1657. Determine if Two Strings Are Close

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# 題目敘述

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"

# 解題思路

# Solution

import java.util.Arrays;

class Solution {

    public boolean closeStrings(String word1, String word2) {

        int[] freq1 = new int[26];

        int[] freq2 = new int[26];

        for (char ch : word1.toCharArray()) {

            freq1[ch - 'a']++;

        }

        for (char ch : word2.toCharArray()) {

            freq2[ch - 'a']++;

        }

        for (int i = 0; i < 26; i++) {

            if ((freq1[i] == 0 && freq2[i] != 0) || (freq1[i] != 0 && freq2[i] == 0)) {

                return false;

            }

        }

        Arrays.sort(freq1);

        Arrays.sort(freq2);

        for (int i = 0; i < 26; i++) {

            if (freq1[i] != freq2[i]) {

                return false;

            }

        }

        return true;

    }

}