Reorder Data in Log Files

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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.

Very basic Idea here

create two lists: letter_logs, digit_logs
sort letter_logs
return letter_logs + digit_logs

That way it would cost us O(MKlogK), with k = len(letter_logs). Of course k can degrade to n in case there are no digit_logs.

Steps :

• The idea in this solution is to create two separate lists that contains letter and digit logs.
• This makes the sort go quicker since the digit logs don't need to be sorted.
• An object is used to store the identifier, the contents, and the full log line of each log entry
• We go through each line in the logs array and populate the letters and digit lists.
• After the letter and digit lists are populated, only the letter logs are sorted according to the requirements.
• Once the sort is complete, the result is returned

Using the above idea implemented custom sort using C# inbuild method Compare from IComparable interface.

String.Compare Method (System)

Use comparison interfaces in Visual C# - C#

Is performance of List.Sort() by Comparison is better than custom IComparer?