计算字谜检查器功能的整体时间复杂度

Question

我具有以下代码库，用于检查单词列表（函数的第二个参数）中特定单词（函数的第一个参数）的字谜：

def anagram_checker(word, words=None):

    anagrams = []

    if not words:  # if the second arg list is empty then there is nothing to check
        return anagrams
    else:
        counter_word = Counter(word)
        for given_word in words:
            if given_word != word:  # Cannot be same as the word in question
                if Counter(given_word) == counter_word:
                    anagrams.append(given_word)
    return anagrams

因此，如果让我找到函数的时间复杂度，我将像这样：

1. Creating a counter dict for my first argument : O(m) , where m is the length of my first word. 

2. Looping through the list of words given and creating counters for each of them is O(n*k) , where n is the number of words in my list and let K be the average length of each word.

3. Comparing two counter objects is O(m).

So if someone asks me to tell them the overall time complexity for me it would O(2m + n*k)

那我怎么说呢？是O（n）还是O（n ^ 2）？

也请告诉我在某处计算复杂度时是否出错。

我在计算复杂度方面非常虚弱，因此非常欢迎任何详细的解释。

谢谢。