Introduction

In this blog post, we will explore how to solve the LeetCode problem "2085" 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

Given two string arrays words1 and words2, return the number of strings that appear exactly once in each of the two arrays. Example 1: Input: words1 = ["leetcode","is","amazing","as","is"], words2 = ["amazing","leetcode","is"] Output: 2 Explanation: - "leetcode" appears exactly once in each of the two arrays. We count this string. - "amazing" appears exactly once in each of the two arrays. We count this string. - "is" appears in each of the two arrays, but there are 2 occurrences of it in words1. We do not count this string. - "as" appears once in words1, but does not appear in words2. We do not count this string. Thus, there are 2 strings that appear exactly once in each of the two arrays. Example 2: Input: words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"] Output: 0 Explanation: There are no strings that appear in each of the two arrays. Example 3: Input: words1 = ["a","ab"], words2 = ["a","a","a","ab"] Output: 1 Explanation: The only string that appears exactly once in each of the two arrays is "ab". Constraints: 1 <= words1.length, words2.length <= 1000 1 <= words1[i].length, words2[j].length <= 30 words1[i] and words2[j] consists only of lowercase English letters.

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Count the frequency of each word in words1 and words2 using dictionaries (hash maps).
  • Iterate through the keys (words) present in the frequency map of words1.
  • Check if the word exists in the frequency map of words2 and if it appears exactly once in both arrays.

• Complexity:

  • Runtime: O(n + m), where n is the length of words1 and m is the length of words2.
  • Storage: O(n + m), to store the frequency maps.

Code

    from collections import Counter

def count_words(words1, words2):
    """
    Counts the number of strings that appear exactly once in each of the two arrays.

    Args:
        words1: A list of strings.
        words2: A list of strings.

    Returns:
        The number of strings that appear exactly once in each of the two arrays.
    """

    count1 = Counter(words1)
    count2 = Counter(words2)

    result = 0
    for word in count1:
        if count1[word] == 1 and word in count2 and count2[word] == 1:
            result += 1

    return result