Determine if Two Strings Are Close

Medium

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Strings

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"

Constraints:

1 <= word1.length, word2.length <= 10⁵
word1 and word2 contain only lowercase English letters.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input strings, word1 and word2, be empty or null?
Are the input strings case-sensitive?
Are there any limitations on the character set used in the input strings (e.g., only lowercase English letters)?
If the strings have different lengths, can they ever be considered close?
Are there any constraints on the length of the input strings?

Brute Force Solution

Approach

The brute force approach for this problem is like trying to rearrange the letters in one string to see if we can make it look like the other. We check all possible rearrangements and transformations to see if the strings can become identical. It’s like exploring every path to see if any leads to the same destination.

Here's how the algorithm would work step-by-step:

First, check if the strings even have the same number of letters. If not, there's no point in continuing.
Next, see if both strings use the exact same set of letters, regardless of how many times each letter appears. If they don't, they can't be 'close'.
Now, consider that you can rearrange the letters in the first string in tons of different ways. Try every single one of those rearrangements.
For each of these rearranged versions of the first string, compare it to the second string.
Also, you can change the number of times a letter appears. For example, if one string has 3 'A's and the other has 5 'B's, you can swap those counts.
Try swapping the counts of each letter in the rearranged first string with the counts of the letters in the second string, looking at every possible combination of swaps.
If, after any of these rearrangements and count swaps, the first string becomes exactly the same as the second string, then the strings are considered 'close'.
If you've tried every possible rearrangement and count swap and still haven't found a match, then the strings are not 'close'.

Code Implementation

def are_strings_close_brute_force(string1, string2):
    if len(string1) != len(string2):
        return False

    string1_characters = set(string1)
    string2_characters = set(string2)

    if string1_characters != string2_characters:
        return False

    import itertools

    for permutation in itertools.permutations(string1):
        rearranged_string = ''.join(permutation)

        # Create frequency maps for comparison
        string1_frequency = {}
        string2_frequency = {}
        for character in rearranged_string:
            string1_frequency[character] = string1_frequency.get(character, 0) + 1
        for character in string2:
            string2_frequency[character] = string2_frequency.get(character, 0) + 1

        string1_counts = sorted(string1_frequency.values())
        string2_counts = sorted(string2_frequency.values())

        # Check if the character counts are the same after rearrangement
        if string1_counts == string2_counts:
            return True

        # Trying count swaps for all permutations of count_indices in string2
        for count_indices in itertools.permutations(range(len(string2_counts))):
            swapped_string1_counts = []
            for count_index in count_indices:
                swapped_string1_counts.append(string2_counts[count_index])

            swapped_string1_counts.sort()

            # Critical check if swapped counts are equal
            if swapped_string1_counts == string1_counts:
                return True

    # Necessary as brute force checks all possibilities
    return False

Big(O) Analysis

Time Complexity

O(∑(n! * n!)) – The most significant cost comes from generating all possible rearrangements of the first string which is n! where n is the length of the string. For each rearrangement, we compare it to the second string, and then consider all possible swaps of character counts, which, in the worst case, is another n! operation for each rearrangement. Since these operations are nested, the total time complexity involves summing the product of the factorials representing the rearrangements and count swaps. This means that the brute-force approach requires factorial time that increases extremely quickly. Therefore, the time complexity can be approximated as O(∑(n! * n!)).

Space Complexity

O(1) – The provided plain English explanation primarily focuses on rearrangement and comparison operations, implying in-place manipulations or simple variable usage. While it speaks of tracking letter counts and swapping them, it doesn't specify data structures that scale with input string length. Therefore, assuming only a fixed number of variables are used to store intermediate counts or indices, the auxiliary space is constant and doesn't depend on the input size N (where N is the length of the input strings).

Optimal Solution

Approach

To determine if two strings are close, we need to check if we can transform one string into the other using two operations: swapping any two characters, or swapping the counts of any two characters. We don't actually need to perform the operations, just check if it's possible.

Here's how the algorithm would work step-by-step:

First, make sure both strings have the same set of characters. If one string has a character that the other doesn't, they can't be close.
Next, count how many times each character appears in both strings.
Then, compare the counts of each character. Even if the characters themselves are different, if the sorted lists of character counts are identical, then the strings are considered close. This is because we can rearrange characters to match counts.

Code Implementation

def areAlmostEquivalent(string1, string2):
    # Check if the strings have the same character set
    if set(string1) != set(string2):
        return False

    string1CharacterCounts = {}
    string2CharacterCounts = {}

    for char in string1:
        string1CharacterCounts[char] = string1CharacterCounts.get(char, 0) + 1

    for char in string2:
        string2CharacterCounts[char] = string2CharacterCounts.get(char, 0) + 1

    # Sort the character counts to compare distributions
    sortedString1Counts = sorted(string1CharacterCounts.values())

    # Sort the character counts to compare distributions
    sortedString2Counts = sorted(string2CharacterCounts.values())

    # Strings are close if the sorted counts are the same
    return sortedString1Counts == sortedString2Counts

Big(O) Analysis

Time Complexity

O(n) – We iterate through each string of length n once to check if they have the same set of characters using a hash set or similar data structure. Then, we iterate through each string again to count character frequencies, which is also O(n). Finally, we sort the frequency counts, which takes O(k log k) time, where k is the number of unique characters. Since k is bounded by the size of the alphabet (a constant), sorting character counts effectively becomes O(1) in the context of overall time complexity. Because of this, the dominant operations are the initial iterations through both strings resulting in O(n).

Space Complexity

O(1) – The algorithm uses two hash maps (or arrays) to store character counts for each string, and these hash maps can have at most 26 entries each (for each letter in the English alphabet). The sorted lists of character counts also require space proportional to the number of unique characters, which is again bounded by the constant 26. Therefore, the auxiliary space required remains constant, regardless of the input string length, and is independent of the length N of the input strings.

Edge Cases

Case	How to Handle
word1 or word2 is null or empty	Return true if both are null or empty, return false if only one is null or empty, as they cannot be transformed into each other.
word1 and word2 have different lengths	Return false immediately because if they are different lengths, you can't transform one into the other.
word1 and word2 are identical	Return true, because no operations are required to transform word1 into word2.
word1 and word2 have the same characters but in different frequencies (e.g., 'abc' and 'abb')	Check if the frequency counts of the characters can be rearranged to be identical; if not, return false.
word1 contains a character that word2 doesn't (or vice versa)	Return false immediately; Operation 2 doesn't create new characters, it only transforms existing ones.
word1 and word2 consist of only one unique character repeated multiple times (e.g., 'aaaa' and 'bbbb')	Return true as all characters can be transformed with operation 2.
Maximum length strings (stress test for performance)	Ensure the solution uses efficient data structures like HashMaps to achieve O(n) time complexity, preventing timeouts.
Large character sets (e.g., extended ASCII or Unicode)	Ensure the chosen character counting data structure (HashMap or array) is large enough to accommodate the entire character set.

word1 or word2 is null or empty

How to Handle:

Return true if both are null or empty, return false if only one is null or empty, as they cannot be transformed into each other.

word1 and word2 have different lengths

How to Handle:

Return false immediately because if they are different lengths, you can't transform one into the other.

word1 and word2 are identical

How to Handle:

Return true, because no operations are required to transform word1 into word2.

word1 and word2 have the same characters but in different frequencies (e.g., 'abc' and 'abb')

How to Handle:

Check if the frequency counts of the characters can be rearranged to be identical; if not, return false.

word1 contains a character that word2 doesn't (or vice versa)

How to Handle:

Return false immediately; Operation 2 doesn't create new characters, it only transforms existing ones.

word1 and word2 consist of only one unique character repeated multiple times (e.g., 'aaaa' and 'bbbb')

How to Handle:

Return true as all characters can be transformed with operation 2.

Maximum length strings (stress test for performance)

How to Handle:

Ensure the solution uses efficient data structures like HashMaps to achieve O(n) time complexity, preventing timeouts.

Large character sets (e.g., extended ASCII or Unicode)

How to Handle:

Ensure the chosen character counting data structure (HashMap or array) is large enough to accommodate the entire character set.