Problem Description

In this problem, we are dealing with an array of strings called logs, where each log entry is composed of an identifier followed by either a string of lowercase English letters (a letter-log) or a string of digits (a digit-log). The identifier and the log content are separated by a space. Our task is to reorder these logs based on certain rules:

1. Letter-logs must be sorted and come before all digit-logs.
2. Letter-logs should be ordered lexicographically (i.e., alphabetically) by their content. If two letter-logs have identical content, they should then be sorted by their identifiers in lexicographic order.
3. Digit-logs should retain their original order as they appeared in the input.

The expected output is the reordered array of the logs following these rules.

Intuition

To solve this, we need a way to differentiate between letter-logs and digit-logs and then sort them according to the given criteria. We can achieve this through a custom sorting function.

When we look at the identifier and the log content, we observe that letter-logs contain alphabetical characters while digit-logs contain numbers. We can exploit this characteristic to split logs into two categories by checking if the first character of the log content is a letter or a digit.

Given the sorting requirements, we know that all letter-logs should be sorted lexicographically by their content, and if these are equal, by their identifiers. Digit-logs do not need sorting among themselves, but they need to be placed after all letter-logs. To keep them in their original order, we can just ignore their content and identifiers during the sorting process.

With these observations, we craft a sorting function that:

1. Splits each log into its identifier and content.
2. Checks whether it's a letter-log or digit-log by inspecting the first character of the content.
3. If it's a letter-log, the function returns a tuple (0, content, identifier). The zero indicates that it's a letter-log for sort precedence, and sorting by content and then identifier follows naturally.
4. If it's a digit-log, the function returns a tuple (1,). The '1' ensures that all digit-logs are ordered after letter-logs, and since no other sorting keys follow, they maintain their relative input order.

Finally, by calling the sorted function and passing our custom comparison function, we obtain a sorted list that meets all the criteria outlined in the problem description. The sorted function rearranges the elements according to the tuples returned by our comparison function cmp, effectively pushing letter-logs to the front and sorting them while keeping digit-logs in their original order at the end of the resulting list.

Learn more about Sorting patterns.

Solution Approach

The implementation of the solution primarily revolves around the Python sorted function and a custom sorting key function, which is the cmp function in the reference solution. The sorted function is an efficient sorting algorithm that uses the TimSort algorithm, which is a hybrid sorting algorithm derived from merge sort and insertion sort.

The cmp function is where the distinction between letter-logs and digit-logs is made and the sorting logic is customized. It takes a log entry as an argument and uses the split method to divide it into two parts, the identifier and the content, with a space being the delimiter. Since we need both parts separately for the sorting criteria, splitting them does the job efficiently.

Here's the step-by-step approach within the cmp function:

1. Split the log into identifier a and content b:

   a, b = x.split(' ', 1)

2. Check whether the log is a letter-log or a digit-log:

   if b[0].isalpha():
       ...
   else:
       ...

3. For letter-logs, return a tuple (0, b, a), where b is the content of the log without the identifier and a is the identifier. The leading 0 ensures these come before any digit-logs.

4. For digit-logs, return a tuple (1,), which only has the digit 1. It ensures digit-logs are placed after letter-logs and since no other values are in the tuple, their original order (relative to other digit-logs) is maintained.

The sorted function uses this cmp key function to sort the entire list of logs. The logs are compared based on the tuples returned by cmp; if the first elements of these tuples are different (which differentiate between letter and digit logs), it sorts based on this. If they are the same (as with all letter-logs), it proceeds to sort based on the subsequent elements in the tuple, which are the content and, if necessary, the identifier.

To sum up, the algorithm utilizes a sorting key function to create sortable tuples for each log entry, which Python's sorted function then utilizes to sort the entire list accordingly. This approach does not require any additional data structures or complex algorithms, making it an efficient and elegant solution to the problem.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Consider the following logs array:

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

We are tasked with reordering these logs based on the rules provided in the problem description.

Step 1: Differentiating logs

Firstly, we want to differentiate the logs into letter-logs and digit-logs. We can see that "let1 art can", "let2 own kit dig", and "let3 art zero" are letter-logs because they contain letters after the first space. "dig1 8 1 5 1" and "dig2 3 6" are digit-logs because they contain numbers after the first space.

Step 2: Applying the custom sorting function

We implement a custom sorting key function, cmp, to sort the logs. This function will handle logs differently based on their type.

Letter-logs

For letter-logs, cmp will return a tuple of the form (0, content, identifier). Here's how it will look for our letter-logs in python code:

# "let1 art can" splits into identifier "let1" and content "art can"
("let1 art can".split(" ", 1)) => ("let1", "art can") => (0, "art can", "let1")

# "let2 own kit dig" splits into identifier "let2" and content "own kit dig"
("let2 own kit dig".split(" ", 1)) => ("let2", "own kit dig") => (0, "own kit dig", "let2")

# "let3 art zero" splits into identifier "let3" and content "art zero"
("let3 art zero".split(" ", 1)) => ("let3", "art zero") => (0, "art zero", "let3")

Digit-logs

For digit-logs, cmp will return a tuple just containing '1', which places all digit-logs after the letter-logs.

# "dig1 8 1 5 1" splits but we only need to recognize it as a digit-log
("dig1 8 1 5 1".split(" ", 1)) => ("dig1", "8 1 5 1") => (1,)

# "dig2 3 6" also splits but we only need to mark it as a digit-log
("dig2 3 6".split(" ", 1)) => ("dig2", "3 6") => (1,)

Step 3: Sorting logs

After applying the cmp function, the sorted function sorts the logs based on these tuples:

# Letter-logs sorted by content and then by identifier if needed
sorted_letter_logs = sorted(["let1 art can", "let2 own kit dig", "let3 art zero"], key=cmp)

# Output for letter-logs is: ["let3 art zero", "let1 art can", "let2 own kit dig"]

# Digit-logs maintain their original order
# Therefore, after letter-logs they are just appended: ["dig1 8 1 5 1", "dig2 3 6"]

# Combined sorted logs
sorted_logs = sorted_letter_logs + ["dig1 8 1 5 1", "dig2 3 6"]

# Final output: ["let3 art zero", "let1 art can", "let2 own kit dig", "dig1 8 1 5 1", "dig2 3 6"]

This final output respects all the sorting rules mentioned in the problem description and uses the power of Python's sorted function along with a custom comparison key to effectively reorder the logs.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(NlogN), where N is the number of logs. This arises from the use of the sorted function, which typically has O(NlogN) time complexity for sorting an array. Specifically, every comparison between two logs requires string comparison, and in the worst case, each comparison can take O(M) time, where M is the maximum length of a log file. However, assuming M is relatively small and therefore the comparison time is constant, the overall time complexity still remains O(NlogN).

The space complexity of the code is O(N). This is because the sorting algorithm used in Python (Timsort) is a hybrid stable sorting algorithm derived from merge sort and insertion sort characterizes O(N) space complexity. This space is needed to hold the intermediate results for the sorted array. Furthermore, the extra space needed for the lambda function's return values is not significant compared to the space needed for these intermediate results as they do not depend on the number of logs N.

Learn more about how to find time and space complexity quickly using problem constraints.