Count Pairs Of Similar Strings

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You are given a 0-indexed string array words.

Two strings are similar if they consist of the same characters.

For example, "abca" and "cba" are similar since both consist of characters 'a', 'b', and 'c'.
However, "abacba" and "bcfd" are not similar since they do not consist of the same characters.

Return the number of pairs (i, j) such that 0 <= i < j <= word.length - 1 and the two strings words[i] and words[j] are similar.

Example 1:

Input: words = ["aba","aabb","abcd","bac","aabc"]
Output: 2
Explanation: There are 2 pairs that satisfy the conditions:
- i = 0 and j = 1 : both words[0] and words[1] only consist of characters 'a' and 'b'. 
- i = 3 and j = 4 : both words[3] and words[4] only consist of characters 'a', 'b', and 'c'.

Example 2:

Input: words = ["aabb","ab","ba"]
Output: 3
Explanation: There are 3 pairs that satisfy the conditions:
- i = 0 and j = 1 : both words[0] and words[1] only consist of characters 'a' and 'b'. 
- i = 0 and j = 2 : both words[0] and words[2] only consist of characters 'a' and 'b'.
- i = 1 and j = 2 : both words[1] and words[2] only consist of characters 'a' and 'b'.

Example 3:

Input: words = ["nba","cba","dba"]
Output: 0
Explanation: Since there does not exist any pair that satisfies the conditions, we return 0.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 100
words[i] consist of 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:

What is the maximum length of the input array of strings, and what is the maximum length of each individual string?
Can the strings in the input array contain characters other than lowercase English letters?
If no similar pairs exist, should I return 0, or is there another specific value I should return?
Are two strings considered similar if they contain the same *set* of characters, regardless of frequency or order?
Could the input array be null or empty?

Brute Force Solution

Approach

The brute force approach involves directly comparing every possible pair of words to see if they are similar. We examine each combination exhaustively to determine if they meet the similarity criteria. This method guarantees finding all similar pairs, even if it's not the most efficient.

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

Take the first word from the list.
Compare that first word to every other word in the list, one at a time.
For each comparison, check if the two words have the same unique letters, irrespective of their count or order.
If the unique letters are the same, it means the pair of words are 'similar', increase a counter to keep track of total similar pairs.
After comparing the first word with all the other words, move to the second word in the list.
Compare the second word to every word after it in the list, (we skip words already considered since order doesn't matter).
Again, check if the two words being compared have the same unique letters.
If they do, increase the counter.
Continue this process, comparing each word to all the words that come after it, until you've gone through all the words in the list.
The final count will give you the total number of 'similar' pairs of words.

Code Implementation

def count_similar_string_pairs(words):
    number_of_similar_pairs = 0

    # Iterate through each word in the list
    for first_word_index in range(len(words)):
        # Compare the first word to all subsequent words
        for second_word_index in range(first_word_index + 1, len(words)):
            first_word = words[first_word_index]
            second_word = words[second_word_index]

            # Convert each word to a set of unique characters
            first_word_unique_characters = set(first_word)

            second_word_unique_characters = set(second_word)

            # Check if the sets of unique characters are equal
            if first_word_unique_characters == second_word_unique_characters:
                # Increment if words are similar
                number_of_similar_pairs += 1

    return number_of_similar_pairs

Big(O) Analysis

Time Complexity

O(n²) – The outer loop iterates through each of the n words in the input list. The inner loop compares the current word with the remaining words in the list, resulting in a nested loop structure. For each pair of words, determining if the unique letters are the same takes O(k) time, where k is the maximum length of a word. However, since the pair checking dominates the runtime and happens for roughly n * (n-1) / 2 pairs, this results in approximately n²/2 comparisons. Therefore, the overall time complexity is O(n²).

Space Complexity

O(1) – The provided brute force approach iterates through pairs of words, comparing them directly. It doesn't create any auxiliary data structures like lists, hash maps, or sets to store intermediate results or track visited words. The algorithm only uses a counter variable to keep track of similar pairs and potentially index variables to iterate through the word list, all of which occupy constant space regardless of the input size N (the number of words). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The key idea is to represent each word by the unique letters it contains. Then, we can easily check if two words are 'similar' by seeing if they have the same unique letters. We can count the pairs of similar words efficiently by using a way to quickly look up how many words have a particular set of unique letters.

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

For each word, identify the unique letters it has and represent it by this set of letters.
Use each unique set of letters as a key in a counter to track how many words have this set.
Once all words are processed, look at each unique set of letters and the count of words associated with it.
For each set, calculate how many pairs of words can be formed from the count of words having that particular set.
Sum up the number of pairs calculated from all unique sets of letters to find the total number of similar pairs.

Code Implementation

def count_similar_string_pairs(words):
    string_signature_counts = {}
    number_of_similar_pairs = 0

    for word in words:
        # Create a unique 'signature' for each word.
        string_signature = "".join(sorted(set(word)))

        if string_signature in string_signature_counts:
            string_signature_counts[string_signature] += 1
        else:
            string_signature_counts[string_signature] = 1

    # Iterate through counts of each unique string signature.
    for string_signature in string_signature_counts:
        count = string_signature_counts[string_signature]

        # Calculate pairs for each string signature.
        # This is nC2: n * (n - 1) / 2
        number_of_similar_pairs += count * (count - 1) // 2

    return number_of_similar_pairs

Big(O) Analysis

Time Complexity

O(n*m + k) – The algorithm iterates through each of the n words to identify the unique letters it contains. In the worst case, each word has m unique letters, so identifying the unique letters takes O(n*m). Then, it iterates through a counter of unique letter sets, where k represents the number of unique letter sets. For each unique set, it performs a constant time calculation. Thus, the total runtime is O(n*m + k). Since k <= n, we can simplify this to O(n*m), and if we assume m is small and constant relative to n, we can further simplify it to O(n). However if we want to keep generality, the time complexity is O(n*m + k).

Space Complexity

O(N) – The dominant space usage comes from the counter that stores the unique sets of letters as keys and their counts as values. In the worst case, if all N words in the input array 'words' have completely unique sets of letters, the counter will store N distinct key-value pairs. Therefore, the space required by the counter grows linearly with the number of words, N, in the input. This auxiliary space approximates to N, thus the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no pairs can exist.
Input array with only one string	Return 0 immediately because at least two strings are needed to form a pair.
All strings in the input array are identical	The number of pairs should be n * (n - 1) / 2, calculate and return.
Strings with very long lengths (potential performance bottleneck)	Ensure that the character set generation and comparison operations are optimized for long strings.
Input array contains duplicate strings that become same after character set reduction	The counting logic should correctly account for the increased frequency of the reduced set of characters.
Case sensitivity in the input strings (e.g., 'abc' vs. 'Abc')	Convert all strings to lowercase or uppercase before comparing them to ensure case-insensitive comparison.
Input array with a large number of strings (scalability considerations)	Use a hash map to efficiently count the occurrences of each unique character set to improve time complexity.
Strings containing non-alphabetic characters (e.g., numbers, symbols)	Handle these characters consistently; either filter them out during character set generation or include them as part of the character set.

Null or empty input array

How to Handle:

Return 0 immediately as no pairs can exist.

Input array with only one string

How to Handle:

Return 0 immediately because at least two strings are needed to form a pair.

All strings in the input array are identical

How to Handle:

The number of pairs should be n * (n - 1) / 2, calculate and return.

Strings with very long lengths (potential performance bottleneck)

How to Handle:

Ensure that the character set generation and comparison operations are optimized for long strings.

Input array contains duplicate strings that become same after character set reduction

How to Handle:

The counting logic should correctly account for the increased frequency of the reduced set of characters.

Case sensitivity in the input strings (e.g., 'abc' vs. 'Abc')

How to Handle:

Convert all strings to lowercase or uppercase before comparing them to ensure case-insensitive comparison.

Input array with a large number of strings (scalability considerations)

How to Handle:

Use a hash map to efficiently count the occurrences of each unique character set to improve time complexity.

Strings containing non-alphabetic characters (e.g., numbers, symbols)

How to Handle:

Handle these characters consistently; either filter them out during character set generation or include them as part of the character set.