Reorder Data in Log Files

Published: Sep 9, 2022

Medium Sorting Array String


This is just a sorting problem. However, some clever comparator is required.

Problem Description

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.


  • 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.


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"]
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"]


This solution relies on Python’s tuple sorting feature. While sorting the tuple, the first element, the second element if the first is the same will be used to compare. The custom comparator returns (1,) for digit logs so that digit logs come later. For letter logs, it returns (0, string except id, id) to comply with the condition. With this custom comparator, just soring the log gives us the answer.


class ReorderDataInLogFiles:
    def reorderLogFiles(self, logs: List[str]) -> List[str]:
        def comp(log):
            id_, rest = log.split(' ', 1)
            if rest[0].isdigit():
                return (1, )
                return (0, rest, id_)
        return sorted(logs, key=comp)


  • Time: O(nlog(n))
  • Space: O(1)