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Problem

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 <= 10^5
  • word1 and word2 contain only lowercase English letters.

Code

  1. Every character that exists in word1 must exist in word2
  2. The frequency of all the characters is always the same
class Solution {
    public boolean closeStrings(String word1, String word2) {
        if(word1.length() != word2.length()) return false;

        Map<Character, Integer> map1 = new HashMap<>();
        Map<Character, Integer> map2 = new HashMap<>();

        for (char c : word1.toCharArray()) {
            map1.put(c, map1.getOrDefault(c, 0) + 1);
        }

        for (char c : word2.toCharArray()) {
            map2.put(c, map2.getOrDefault(c, 0) + 1);
        }

        if (!map1.keySet().equals(map2.keySet())) {
            return false;
        }

        List<Integer> freq1 = new ArrayList<>(map1.values());
        List<Integer> freq2 = new ArrayList<>(map2.values());

        Collections.sort(freq1);
        Collections.sort(freq2);

        return freq1.equals(freq2);
    }
}