# Solving Leetcode Interviews in Seconds with AI: Reorder Data in Log Files


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "937" 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
	> You are given an array of logs. Each log is a space-delimited string of words, where the first word is the identifier. There are two types of logs:  Letter-logs: All words (except the identifier) consist of lowercase English letters. Digit-logs: All words (except the identifier) consist of digits.  Reorder these logs so that:  The letter-logs come before all digit-logs. The letter-logs are sorted lexicographically by their contents. If their contents are the same, then sort them lexicographically by their identifiers. The digit-logs maintain their relative ordering.  Return the final order of the logs.   Example 1:  Input: logs = ["dig1 8 1 5 1","let1 art can","dig2 3 6","let2 own kit dig","let3 art zero"] Output: ["let1 art can","let3 art zero","let2 own kit dig","dig1 8 1 5 1","dig2 3 6"] Explanation: The letter-log contents are all different, so their ordering is "art can", "art zero", "own kit dig". The digit-logs have a relative order of "dig1 8 1 5 1", "dig2 3 6".  Example 2:  Input: logs = ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"] Output: ["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]    Constraints:  1 <= logs.length <= 100 3 <= logs[i].length <= 100 All the tokens of logs[i] are separated by a single space. logs[i] is guaranteed to have an identifier and at least one word after the identifier.  

	# Explanation
	*   **Identify Log Types:** Separate the logs into letter-logs and digit-logs based on the content after the identifier.
*   **Sort Letter-Logs:** Sort the letter-logs lexicographically by their content, and then by their identifier if the content is the same.
*   **Combine and Return:** Concatenate the sorted letter-logs with the original order of digit-logs.

*   **Runtime Complexity:** O(N log N), where N is the number of logs due to the sorting of letter logs. **Storage Complexity:** O(N), for storing letter and digit logs in separate lists.

	
	# Code
	```python
	def reorderLogFiles(logs: list[str]) -> list[str]:
    letter_logs = []
    digit_logs = []

    for log in logs:
        parts = log.split(" ", 1)
        if parts[1][0].isdigit():
            digit_logs.append(log)
        else:
            letter_logs.append(log)

    letter_logs.sort(key=lambda log: (log.split(" ", 1)[1], log.split(" ", 1)[0]))

    return letter_logs + digit_logs
	```