Problem

Solution

Intuition

This problem requires us to find the total number of elements which have the maximum frequency in a list. For example, if a list contains [1, 1, 2, 2, 3, 4], the answer would be 4 as there are 2 1’s and 2 2’s which is the maximum frequency. There are thus, 4 total elements that are part of the numbers that have maximum frequency.

To solve this problem we utilize a hashmap to store the frequencies of each element and then find the total unique number of elements which have a maximal frequency. We then multiply the unique number of elements with maximal frequency by the maximal frequency for the answer.

Approach

1. Either iterate through nums to populate a hashmap of frequencies called counts with the unique element as the key and the frequency as the value, or utilize Counter library.
2. Initialize max_freq and uniq_elem variables to store the maximum found frequency thus far and the number of unique elements with this frequency thus far.
3. Iterate through our hashmap to count max_freq and uniq_elem. We reset uniq_elem when we find a new higher frequency.
4. Return the product of max_freq and uniq_elem for the total number of elements that have maximal frequency.

Complexity

• Time complexity: $O(n)$. We iterate through the array to populate our hashmap and then iterate through the hashmap to check the number of unique elements with maximal frequency. These iterations both take linear time.

• Space complexity: $O(n)$. We utilize a hashmap for counting frequencies which takes linear space complexity.

Code

class Solution:
    def maxFrequencyElements(self, nums: List[int]) -> int:
        counts = Counter(nums)
        
        max_freq, uniq_elem = 0, 0

        for count in counts:
            if counts[count] == max_freq:
                uniq_elem += 1
            elif counts[count] > max_freq:
                max_freq = counts[count]
                uniq_elem = 1
        
        return max_freq * uniq_elem