Here’s a solution to the “Anagram Pairs” problem:

from typing import List
from collections import defaultdict

def count_anagram_pairs(strings: List[str]) -> int:
    """
    Find the number of pairs of strings that are anagrams.

    Parameters:
    - strings: A list of strings.

    Returns:
    - int: The number of anagram pairs.
    """
    # Dictionary to store the frequency of sorted characters in each string
    char_freq_count = defaultdict(int)
    # Counter for tracking the number of anagram pairs
    anagram_pairs_count = 0

    for string in strings:
        # Sort the characters in the string and convert it to a tuple
        sorted_tuple = tuple(sorted(string))

        # Update the dictionary with the sorted character tuple
        char_freq_count[sorted_tuple] += 1

    # Count the number of anagram pairs based on the frequency of sorted character tuples
    for count in char_freq_count.values():
        anagram_pairs_count += count * (count - 1) // 2

    return anagram_pairs_count

# Example usage:
result = count_anagram_pairs(["abc", "bca"])
print(result)
assert result == 1

result = count_anagram_pairs(["abc", "Bca"])
print(result)
assert result == 0

result = count_anagram_pairs(["abc"])
print(result)
assert result == 0

result = count_anagram_pairs(["abc", "cba", "def", "fed", "12", "21"])
print(result)
assert result == 3

result = count_anagram_pairs(["listen", "silent", "enlist", "hello", "world"])
print(result)
assert result == 3

result = count_anagram_pairs(["hello", "world", "dog", "god", "cat", "tac"])
print(result)
assert result == 2

In this solution, the characters in each string are sorted before creating a tuple, ensuring that anagrams will have the same sorted character tuples regardless of their original order.

Time Complexity:

The time complexity of the revised solution is O(N * M * log(M)), where N is the number of strings, and M is the average length of the strings. Here’s the breakdown:

  1. Iterating through each string:
    • We iterate through each string in the input list once, which gives us a factor of O(N).
  2. Sorting characters in each string:
    • For each string, we sort its characters. The time complexity of sorting is O(M * log(M)), where M is the length of the string. Since we do this for each of the N strings, the overall complexity for sorting becomes O(N * M * log(M)).
  3. Counting anagram pairs:
    • After sorting, we count the occurrences of each sorted character tuple in a defaultdict. The insertion and lookup operations in a defaultdict are generally considered to be O(1). However, since we iterate through each string, and hash a tuple of length M, the overall complexity for updating the defaultdict becomes O(N * M).
  4. Final Time Complexity:
    • Combining the complexities mentioned above, the overall time complexity is O(N * M * log(M)).

In summary, the dominant factor in the time complexity is the sorting operation, and the overall complexity is determined by the product of the number of strings (N), the average length of the strings (M), and the logarithmic factor due to sorting.

Space Complexity:

The space complexity of the solution is O(N), where N is the number of strings. Here’s the breakdown:

  1. char_freq_count Dictionary:
    • We use a defaultdict named char_freq_count to store the frequency of sorted character tuples. The keys in this dictionary correspond to the sorted character tuples, and the values represent the count of occurrences.
    • In the worst case, all strings are unique, and there are N distinct sorted tuples.
    • Therefore, the space complexity for storing the frequencies is O(N).
  2. anagram_pairs_count Variable:
    • We use a single variable anagram_pairs_count to keep track of the total number of anagram pairs. This variable is independent of the input size and requires constant space, O(1).
  3. Other Variables:
    • The other variables used in the function, such as the string and sorted_tuple variables within the loops, are of constant size and do not depend on the input size.
  4. Final Space Complexity:
    • Combining the complexities mentioned above, the overall space complexity is O(N).

In summary, the dominant factor in the space complexity is the storage of frequencies in the char_freq_count dictionary. The space required is proportional to the the number of strings (N).