Introduction

In this blog post, we will explore how to solve the LeetCode problem "2451" 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 an array of equal-length strings words. Assume that the length of each string is n. Each string words[i] can be converted into a difference integer array difference[i] of length n - 1 where difference[i][j] = words[i][j+1] - words[i][j] where 0 <= j <= n - 2. Note that the difference between two letters is the difference between their positions in the alphabet i.e. the position of 'a' is 0, 'b' is 1, and 'z' is 25. For example, for the string "acb", the difference integer array is [2 - 0, 1 - 2] = [2, -1]. All the strings in words have the same difference integer array, except one. You should find that string. Return the string in words that has different difference integer array. Example 1: Input: words = ["adc","wzy","abc"] Output: "abc" Explanation: - The difference integer array of "adc" is [3 - 0, 2 - 3] = [3, -1]. - The difference integer array of "wzy" is [25 - 22, 24 - 25]= [3, -1]. - The difference integer array of "abc" is [1 - 0, 2 - 1] = [1, 1]. The odd array out is [1, 1], so we return the corresponding string, "abc". Example 2: Input: words = ["aaa","bob","ccc","ddd"] Output: "bob" Explanation: All the integer arrays are [0, 0] except for "bob", which corresponds to [13, -13]. Constraints: 3 <= words.length <= 100 n == words[i].length 2 <= n <= 20 words[i] consists of lowercase English letters.

Explanation

• Calculate Differences: For each word, calculate its difference integer array.
  • Hashing for Comparison: Use the difference array as a key in a dictionary to count occurrences of each unique difference array.
  • Identify the Odd One Out: Iterate through the dictionary and find the difference array that appears only once. Return the corresponding word.

• Runtime Complexity: O(N M), where N is the number of words and M is the length of each word. *Storage Complexity: O(N), where N is the number of words, in the worst case where all difference arrays are unique.

Code

    def odd_string(words):
    """
    Finds the string in words that has a different difference integer array.

    Args:
        words: A list of strings.

    Returns:
        The string in words that has a different difference integer array.
    """

    diff_counts = {}
    word_to_diff = {}

    for word in words:
        diff = []
        for i in range(len(word) - 1):
            diff.append(ord(word[i + 1]) - ord(word[i]))

        diff_tuple = tuple(diff)  # Convert list to tuple for hashing
        word_to_diff[word] = diff_tuple

        if diff_tuple in diff_counts:
            diff_counts[diff_tuple] += 1
        else:
            diff_counts[diff_tuple] = 1

    for word, diff_tuple in word_to_diff.items():
        if diff_counts[diff_tuple] == 1:
            return word