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937. Reorder Data in Log Files

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.

 

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.

Solutions

  • class Solution {
        public String[] reorderLogFiles(String[] logs) {
            Arrays.sort(logs, this::cmp);
            return logs;
        }
    
        private int cmp(String a, String b) {
            String[] t1 = a.split(" ", 2);
            String[] t2 = b.split(" ", 2);
            boolean d1 = Character.isDigit(t1[1].charAt(0));
            boolean d2 = Character.isDigit(t2[1].charAt(0));
            if (!d1 && !d2) {
                int v = t1[1].compareTo(t2[1]);
                return v == 0 ? t1[0].compareTo(t2[0]) : v;
            }
            if (d1 && d2) {
                return 0;
            }
            return d1 ? 1 : -1;
        }
    }
    
  • class Solution:
        def reorderLogFiles(self, logs: List[str]) -> List[str]:
            def cmp(x):
                a, b = x.split(' ', 1)
                return (0, b, a) if b[0].isalpha() else (1,)
    
            return sorted(logs, key=cmp)
    
    
  • function reorderLogFiles(logs: string[]): string[] {
        const isDigit = (c: string) => c >= '0' && c <= '9';
        return logs.sort((a, b) => {
            const end1 = a[a.length - 1];
            const end2 = b[b.length - 1];
            if (isDigit(end1) && isDigit(end2)) {
                return 0;
            }
            if (isDigit(end1)) {
                return 1;
            }
            if (isDigit(end2)) {
                return -1;
            }
            const content1 = a.split(' ').slice(1).join(' ');
            const content2 = b.split(' ').slice(1).join(' ');
            if (content1 === content2) {
                return a < b ? -1 : 1;
            }
            return content1 < content2 ? -1 : 1;
        });
    }
    
    
  • impl Solution {
        pub fn reorder_log_files(mut logs: Vec<String>) -> Vec<String> {
            logs.sort_by(|s1, s2| {
                let (start1, content1) = s1.split_once(' ').unwrap();
                let (start2, content2) = s2.split_once(' ').unwrap();
                match
                    (
                        content1.chars().nth(0).unwrap().is_digit(10),
                        content2.chars().nth(0).unwrap().is_digit(10),
                    )
                {
                    (true, true) => std::cmp::Ordering::Equal,
                    (true, false) => std::cmp::Ordering::Greater,
                    (false, true) => std::cmp::Ordering::Less,
                    (false, false) => content1.cmp(&content2).then(start1.cmp(&start2)),
                }
            });
            logs
        }
    }
    
    

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