Groups of Special-Equivalent Strings

Medium

Asked by:

8 views

Topics:

ArraysStrings

You are given an array of strings of the same length words.

In one move, you can swap any two even indexed characters or any two odd indexed characters of a string words[i].

Two strings words[i] and words[j] are special-equivalent if after any number of moves, words[i] == words[j].

For example, words[i] = "zzxy" and words[j] = "xyzz" are special-equivalent because we may make the moves "zzxy" -> "xzzy" -> "xyzz".

A group of special-equivalent strings from words is a non-empty subset of words such that:

Every pair of strings in the group are special equivalent, and
The group is the largest size possible (i.e., there is not a string words[i] not in the group such that words[i] is special-equivalent to every string in the group).

Return the number of groups of special-equivalent strings from words.

Example 1:

Input: words = ["abcd","cdab","cbad","xyzz","zzxy","zzyx"]
Output: 3
Explanation: 
One group is ["abcd", "cdab", "cbad"], since they are all pairwise special equivalent, and none of the other strings is all pairwise special equivalent to these.
The other two groups are ["xyzz", "zzxy"] and ["zzyx"].
Note that in particular, "zzxy" is not special equivalent to "zzyx".

Example 2:

Input: words = ["abc","acb","bac","bca","cab","cba"]
Output: 3

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 20
words[i] consist of lowercase English letters.
All the strings are of the same length.

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:

What are the possible characters within the strings, and are they case-sensitive?
Can the input array `A` be empty or contain null/empty strings?
Are there any length constraints on the strings within the input array?
If two strings are special-equivalent, but are present multiple times in the input, should those duplicates be counted as distinct groups?
What is the expected behavior if any of the strings in the input are null or invalid (e.g., contain characters outside the expected range)?

Brute Force Solution

Approach

We want to find how many groups of words are considered 'special-equivalent'. A brute force method means we'll check every possible pair of words to see if they belong to the same group. We'll keep track of the unique groups we find along the way.

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

Start with the first word in the list.
Compare this first word to every other word in the list.
For each comparison, check if the two words are 'special-equivalent'. This means their characters at even positions are the same if rearranged, and their characters at odd positions are the same if rearranged.
If they are special-equivalent, consider them part of the same group as the first word.
Move to the next word in the list that hasn't been assigned to a group yet.
Repeat the process of comparing this word to every other word, checking for special-equivalence.
If a word is special-equivalent to the current word, they belong to the same group.
Continue until every word in the list has been assigned to a group.
Count the number of unique groups you've found. This is your answer.

Code Implementation

def groups_of_special_equivalent_strings(words):
    number_of_words = len(words)
    groups = []
    word_group_assignments = [None] * number_of_words

    for first_word_index in range(number_of_words):
        if word_group_assignments[first_word_index] is not None:
            continue

        # Create a new group for the current word
        groups.append([words[first_word_index]])
        word_group_assignments[first_word_index] = len(groups) - 1

        for second_word_index in range(first_word_index + 1, number_of_words):
            if word_group_assignments[second_word_index] is not None:
                continue

            first_word = words[first_word_index]
            second_word = words[second_word_index]

            if is_special_equivalent(first_word, second_word):
                # Assign second word to first word's group
                groups[-1].append(words[second_word_index])
                word_group_assignments[second_word_index] = len(groups) - 1

    return len(groups)

def is_special_equivalent(first_word, second_word):
    if len(first_word) != len(second_word):
        return False

    first_word_even_characters = sorted(first_word[::2])
    first_word_odd_characters = sorted(first_word[1::2])
    second_word_even_characters = sorted(second_word[::2])
    second_word_odd_characters = sorted(second_word[1::2])

    # Check if even and odd characters are same when sorted
    if first_word_even_characters != second_word_even_characters:
        return False

    if first_word_odd_characters != second_word_odd_characters:
        return False

    return True

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n strings in the input list. For each string, it compares it to every other string in the list to check for special-equivalence. The special-equivalence check itself involves sorting the characters at even and odd indices, which takes O(1) time since the string length is limited to 20 (a constant). Therefore, the dominant factor is comparing each of the n strings to the remaining n-1 strings, leading to approximately n * (n-1) comparisons. This simplifies to O(n²).

Space Complexity

O(N) – The algorithm implicitly maintains a list of groups, where each word is assigned to a specific group. In the worst case, each word could belong to its own unique group. Therefore, we might need to store group assignments for each of the N words in the input list, requiring auxiliary space proportional to N. This leads to a space complexity of O(N).

Optimal Solution

Approach

The goal is to count groups of strings that are 'special-equivalent'. Two strings are special-equivalent if, after sorting the characters at even positions and the characters at odd positions, the resulting strings are identical. To efficiently solve this, we need to come up with a standard way to represent each string so that special-equivalent strings have the same representation, allowing us to easily count the unique groups.

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

For each string, separate the characters at even positions from those at odd positions.
Sort the characters at even positions alphabetically, and then separately sort the characters at odd positions alphabetically.
Combine the sorted even characters and the sorted odd characters back into a single string, forming a unique 'signature' for that string. This signature represents the special-equivalence class of the original string.
Store each signature we generate in a collection (like a set). This ensures we only keep track of unique special-equivalence classes.
After processing all the strings, the number of unique signatures in our collection represents the number of groups of special-equivalent strings.

Code Implementation

def groups_of_special_equivalent_strings(words):
    signatures = set()

    for word in words:
        even_characters = sorted(word[::2])
        odd_characters = sorted(word[1::2])

        # Create a unique signature
        signature = "".join(even_characters) + "".join(odd_characters)

        # Only store unique signatures
        signatures.add(signature)

    # The number of unique signatures represents the number of groups
    return len(signatures)

Big(O) Analysis

Time Complexity

O(n*m log m) – Let n be the number of strings in the input array and m be the maximum length of any string. For each of the n strings, we separate the characters into even and odd positions, which takes O(m) time. Sorting the even and odd characters each takes O(m log m) time. Combining the sorted characters takes O(m) time. Since we perform these operations for each of the n strings, the overall time complexity is dominated by the sorting step for each string repeated for all n strings, resulting in O(n*m log m).

Space Complexity

O(N) – The primary auxiliary space usage comes from the set used to store the unique signatures. In the worst-case scenario, all N strings are special-equivalent to each other, resulting in N unique signatures being stored in the set. Additionally, each signature, which is derived from sorting even and odd positioned characters, can be as long as the original string, potentially requiring storage proportional to the string's length. Therefore, the space complexity is dominated by the size of the set, which can grow up to N, where N is the number of strings in the input.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 as there are no strings to form groups.
Array containing an empty string	An empty string should be considered special-equivalent to itself and other empty strings.
Array containing only one string	Return 1 as a single string forms one group.
Strings with different lengths	Strings of different lengths cannot be special-equivalent, so they should always be in different groups.
Strings with a large length (scalability)	Sorting character counts should be done efficiently (e.g., using a fixed-size array) to avoid performance bottlenecks.
Strings with very long sequences of the same character at even or odd indices	The character count method must correctly handle strings with long repeated sequences, not overflowing or miscounting.
Large input array size	Use appropriate data structures such as hash sets to ensure efficient comparisons and prevent time limit exceeded errors.
Array containing strings with only special characters	The program should handle non-alphanumeric special characters when doing character counting.

Null or empty input array

How to Handle:

Return 0 as there are no strings to form groups.

Array containing an empty string

How to Handle:

An empty string should be considered special-equivalent to itself and other empty strings.

Array containing only one string

How to Handle:

Return 1 as a single string forms one group.

Strings with different lengths

How to Handle:

Strings of different lengths cannot be special-equivalent, so they should always be in different groups.

Strings with a large length (scalability)

How to Handle:

Sorting character counts should be done efficiently (e.g., using a fixed-size array) to avoid performance bottlenecks.

Strings with very long sequences of the same character at even or odd indices

How to Handle:

The character count method must correctly handle strings with long repeated sequences, not overflowing or miscounting.

Large input array size

How to Handle:

Use appropriate data structures such as hash sets to ensure efficient comparisons and prevent time limit exceeded errors.

Array containing strings with only special characters

How to Handle:

The program should handle non-alphanumeric special characters when doing character counting.