How do you solve Reorder Data in Log Files on LeetCode?

Reorder Data in Log Files (LeetCode #937) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Letter-logs must be sorted based on content and identifier.

What is the time complexity of Reorder Data in Log Files?

The optimal solution for Reorder Data in Log Files runs in O(n log n) time and uses O(n) space. The sorting of letter-logs is O(n log n), and since we are using additional space for storing logs, the space complexity is O(n).

What is the brute force solution for Reorder Data in Log Files?

The brute force approach for Reorder Data in Log Files is Brute Force, with O(n²) time complexity and O(n) space complexity. In this approach, we will separate the logs into letter-logs and digit-logs. Then, we will sort the letter-logs and concatenate them with the digit-logs, maintaining their original order. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Reorder Data in Log Files?

Reorder Data in Log Files uses the following concepts: Array, String, Sorting. The recommended patterns to study are: Sorting, Array.

What are the interview tips for Reorder Data in Log Files?

Clarify the problem requirements before coding. Think about edge cases, such as all logs being digit-logs. Practice writing clean and concise code.

What are common mistakes when solving Reorder Data in Log Files?

Confusing the sorting criteria for letter-logs. Not maintaining the relative order of digit-logs.

#937

Reorder Data in Log Files

Medium

ArrayStringSortingSortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(n)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

We can achieve a more efficient solution by using a single pass to classify the logs and then sorting only the letter-logs. This reduces time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Create two lists for letter-logs and digit-logs.
2Step 2: Iterate through the logs, classifying each log as a letter-log or digit-log.
3Step 3: Sort the letter-logs based on their contents and identifiers using a custom sorting function.
4Step 4: Concatenate the sorted letter-logs with the digit-logs.
5Step 5: Return the final ordered list of logs.

solution.py12 lines

1def reorderLogs(logs):
2    letter_logs = []
3    digit_logs = []
4
5    for log in logs:
6        if log.split()[1].isdigit():
7            digit_logs.append(log)
8        else:
9            letter_logs.append(log)
10
11    letter_logs.sort(key=lambda x: (x.split()[1:], x.split()[0]))
12    return letter_logs + digit_logs

ℹ

Complexity note: The sorting of letter-logs is O(n log n), and since we are using additional space for storing logs, the space complexity is O(n).

1Letter-logs must be sorted based on content and identifier.
2Digit-logs retain their original order.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.