# Solving Leetcode Interviews in Seconds with AI: Count Common Words With One Occurrence


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```