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;
    }
}