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?
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:
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:
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
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:
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
Case | How to Handle |
---|---|
Null or empty input array | Return 0 if the input array is null or empty, as no elements exist. |
Array with only one element | The frequency of the single element is 1, which is also the maximum frequency, so return 1. |
Array with all elements identical | All 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 numbers | Hash map can handle negative numbers as keys, so no special handling is needed. |
Array with zeros | Zeros are treated as regular integers, so no special handling is required. |
Array containing Integer.MAX_VALUE or Integer.MIN_VALUE | The frequency count should not overflow for maximum and minimum integer values. |
Multiple elements sharing the same maximum frequency | The algorithm correctly counts all elements with the maximum frequency. |