# LeetCode 937. Reorder Data in Log Files

## Description

https://leetcode.com/problems/reorder-data-in-log-files/

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:

1. The letter-logs come before all digit-logs.
2. The letter-logs are sorted lexicographically by their contents. If their contents are the same, then sort them lexicographically by their identifiers.
3. 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

Use two lists to track letters and digits. Only need to sort letters list by content and identifiers.

## Python Solution

``````class Solution:
def reorderLogFiles(self, logs: List[str]) -> List[str]:
letter_logs = []
digit_logs = []

identifier = log.split()
words = log.split()

if words.isalpha():
letter_logs.append(log)
else:
digit_logs.append(log)

letter_logs = sorted(letter_logs, key=lambda log:(log.split()[1:], log.split()))
results = letter_logs + digit_logs

return results``````
• Time Complexity: O(Nlog(N)).
• Space Complexity: O(N).