Frequency Of The Most Frequent Element - Leetcode Solution

Problem Statement:

Given an array of integers nums and an integer k, return the maximum frequency of any element in the array.

Example:

Input: nums = [1,2,4], k = 5 Output: 1 Explanation: There is no element in nums that appears k times or more.

Solution:

To solve this problem, we need to find the maximum frequency of any element in the given array. To do this, we can follow these steps:

1. Initialize a dictionary "count" to keep track of the frequency of each element in the array.
2. Iterate through the array "nums" and update the count of each element in the dictionary "count".
3. Initialize two variables "max_count" and "max_element" to keep track of the maximum frequency of any element and the element which has the maximum frequency respectively.
4. Iterate through the dictionary "count" and if the frequency of any element is greater than "max_count", update the value of "max_count" and "max_element" accordingly.
5. Return the value of "max_count".

Here is the Python code to implement the above solution:

def maxFrequency(nums: List[int], k: int) -> int:

count = {}

for n in nums:
    count[n] = count.get(n, 0) + 1
    
max_count, max_element = 0, 0

for key, value in count.items():
    if value > max_count:
        max_count, max_element = value, key

return max_count

The time complexity of this code is O(n), where n is the length of the input array "nums". The space complexity is also O(n) because in the worst case, all elements of the array will have different values.

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