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

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Reorder Data in Log Files?

This problem uses Array, String, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Reorder Data in Log Files?

Insight 1: Identifying letter-logs vs digit-logs is the first crucial step. This is done by checking if the first character after the identifier is a digit. Insight 2: Python's `sorted()` function with a custom `key` is a clean and efficient way to sort based on multiple criteria. The tuple returned by the key function dictates the sorting order.

Which companies ask LeetCode 937. Reorder Data in Log Files in interviews?

This problem is frequently asked at Snap.

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Reorder Data in Log Files - LeetCode 937 Solution

Reorder Data in Log Files - Complete Solution Guide

Reorder Data in Log Files is LeetCode problem 937, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 lexicographica

Detailed Explanation

The problem requires reordering an array of log files. Each log file has an identifier followed by a space and then either words (letter-logs) or numbers (digit-logs). The goal is to arrange the logs such that all letter-logs come before all digit-logs. The letter-logs should be sorted lexicographically by their contents (the part after the identifier). If two letter-logs have the same content, they should be sorted lexicographically by their identifiers. The digit-logs should maintain their original order relative to each other.

Solution Approach

The solution involves defining a custom sorting key that encapsulates the sorting logic. This key function assigns a tuple to each log file. The first element of the tuple indicates whether the log is a letter-log (0) or a digit-log (1). For letter-logs, the second element of the tuple is the content of the log, and the third is the identifier. For digit-logs, only the '1' to categorize the log type is needed. The standard sorting algorithm then uses these tuples to compare and sort the log files according to the problem requirements.

Step-by-Step Algorithm

Step 1: Define a helper function (or lambda in C++) `get_sort_key(log)` that takes a log string as input.
Step 2: Inside `get_sort_key(log)`, split the log into the identifier and the content.
Step 3: Check if the first character of the content is a digit. If so, return `(1,)`. This marks it as a digit-log and places it after letter logs because 1 > 0.
Step 4: If the first character of the content is not a digit, return `(0, content, identifier)`. This marks it as a letter-log and provides the content and identifier for lexicographical sorting.
Step 5: Use the `sorted()` function (or equivalent in other languages) with the `get_sort_key` function as the key to sort the logs.
Step 6: Return the sorted list of logs.

Key Insights

Insight 1: Identifying letter-logs vs digit-logs is the first crucial step. This is done by checking if the first character after the identifier is a digit.
Insight 2: Python's `sorted()` function with a custom `key` is a clean and efficient way to sort based on multiple criteria. The tuple returned by the key function dictates the sorting order.
Insight 3: Maintaining the relative order of digit-logs requires using a stable sorting algorithm, which Python's `sorted()` and Java's `Arrays.sort()` are. This also means that within the `key` function, digit logs should be compared as equal with respect to the digit portion, enabling the stable sort to maintain relative order.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Sorting.

Companies

Asked at: Snap.

Reorder Data in Log Files | LeetCode 937

Medium

Array String Sorting

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        Arrays.sort(logs, new Comparator<String>() {
            @Override
            public int compare(String log1, String log2) {
                String[] split1 = log1.split(" ", 2);
                String[] split2 = log2.split(" ", 2);

                boolean isDigit1 = Character.isDigit(split1[1].charAt(0));
                boolean isDigit2 = Character.isDigit(split2[1].charAt(0));

                if (!isDigit1 && !isDigit2) {
                    int contentCompare = split1[1].compareTo(split2[1]);
                    if (contentCompare != 0) {
                        return contentCompare;
                    }
                    return split1[0].compareTo(split2[0]);
                } else if (isDigit1 && isDigit2) {
                    return 0;
                } else if (!isDigit1 && isDigit2) {
                    return -1;
                } else {
                    return 1;
                }
            }
        });
        return logs;
    }
}

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Problem Info

Description

Problem 937: Reorder Data in Log Files - Detailed explanation and solution approaches.

Topics

Array String Sorting

Companies

Snap

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:937

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Define a helper function (or lambda in C++) `get_sort_key(log)` that takes a log string as input.
2
Step 2: Inside `get_sort_key(log)`, split the log into the identifier and the content.
3
Step 3: Check if the first character of the content is a digit. If so, return `(1,)`. This marks it as a digit-log and places it after letter logs because 1 > 0.
4
Step 4: If the first character of the content is not a digit, return `(0, content, identifier)`. This marks it as a letter-log and provides the content and identifier for lexicographical sorting.
5
Step 5: Use the `sorted()` function (or equivalent in other languages) with the `get_sort_key` function as the key to sort the logs.
6
Step 6: Return the sorted list of logs.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the sorting operation. Python's `sorted()` and Java's `Arrays.sort()` typically use comparison-based sorting algorithms like Timsort or Merge Sort, which have a time complexity of O(n log n). The space complexity is O(n) because, in most cases, the sorting algorithm needs to create a copy of the data during the sorting process.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Two-Pass Approach

Instead of using a custom sorting key, you could first iterate through the logs and separate them into two lists: letter-logs and digit-logs. Then, sort the letter-logs using a standard sorting algorithm with a custom comparator. Finally, concatenate the sorted letter-logs with the digit-logs.

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

Trade-offs: This approach might be more readable for some, but it involves creating and managing multiple lists, which can be less efficient than the single-pass approach with a custom sorting key, especially for large input sizes.

✨

Key Insights

1Insight 1: Identifying letter-logs vs digit-logs is the first crucial step. This is done by checking if the first character after the identifier is a digit.
2Insight 2: Python's `sorted()` function with a custom `key` is a clean and efficient way to sort based on multiple criteria. The tuple returned by the key function dictates the sorting order.
3Insight 3: Maintaining the relative order of digit-logs requires using a stable sorting algorithm, which Python's `sorted()` and Java's `Arrays.sort()` are. This also means that within the `key` function, digit logs should be compared as equal with respect to the digit portion, enabling the stable sort to maintain relative order.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty log file. The code assumes there is always at least one space. The provided codes handle the general format correctly, but should check the edge cases to make the solution more robust.
Edge case 2: Log file with no content after the identifier (e.g., "id"). The provided codes should handle this gracefully as the splitting will result in an empty string for the content, and the first character check will still work.
Edge case 3: Identifiers containing spaces. The splitting logic is designed to split on the *first* space, so identifiers with spaces will be treated correctly.
Edge case 4: All logs are digit logs. The relative order of digit logs must be preserved. Because the comparison within the key function makes digit logs equal to each other (returning 0), this is handled due to the stable sort implementation of sort methods in most languages.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the relative order of digit-logs correctly. If a non-stable sorting algorithm is used or the comparison logic for digit-logs is not correct, the digit-logs might not maintain their original order.

✗Common mistake 2: Incorrectly identifying letter-logs and digit-logs. Failing to check if the character after the identifier is a digit can lead to incorrect sorting.

✗Common mistake 3: Forgetting to handle the case where letter-logs have the same content. The identifier must be used as a tie-breaker in this scenario.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Log file analysis. In systems monitoring and debugging, log files are often generated with identifiers and messages. Reordering and sorting these logs can make it easier to identify patterns and diagnose problems.

🏦Real-world scenario 2: Data processing and organization. The general principle of sorting data based on multiple criteria is applicable in various data processing tasks, such as sorting customer records by name, then by purchase history, or sorting search results by relevance, then by date.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Reorder Data in Log Files"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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