1657. Determine if Two Strings Are Close

1. Description

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.

2. Example

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”

3. Constraints

• 1 <= word1.length, word2.length <= $10^5$
• word1 and word2 contain only lowercase English letters.

4. Solutions

Hash Table

n1 = word1.size(), n2 = word2.size()
Time complexity: O(n1 + n2)
Space complexity: O(1)

class Solution {
public:
    bool closeStrings(const string &word1, const string &word2) {
        if (word1.size() != word2.size()) {
            return false;
        }

        vector<int> letter_count1(alphabet_letters), letter_count2(alphabet_letters);
        for (auto letter : word1) {
            ++letter_count1[letter - 'a'];
        }
        for (auto letter : word2) {
            ++letter_count2[letter - 'a'];
        }

        for (int i = 0; i < alphabet_letters; ++i) {
            if ((letter_count1[i] == 0) ^ (letter_count2[i] == 0)) {
                return false;
            }
        }

        sort(letter_count1.begin(), letter_count1.end());
        sort(letter_count2.begin(), letter_count2.end());
        return letter_count1 == letter_count2;
    }

private:
    const int alphabet_letters = 26;
};