O(n) time and O(n) space using hashmap(Counter)

Author: 149ps

1657. Determine if Two Strings Are Close/1657. Determine if Two Strings Are Close.py

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+"""
+Two strings are considered close if you can attain one from the other using the following operations:
+
+Operation 1: Swap any two existing characters.
+For example, abcde -> aecdb
+Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
+For example, aacabb -> bbcbaa (all a's turn into b's, and all b's turn into a's)
+You can use the operations on either string as many times as necessary.
+
+Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.
+
+
+
+Example 1:
+
+Input: word1 = "abc", word2 = "bca"
+Output: true
+Explanation: You can attain word2 from word1 in 2 operations.
+Apply Operation 1: "abc" -> "acb"
+Apply Operation 1: "acb" -> "bca"
+Example 2:
+
+Input: word1 = "a", word2 = "aa"
+Output: false
+Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.
+Example 3:
+
+Input: word1 = "cabbba", word2 = "abbccc"
+Output: true
+Explanation: You can attain word2 from word1 in 3 operations.
+Apply Operation 1: "cabbba" -> "caabbb"
+Apply Operation 2: "caabbb" -> "baaccc"
+Apply Operation 2: "baaccc" -> "abbccc"
+
+
+Constraints:
+
+1 <= word1.length, word2.length <= 105
+word1 and word2 contain only lowercase English letters.
+"""
+class Solution:
+    def closeStrings(self, word1: str, word2: str) -> bool:
+        count_word1 = collections.Counter(word1)
+        count_word2 = collections.Counter(word2)
+        counter1 = collections.Counter(count_word1.values())
+        counter2 = collections.Counter(count_word2.values())
+        return len(count_word1.keys() - count_word2.keys()) == 0  and len(counter1 - counter2) == 0