Own Code:

1. create dictionary which counts the occurrence of every characters.
2. Check if there is a character where another dictionary don't have
3. sort the values and check if all the values are same

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        if len(word1) != len(word2): return False

        dict_word1, dict_word2 = {}, {}
        for item in word1:
            dict_word1[item] = dict_word1.get(item, 0) + 1
        for item in word2:
            dict_word2[item] = dict_word2.get(item, 0) + 1
            if not dict_word1.get(item):
                return False

        values_word1 = sorted(list(dict_word1.values()))
        values_word2 = sorted(list(dict_word2.values()))

        index = 0
        while index < len(values_word1):
            if values_word1[index] != values_word2[index]:
                return False
            index += 1
        return True

Better Code:
那么我们可以手动统计每个字符出现的次数，并比较它们的频次集合是否相同。同时，我们还需要比较两个单词的字符集合是否相同。

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        if len(word1) != len(word2):
            return False
        
        # 统计字符频次
        freq1 = {}
        freq2 = {}
        for char in word1:
            freq1[char] = freq1.get(char, 0) + 1
        for char in word2:
            freq2[char] = freq2.get(char, 0) + 1
        
        # 检查字符集合是否相同
        if set(freq1.keys()) != set(freq2.keys()):
            return False
        
        # 检查频次列表是否相同
        return sorted(freq1.values()) == sorted(freq2.values())

If you use Counter, running time will decrease, but the main idea is same.

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        counter_word1 = Counter(word1)
        counter_word2 = Counter(word2)

        return counter_word1.keys() == counter_word2.keys() and sorted(counter_word1.values()) == sorted(counter_word2.values())