# 資訊

:::info
- Question: 3005. Count Element With Maximum Frequency
- From: Leetcode Daily Challenge 2024.03.08
- Difficulty: Easy
:::

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# 目錄

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[TOC]
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# 題目

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.

> Example 1:
:::success
- Input: `nums = [1,2,2,3,1,4]`
- Output: 4
- Explanation: 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.
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> Example 2:
:::success
- Input: `nums = [1,2,3,4,5]`
- Output: 5
- Explanation: 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.
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> Constraints:
:::success
- 1 <= `nums.length` <= 100
- 1 <= `nums[i]` <= 100
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# 解法

## 概念

以後看到 hash table 跟 counting 就要想到 Python 黑科技，記得以後真的使用前要寫上 `import collections` 這一行！

這題需要的大概就是紀錄頻率跟比大小，所以如果有能快速幫我找出 maximum 的對應的會比較好實作，這時候就想到了 python 的 collections，因為裡面有個功能是 counter，counter 基本上就是所有計數器該有的功能都先幫你包得好好的，然後用字典讓你做查詢、運算，這次就拿它來使用

* 第 3 行：創一個 counter
* 第 4 - 5 行：用 counter 紀錄元素出現頻率
* 第 6 行：利用 `counter.most_common()` 功能把出現頻率依照大小排序，回傳一個 list
* 第 7 - 12 行：尋找最大頻率的所有解，把他們加起來
* 第 13 行：回傳答案

## 程式碼

class Solution:
    def maxFrequencyElements(self, nums: List[int]) -> int:
        c = Counter()
        for e in nums:
            c[str(e)] += 1
        c = c.most_common()
        ans = c[0][1]
        for i in range( 1, len(c), 1 ):
            if c[i][1] != c[0][1]:
                break
            else:
                ans += c[i][1]
        return ans

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# 複雜度

## 時間複雜度

走訪過整條 list 紀錄頻率，所以是 $O(n)$


## 空間複雜度

創了 counter，後面又為了紀載 cost_common 而多了一個 list，算是 $O(2n) = O(n)$