How do you solve Count Elements With Maximum Frequency on LeetCode?

Count Elements With Maximum Frequency (LeetCode #3005) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient frequency counting and retrieval.

What is the brute force solution for Count Elements With Maximum Frequency?

The brute force approach for Count Elements With Maximum Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will count the occurrences of each number in the array by iterating through the array multiple times. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Elements With Maximum Frequency?

Count Elements With Maximum Frequency uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Elements With Maximum Frequency?

Always consider edge cases, such as arrays where all elements are unique. Explain your thought process clearly while coding; it shows your understanding. Practice writing both brute force and optimal solutions to build problem-solving skills.

What are common mistakes when solving Count Elements With Maximum Frequency?

Not using a hash map and trying to count frequencies with nested loops, leading to inefficiency. Forgetting to check if the frequency of an element is equal to the maximum frequency when summing the total.

#3005

Count Elements With Maximum Frequency

Easy

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

In the optimal approach, we will use a hash map to count the frequencies of each element in a single pass. This allows us to efficiently find the maximum frequency and count the total occurrences of elements with that frequency.

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Algorithm

4 steps

1Step 1: Create a hash map to store the frequency of each element.
2Step 2: Iterate through the array and populate the hash map with frequencies.
3Step 3: Determine the maximum frequency from the hash map.
4Step 4: Sum the frequencies of all elements that have the maximum frequency and return the total.

solution.py10 lines

1# Full working Python code
2
3def count_max_frequency(nums):
4    from collections import Counter
5    freq_map = Counter(nums)
6    max_freq = max(freq_map.values())
7    return sum(count for count in freq_map.values() if count == max_freq)
8
9# Example usage
10print(count_max_frequency([1, 2, 2, 3, 1, 4]))

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Complexity note: The time complexity is O(n) because we only make a single pass through the array to count frequencies and another pass through the hash map to find the maximum frequency. The space complexity is O(n) due to the hash map storing frequencies.

1Using a hash map allows for efficient frequency counting and retrieval.
2The maximum frequency can be determined in a single pass after counting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.