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.
Wrong Solution
class Solution:
def reorderLogFiles(self, logs):
return sorted(logs, key = lambda l: '0'+" ".join(l.split()[1:])+l.split()[0] if l.split()[1].isalpha() else '1')
Probably not the most efficient because we split the same line multiple times, but it looks like Python does it pretty efficiently …
Using the standard approach, with a lambda function:
check the l[1] element, if it’s numeric, return '0' + l[1:] + l[0]
, otherwise returns '1'
.
That way the text entries will be shown first sorted based on l[1:] + l[0]
and the other entries (presumably numeric) will follow in the original order.
Solution
class Solution:
def reorderLogFiles(self, logs):
l = filter(lambda l: l[l.find(" ") + 1].isalpha(), logs)
d = filter(lambda l: l[l.find(" ") + 1].isdigit(), logs)
return sorted(l, key = lambda x: (x[x.find(" "):], x[:x.find(" ")])) + list(d)