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Count Elements With Maximum Frequency

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Arrays

You are given an array nums consisting of positive integers. Return the total frequencies of elements in nums such that those elements all have the maximum frequency. The frequency of an element is the number of occurrences of that element in the array.

For example:

  • nums = [1, 2, 2, 3, 1, 4] The elements 1 and 2 have a frequency of 2 which is the maximum frequency in the array. So the number of elements in the array with maximum frequency is 4.
  • nums = [1, 2, 3, 4, 5] All elements of the array have a frequency of 1 which is the maximum. So the number of elements in the array with maximum frequency is 5.

Write a function to solve this problem. Consider edge cases such as when the array is empty, contains only one element, when all elements are the same, and when all elements are unique. What is the time and space complexity of your solution?

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What is the expected range of values within the integer array `nums`?
  2. Is it possible for the input array `nums` to be empty or null?
  3. If multiple numbers share the same maximum frequency, do I count each of those numbers?
  4. Are there any constraints on the size of the input array `nums`?
  5. Should I be concerned about integer overflow when calculating frequencies?

Brute Force Solution

Approach

The brute force approach to this problem involves checking the frequency of each number in the given group. We will count the occurrences of each number and then identify the numbers with the highest count.

Here's how the algorithm would work step-by-step:

  1. Go through the entire group of numbers, one number at a time.
  2. For each number, count how many times it appears in the group.
  3. After counting how many times each number appears, find the highest count (the maximum frequency).
  4. Finally, count how many numbers have that highest count. These are the numbers that occur the most frequently.

Code Implementation

def count_elements_with_maximum_frequency(number_group):
    number_frequencies = {}
    
    for number in number_group:
        if number in number_frequencies:
            number_frequencies[number] += 1
        else:
            number_frequencies[number] = 1

    # Determine the maximum frequency
    maximum_frequency = 0
    for number in number_frequencies:
        if number_frequencies[number] > maximum_frequency:
            maximum_frequency = number_frequencies[number]

    # Count how many numbers have the maximum frequency
    element_count_with_maximum_frequency = 0
    # Count each element with max frequency.
    for number in number_frequencies:
        if number_frequencies[number] == maximum_frequency:
            element_count_with_maximum_frequency += 1

    return element_count_with_maximum_frequency

Big(O) Analysis

Time Complexity
O(n²)The algorithm iterates through each of the n elements in the input array. For each of these n elements, it then iterates through the entire array again to count the frequency of that specific element. This results in a nested loop structure. Therefore, the total number of operations scales with n * n, or n squared. Hence, the time complexity is O(n²).
Space Complexity
O(1)The brute force approach, as described, involves iterating through the input array and counting the frequency of each number. It uses a variable to keep track of the current number's count and another variable to store the maximum frequency seen so far. These variables use a constant amount of space regardless of the input size N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

To efficiently count elements with the maximum frequency, we first need to determine how often each number appears. After identifying the most frequent count, we simply count how many numbers actually have that frequency.

Here's how the algorithm would work step-by-step:

  1. Go through the list of numbers and keep track of how many times each number appears. Think of this like making a tally for each unique number.
  2. Find the highest tally among all the numbers. This tells us the maximum frequency any number has.
  3. Go through the tallies again and count how many numbers have a tally that matches the maximum frequency you found earlier.
  4. The final count you get is the answer: the number of elements with the maximum frequency.

Code Implementation

def count_elements_with_max_frequency(number_list):
    element_counts = {}
    for number in number_list:
        element_counts[number] = element_counts.get(number, 0) + 1

    # Find the maximum frequency in the dictionary.
    max_frequency = 0
    for frequency in element_counts.values():
        if frequency > max_frequency:
            max_frequency = frequency

    # Count elements with the maximum frequency.
    elements_with_max_frequency_count = 0
    # Iterate to check against max frequency
    for frequency in element_counts.values():
        if frequency == max_frequency:
            elements_with_max_frequency_count += 1

    return elements_with_max_frequency_count

Big(O) Analysis

Time Complexity
O(n)The algorithm first iterates through the input array of size n to count the frequency of each number using a hash map or similar data structure. This step takes O(n) time. Then, it iterates through the frequency counts (at most n unique numbers) to find the maximum frequency, taking O(n) time. Finally, it iterates through the frequency counts again to count the elements with the maximum frequency, which also takes O(n) time. Since all operations are linear, the overall time complexity is O(n).
Space Complexity
O(N)The algorithm utilizes a data structure (implicitly described as tallies) to store the frequency of each number. In the worst-case scenario, where all N numbers in the input list are unique, this data structure will need to store N distinct frequencies. Thus, the auxiliary space used is proportional to the number of elements in the input, resulting in a space complexity of O(N).

Edge Cases

CaseHow to Handle
Null or empty input arrayReturn 0 if the input array is null or empty, as no elements exist.
Array with only one elementThe frequency of the single element is 1, which is also the maximum frequency, so return 1.
Array with all elements identicalAll elements will have the same maximum frequency, so return the length of the array.
Array with highly skewed frequency distribution (one element dominates)The algorithm should correctly identify the maximum frequency and count elements with that frequency regardless of the distribution.
Array with negative numbersHash map can handle negative numbers as keys, so no special handling is needed.
Array with zerosZeros are treated as regular integers, so no special handling is required.
Array containing Integer.MAX_VALUE or Integer.MIN_VALUEThe frequency count should not overflow for maximum and minimum integer values.
Multiple elements sharing the same maximum frequencyThe algorithm correctly counts all elements with the maximum frequency.