Code Implementation

class Solution:
    def reorderLogFiles(self, logs):
        letter_logs = []
        digit_logs = []
        for log in logs:
            id_, rest = log.split(" ", 1)
            if rest[0].isdigit():
                digit_logs.append(log)
            else:
                letter_logs.append((rest, id_, log))
        # Sort by content, then identifier
        letter_logs.sort(key=lambda x: (x[0], x[1]))
        return [log[2] for log in letter_logs] + digit_logs
      

class Solution {
public:
    vector<string> reorderLogFiles(vector<string>& logs) {
        vector<pair<string, string>> letter_logs;
        vector<string> digit_logs;
        for (const string& log : logs) {
            int pos = log.find(' ');
            string id = log.substr(0, pos);
            string rest = log.substr(pos + 1);
            if (isdigit(rest[0])) {
                digit_logs.push_back(log);
            } else {
                letter_logs.emplace_back(rest, id);
            }
        }
        sort(letter_logs.begin(), letter_logs.end(), [](const pair<string, string>& a, const pair<string, string>& b) {
            if (a.first == b.first)
                return a.second < b.second;
            return a.first < b.first;
        });
        vector<string> result;
        for (const auto& p : letter_logs) {
            result.push_back(p.second + " " + p.first);
        }
        result.insert(result.end(), digit_logs.begin(), digit_logs.end());
        return result;
    }
};
      

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        List<String> letterLogs = new ArrayList<>();
        List<String> digitLogs = new ArrayList<>();
        for (String log : logs) {
            int idx = log.indexOf(' ');
            if (Character.isDigit(log.charAt(idx + 1))) {
                digitLogs.add(log);
            } else {
                letterLogs.add(log);
            }
        }
        Collections.sort(letterLogs, (a, b) -> {
            int idxA = a.indexOf(' ');
            int idxB = b.indexOf(' ');
            String contentA = a.substring(idxA + 1);
            String contentB = b.substring(idxB + 1);
            if (contentA.equals(contentB)) {
                return a.substring(0, idxA).compareTo(b.substring(0, idxB));
            }
            return contentA.compareTo(contentB);
        });
        String[] result = new String[logs.length];
        int i = 0;
        for (String log : letterLogs) result[i++] = log;
        for (String log : digitLogs) result[i++] = log;
        return result;
    }
}
      

var reorderLogFiles = function(logs) {
    const letterLogs = [];
    const digitLogs = [];
    for (const log of logs) {
        const idx = log.indexOf(' ');
        const id = log.substring(0, idx);
        const rest = log.substring(idx + 1);
        if (/\d/.test(rest[0])) {
            digitLogs.push(log);
        } else {
            letterLogs.push([rest, id, log]);
        }
    }
    letterLogs.sort((a, b) => {
        if (a[0] === b[0]) {
            return a[1] < b[1] ? -1 : 1;
        }
        return a[0] < b[0] ? -1 : 1;
    });
    return letterLogs.map(x => x[2]).concat(digitLogs);
};
      

Problem Description

You are given an array of log strings, logs. Each log is a space-delimited string with two parts:

• An identifier (the first word)
• The rest of the log, which may contain either all letters (for letter-logs) or all digits (for digit-logs)

1. All letter-logs come before any digit-log.
2. Letter-logs are sorted lexicographically by their content. If contents are the same, sort by identifier.
3. Digit-logs should remain in their original order.

Thought Process

To solve this problem, we need to distinguish between two types of logs: letter-logs and digit-logs. The main challenge is to sort the letter-logs according to their content and identifier, while keeping the digit-logs in their original order. A brute-force approach might try to sort all logs together, but this would not respect the special ordering rules for digit-logs. Instead, it's better to separate the logs into two groups, process each group as required, and then combine them.

The key insight is to treat letter-logs and digit-logs differently:

• Letter-logs need custom sorting.
• Digit-logs only need to be appended after the letter-logs, preserving their input order.

Solution Approach

Let's break down the solution step by step:

1. Separate logs into letter-logs and digit-logs:
   • For each log, split it into the identifier and the rest of the log.
   • If the first character of the rest is a digit, it's a digit-log; otherwise, it's a letter-log.
2. Sort the letter-logs:
   • Sort primarily by the content (everything after the identifier).
   • If contents are the same, sort by identifier.
   • This can be done using a tuple or custom comparator.
3. Combine the results:
   • Place all sorted letter-logs first.
   • Then append all digit-logs in their original order.

We use a list or array to store each group. Sorting is efficient because we only sort letter-logs, and digit-logs are simply appended.

Example Walkthrough

Let's use the following example input:

logs = [
  "dig1 8 1 5 1",
  "let1 art can",
  "dig2 3 6",
  "let2 own kit dig",
  "let3 art zero"
]
  

1. Separate into letter-logs and digit-logs:
   • Letter-logs: ["let1 art can", "let2 own kit dig", "let3 art zero"]
   • Digit-logs: ["dig1 8 1 5 1", "dig2 3 6"]
2. Sort letter-logs by content and identifier:
   • "let1 art can" (content: "art can")
   • "let3 art zero" (content: "art zero")
   • "let2 own kit dig" (content: "own kit dig")

   After sorting by content, we get:

   1. "let1 art can"
   2. "let3 art zero"
   3. "let2 own kit dig"

3. Combine sorted letter-logs with digit-logs:
   • Result: ["let1 art can", "let3 art zero", "let2 own kit dig", "dig1 8 1 5 1", "dig2 3 6"]

This ordering meets all problem requirements.

Time and Space Complexity

• Brute-force approach:
  • If you tried to sort all logs together and checked types during comparison, the complexity would be at least O(N log N), but with more overhead for custom logic.
• Optimized approach:
  • Separating into letter-logs and digit-logs takes O(N).
  • Sorting letter-logs is O(M log M), where M is the number of letter-logs (M ≤ N).
  • Combining the lists is O(N).
  • Total time complexity: O(N log N), dominated by the sorting step.
  • Space complexity: O(N), for storing the separated logs.

Summary

The key to solving "Reorder Data in Log Files" is to separate the problem into manageable parts: identify letter-logs and digit-logs, sort only the letter-logs with a custom comparator, and maintain the original order of digit-logs. This approach leverages efficient sorting and simple list manipulations, resulting in a clean and effective solution. The elegance comes from reducing the problem to sorting with a custom key and combining the results in a straightforward way.