Introduction

In this blog post, we will explore how to solve the LeetCode problem "2506" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the solution to the problem, with explanations:

• High-level approach:

  • Convert each word into a canonical representation based on its unique characters (e.g., a sorted string of unique characters or a bitmask).
  • Iterate through all possible pairs of words.
  • Compare the canonical representations of each pair; if they're equal, increment the count of similar pairs.

• Complexity:

  • Runtime: O(N^2 * M), where N is the number of words and M is the average length of a word. This is due to the nested loops to compare pairs and the potential O(M) work to generate each word's representation.
  • Storage: O(N), primarily to store the canonical representations of the words.

Code

    def similarPairs(words):
    """
    Finds the number of pairs of similar strings in the input list.

    Args:
        words: A list of strings.

    Returns:
        The number of similar pairs.
    """

    n = len(words)
    count = 0

    def get_canonical_representation(word):
        """
        Converts a word into a canonical representation.
        """
        return "".join(sorted(set(word)))

    canonical_representations = [get_canonical_representation(word) for word in words]

    for i in range(n):
        for j in range(i + 1, n):
            if canonical_representations[i] == canonical_representations[j]:
                count += 1

    return count