Problem

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.

My Code

class Solution:
    def reorderLogFiles(self, logs: List[str]) -> List[str]:
        letter=[]
        digit=[]
        
        for i in logs:
            if i.split()[1].isdigit():
                digit.append(i)
            else:
                letter.append(i)
        
        a=sorted(letter, key=lambda x: (x.split()[1:], x.split()[0]))
        return a+digit

Review

우선 순위에 따른 차순 정리 부분이 너무 어려웠어서 다른 분들 걸 참고하고야 말았....
lambda x: x.split()[1]만 써줬다가 빠꾸, 다른거 안썼다고 빠꾸, :안써서 다시 빠꾸....

문제 어려웠어 ㅠ...