Given an array of strings, find the number of pairs of strings that are anagrams of each other. Two strings are considered anagrams if they have the same characters, but in a different order.

Strings in the pairs can be in any order and each pair should be counted only once.

The case of characters should be considered (e.g., “listen” and “silent” are anagrams, but “Listen” and “Silent” are not).

Function Signature:

from typing import List

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.
    """
    # Your code here

Example:

result = count_anagram_pairs(["abc", "bca"])
print(result)
assert result == 1
# Explanation: There is only one anagram pair ("abc", "bca").

result = count_anagram_pairs(["abc", "Bca"])
print(result)
assert result == 0
# Explanation: "abc" and "Bca" are not an anagram pair because the cases ('b' and 'B') are different.

result = count_anagram_pairs(["abc"])
print(result)
assert result == 0
# Explanation: There is only one element. There is no pair.

result = count_anagram_pairs(["abc", "cba", "def", "fed", "12", "21"])
print(result)
assert result == 3
# Explanation: Anagram pairs are ("abc", "cba"), ("def", "fed"), ("12", "21").

result = count_anagram_pairs(["listen", "silent", "enlist", "hello", "world"])
print(result)
assert result == 3
# Explanation: Anagram pairs are ("listen", "silent"), ("listen", "enlist"), and ("silent", "enlist").

result = count_anagram_pairs(["hello", "world", "dog", "god", "cat", "tac"])
print(result)
assert result == 2
# Explanation: Anagram pairs are ("dog", "god"), ("cat", "tac").

Note:

  • Assume that there is no duplicate element in the list.

Instructions:

  • Write the count_anagram_pairs function to solve the problem.
  • Implement your solution in Python.
  • Provide clear comments in your code.
  • Discuss the time and space complexity of your solution.

As always, we’ll share our solution to this problem tomorrow. Stay tuned 😊