LeetCode - Reorder Data in Log Files

You have an array of logs.  Each log is a space delimited string of words.

For each log, the first word in each log is an alphanumeric identifier.  Then, either:

• Each word after the identifier will consist only of lowercase letters, or;
• Each word after the identifier will consist only of digits.

We will call these two varieties of logs letter-logs and digit-logs.  It is guaranteed that each log has at least one word after its identifier.

Reorder the logs so that all of the letter-logs come before any digit-log.  The letter-logs are ordered lexicographically ignoring identifier, with the identifier used in case of ties.  The digit-logs should be put in their original order.

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"]

Constraints:

1. 0 <= logs.length <= 100
2. 3 <= logs[i].length <= 100
3. logs[i] is guaranteed to have an identifier, and a word after the identifier.

Solution :

Java Solution :

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        Arrays.sort(logs,(log1,log2)->{
            String[] set1 = log1.split(" ",2);
            String[] set2 = log2.split(" ",2);
            boolean isCharacter1 = Character.isLetter(set1[1].charAt(0));
            boolean isCharacter2 = Character.isLetter(set2[1].charAt(0));
            if(isCharacter1 && isCharacter2){
                int cmp = set1[1].compareTo(set2[1]);
                if(cmp !=0) return cmp;
                return set1[0].compareTo(set2[0]);
            }
            return !isCharacter1 ? (!isCharacter2 ? 0 : 1) : -1;
        });
        return logs;
    }
}