1657 Determine if Two Strings Are Close

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"

Example 4:

Input: word1 = "cabbba", word2 = "aabbss"
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any amount of operations.

Constraints:

• 1 <= word1.length, word2.length <= 105

• word1 and word2 contain only lowercase English letters.

这题，看了题目以后，想了半天。在怀疑是不是用dp但又想不出递推式。后来看了hint发现，这个op1和op2都是障眼法。op1可以让任意字母互换，op2能让任意字母对应的频率互换。然后就想出了hashmap了。这里实现的时候需要注意一点，hashmap的values()方法返回的是collection，直接equals是不等的（即使同样位置同样数字），要用containsAll()来判断两个collection是不是互相包含。

public boolean closeStrings(String word1, String word2) {
    if (word1 == null && word2 ==  null) {
        return true;
    } else if (word2 == null || word1 == null) {
        return false;
    } if (word1.length() != word2.length()) {
        return false;
    }

    Map<Character, Integer> freq1 = new HashMap<>();
    Map<Character, Integer> freq2 = new HashMap<>();

    for (char c : word1.toCharArray()) {
        freq1.put(c, freq1.getOrDefault(c, 0) + 1);
    }

    for (char c : word2.toCharArray()) {
        freq2.put(c, freq2.getOrDefault(c, 0) + 1);
    }

    if (freq1.keySet().equals(freq2.keySet()) 
     && freq1.values().containsAll(freq2.values()) 
     && freq2.values().containsAll(freq1.values())) {
        return true;
    } else {
        return false;
    }