1657. Determine if Two Strings Are Close

[songyy5517]

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"

Intuition The problem is to determine whether two strings are close. From the description, we know there are two conditions to determine whether two strings are close. For the first one, we can infer that close strings are those which have the same character set and the same occurence of each character. For the other case, close strings must have the same character set and occurence set, but the mapping between the character set and the ocurrence set can be different. Therefore, we can use two hashmaps to record the characters and their corresponding ocurrences in each string. If the key set and the value set of two hashmaps contain same elements, then the two strings are close; vice versa.

[songyy5517]

Approach: HashMap & Comparison of keySet & values

1. Exception Handling;
2. Define two hashmaps;
3. For each string, record every character and its corresponding occurences;
4. Compare the key set and the value set of two hashmaps;
   • keyset: hm1.keySet().equals(hm2.keySet());
   • values: values -> array -> sort -> Arrays.euqals.
5. If they are equal, then return true; vice versa.

Complexity Analysis

• Time complexity: $O(N)$ , $N$ is the length of both strings, we iterate over two strings when constructing two hashmaps, which takes $O(N)$ time; The sort of the value set of two hashmaps costs $O(26\log 26)$ time.
• Space complexity: $O(1)$ , the maximum size of the hashmap is $O(26) = O(1)$ .

[songyy5517]

import java.util.*;
class Solution {
    public boolean closeStrings(String word1, String word2) {
        // Intuition: HashMap, comparision of keySet & values.
        // 1. Exception handling
        if (word1 == null || word2 == null || word1.length() != word2.length())
            return false;

        // 2. Define HashMap
        HashMap<Character, Integer> hm1 = new HashMap();
        HashMap<Character, Integer> hm2 = new HashMap();

        for (int i = 0; i < word1.length(); i++){
            hm1.put(word1.charAt(i), hm1.getOrDefault(word1.charAt(i), 0) + 1);
            hm2.put(word2.charAt(i), hm2.getOrDefault(word2.charAt(i), 0) + 1);
        }

        // 3. Compare keySet and values of two HashMap
        // values -> Interger[] -> Sort -> Arrays.equal
        Object[] hm1_ = hm1.values().toArray();
        Integer[] hm2_ = hm2.values().toArray(new Integer[hm2.size()]);
        Arrays.sort(hm1_);
        Arrays.sort(hm2_);

        return hm1.keySet().equals(hm2.keySet()) && Arrays.equals(hm1_, hm2_);        
    }
}

[songyy5517]

2024/5/17

• Comparison of keyset: hm1.keySet().equals(hm2.keySet());
• Comparison of values: values() -> array (Wrapper class) -> sort -> arrays.equals
• hashmap. values -> array: hm.values().toArray(); hm.values().toArray(new Integer[hm.size()]);