinct elements with maximum frequency rather than the sum of their frequencies.

# Wrong interpretation - returns count of distinct elements
return sum(1 for x in cnt.values() if x == mx)  # This would return 2 for [1,2,2,3,3]

# Correct - returns sum of frequencies
return sum(x for x in cnt.values() if x == mx)  # This returns 4 for [1,2,2,3,3]

4. Not Considering Integer Overflow (Language-Specific)

Pitfall: While Python handles arbitrary precision integers automatically, in other languages like Java or C++, summing large frequencies could cause integer overflow.

Solution for other languages: Use appropriate data types (e.g., long in Java, long long in C++) when dealing with potentially large sums.