937. Reorder Data in Log Files

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Sort an array of logs where letter-logs go first (lexicographic by content then identifier) and digit-logs keep relative order. A custom-key sort with a tuple tiebreaker.

By Sam K., Founder, InterviewChamp.AI · Last verified 2026-05-26

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 after the identifier consist of lowercase English letters) and digit-logs (all words after the identifier consist of digits). Reorder these logs so that: (1) all letter-logs come before any digit-log; (2) letter-logs are sorted lexicographically by their contents — if contents are the same, then sort by identifier; (3) digit-logs maintain their relative ordering. Return the final order of the logs.

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.

Examples

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"]

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Output

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Hints

Progressive — try the first before opening the next.

Hint 1

Split each log into identifier and content. Classify by whether the first content character is a digit.

Hint 2

Use a stable sort with a key that puts digit-logs after letter-logs.

Hint 3

Key for letter-logs: (0, content, identifier). Key for digit-logs: (1,) so they all tie and the stable sort preserves their order.

Solution approach

Reveal approach

Stable sort with a tuple key. For each log, split into identifier and rest at the first space. If rest[0] is a digit, the key is (1,) — all digit-logs tie and the stable sort keeps their original order. Otherwise the key is (0, rest, identifier) — letter-logs come first, sorted by content with identifier as tiebreaker. O(n * m log n) time where n = number of logs and m = average log length (string comparison cost). O(n * m) space for the keys and the output.

Complexity

Time: O(n * m log n)
Space: O(n * m)

Related patterns

custom-sort
string-parsing
stable-sort

Asked at

Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).

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