Determine if Two Strings Are Close

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.

Examples

Example 1:

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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:

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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:

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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"

Solution

Method 1 - Using map and sorting

To determine if two strings are close, we need to check

Both strings must have the same set of characters.
The frequency of each character must match, but the specific characters can be permuted.

So, here si the approach:

Check if both strings have the same length.
Count the frequency of each character in both strings.
Check if the sets of characters in both strings are identical.
Check if the sorted frequency lists of both strings are identical.

Code

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public class Solution {
    public boolean closeStrings(String word1, String word2) {
        if (word1.length() != word2.length()) {
            return false;
        }

        int[] count1 = new int[26];
        int[] count2 = new int[26];

        for (char c : word1.toCharArray()) {
            count1[c - 'a']++;
        }

        for (char c : word2.toCharArray()) {
            count2[c - 'a']++;
        }

        // Check if both strings have the same set of characters
        for (int i = 0; i < 26; i++) {
            if ((count1[i] == 0 && count2[i] != 0) || (count1[i] != 0 && count2[i] == 0)) {
                return false;
            }
        }

        Arrays.sort(count1);
        Arrays.sort(count2);

        return Arrays.equals(count1, count2);
    }
}

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class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        if len(word1) != len(word2):
            return False
        
        count1 = [0] * 26
        count2 = [0] * 26
        
        for ch in word1:
            count1[ord(ch) - ord('a')] += 1
        
        for ch in word2:
            count2[ord(ch) - ord('a')] += 1
        
        # Check if the sets of characters are identical
        for i in range(26):
            if (count1[i] == 0 and count2[i] != 0) or (count1[i] != 0 and count2[i] == 0):
                return False
        
        count1.sort()
        count2.sort()
        
        return count1 == count2

Complexity

⏰ Time complexity: O(n + m) where n is the length of word1 and m is the length of word2.
🧺 Space complexity: O(1) since we use fixed space for character counts (assuming a constant alphabet size of 26).

Problem#

Examples#

Solution#

Method 1 - Using map and sorting#

Code#

Complexity#