Problem Description

This problem asks you to design an encoding and decoding system for a list of strings. You need to implement two functions:

1. encode: Takes a list of strings and converts it into a single string
2. decode: Takes the encoded string and converts it back to the original list of strings

The key challenge is that the strings in the list can contain any characters, including special characters, spaces, or even characters that might typically be used as delimiters. Your encoding scheme must be able to handle all possible string contents and accurately reconstruct the original list.

Example workflow:

• Machine 1 has a list of strings like ["Hello", "World"]
• It encodes this list into a single string using your encode function
• This encoded string is sent over the network to Machine 2
• Machine 2 uses your decode function to reconstruct the exact original list ["Hello", "World"]

Solution Approach:

The solution uses a length-prefixing technique:

Encoding Process:

• For each string in the list, first write its length as a fixed 4-digit number (padded with leading zeros if necessary)
• Then append the actual string content
• Concatenate all these encoded pieces together

For example, ["Hello", "World"] becomes "0005Hello0005World" where:

• "0005" indicates the next string has length 5
• "Hello" is the first string
• "0005" indicates the next string has length 5
• "World" is the second string

Decoding Process:

• Read the first 4 characters to get the length of the next string
• Extract that many characters as one string from the original list
• Repeat until the entire encoded string is processed

This approach works because:

• The length prefix tells us exactly how many characters belong to each string
• We don't need to worry about delimiter conflicts since we're using length information rather than special separator characters
• The fixed 4-digit format ensures we always know where each length value ends

Intuition

The main challenge in this problem is finding a way to combine multiple strings into one while preserving the boundaries between them. Let's think through why common approaches might fail:

Why simple delimiters don't work: If we try to use a delimiter like "," or "|" to separate strings, what happens when the original strings contain these delimiters? For example, if we encode ["hello,world", "test"] using commas, we'd get "hello,world,test". When decoding, we can't tell if this should be ["hello", "world", "test"] or ["hello,world", "test"].

Why escape sequences are complex: We could use escape sequences (like replacing "," with "\," in the strings), but this requires careful handling of the escape character itself and makes the implementation more error-prone.

The key insight - use metadata: Instead of trying to find a "safe" delimiter that won't appear in the strings, we can store metadata about each string. The most useful metadata is the string's length. If we know a string is exactly 5 characters long, we can read exactly 5 characters without worrying about what those characters are.

Why fixed-width length encoding: We encode the length as a fixed 4-digit number because:

• If we used variable-length numbers, we'd face the same delimiter problem (how do we know where the length ends and the string begins?)
• With 4 digits, we can handle strings up to length 9999, which is sufficient for most practical cases
• We always know to read exactly 4 characters to get the length, then read exactly that many more characters to get the string

This approach elegantly sidesteps the delimiter problem entirely. Each string is self-contained with its length prefix, and we can decode the entire list by repeatedly reading length-value pairs until we've consumed the entire encoded string.

Solution Approach

Let's walk through the implementation step by step:

Encoding Algorithm

The encoding process transforms a list of strings into a single string by prefixing each string with its length:

1. Initialize an empty list ans to collect encoded parts
2. For each string s in the input list:
   • Calculate the length of the string using len(s)
   • Format this length as a 4-digit string using "{:4}".format(len(s))
     • This creates a string with exactly 4 characters, padding with spaces if needed
     • For example: length 5 becomes " 5", length 123 becomes " 123"
   • Concatenate the 4-digit length with the actual string content
   • Append this to the ans list
3. Join all parts using "".join(ans) to create the final encoded string

Example encoding process:

• Input: ["Hello", "World"]
• Process string "Hello": length is 5 → " 5Hello"
• Process string "World": length is 5 → " 5World"
• Final encoded string: " 5Hello 5World"

Decoding Algorithm

The decoding process extracts strings from the encoded format by reading length prefixes:

1. Initialize variables:

   • ans = empty list to store decoded strings
   • i = 0 (current position in the encoded string)
   • n = length of the encoded string

2. While we haven't processed the entire string (i < n):

   • Read the length prefix: Extract 4 characters starting at position i using s[i : i + 4]
   • Convert to integer: Parse these 4 characters as an integer to get the string length
   • Move pointer: Advance i by 4 to skip past the length prefix
   • Extract the string: Read size characters starting at position i using s[i : i + size]
   • Store the string: Append the extracted string to ans
   • Move pointer: Advance i by size to move to the next encoded string

3. Return the list of decoded strings

Example decoding process:

• Input: " 5Hello 5World"
• Read first 4 chars: " 5" → length = 5
• Read next 5 chars: "Hello" → first string
• Read next 4 chars: " 5" → length = 5
• Read next 5 chars: "World" → second string
• Result: ["Hello", "World"]

Key Design Decisions

• Fixed-width length encoding (4 digits): Ensures we always know exactly how many characters represent the length
• No delimiter characters: Completely avoids the problem of strings containing delimiter characters
• Sequential processing: Both encoding and decoding process strings in order, maintaining the original sequence
• String slicing: Python's slice notation s[i:j] efficiently extracts substrings without additional memory allocation for intermediate results

This approach has O(n) time complexity for both encoding and decoding, where n is the total length of all strings combined.

Example Walkthrough

Let's walk through encoding and decoding the list ["Hi", "I'm", "Sam!"]:

Encoding Process:

1. Start with empty result: ""

2. Process "Hi" (length = 2):

   • Format length as 4 digits: " 2" (2 with 3 leading spaces)
   • Append string: " 2Hi"
   • Result so far: " 2Hi"

3. Process "I'm" (length = 3):

   • Format length as 4 digits: " 3"
   • Append string: " 3I'm"
   • Result so far: " 2Hi 3I'm"

4. Process "Sam!" (length = 4):

   • Format length as 4 digits: " 4"
   • Append string: " 4Sam!"
   • Final encoded result: " 2Hi 3I'm 4Sam!"

Decoding Process:

Starting with encoded string: " 2Hi 3I'm 4Sam!" (total length = 18)

1. Position i = 0:

   • Read 4 chars at position 0-3: " 2" → length = 2
   • Move to position 4
   • Read 2 chars at position 4-5: "Hi"
   • Add "Hi" to result list
   • Move to position 6

2. Position i = 6:

   • Read 4 chars at position 6-9: " 3" → length = 3
   • Move to position 10
   • Read 3 chars at position 10-12: "I'm"
   • Add "I'm" to result list
   • Move to position 13

3. Position i = 13:

   • Read 4 chars at position 13-16: " 4" → length = 4
   • Move to position 17
   • Read 4 chars at position 17-20: "Sam!"
   • Add "Sam!" to result list
   • Move to position 21

4. Position i = 21 ≥ 18 (string length), so stop

Final decoded result: ["Hi", "I'm", "Sam!"]

Notice how the special characters (apostrophe and exclamation mark) are handled naturally without any escaping needed, since we rely on length prefixes rather than delimiters.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the total length of all strings combined.

• For encode(): We iterate through each string once and perform constant-time operations (formatting length and concatenation to list). The string join operation at the end is O(n) where n is the total length of all strings.

• For decode(): We traverse the encoded string once from start to end. Each character is visited exactly once, and substring extraction s[i : i + size] takes O(size) time. The total time across all substrings is O(n).

Space Complexity: O(n) where n is the total length of all strings.

• For encode(): The ans list stores formatted strings that together have length proportional to the input (4 extra characters per string for length encoding plus the original strings). The final joined string also takes O(n) space.

• For decode(): The ans list stores all decoded strings which together have the same total length as the original input, taking O(n) space.

Common Pitfalls

1. String Length Exceeding Fixed Width

The most critical pitfall in this implementation is that it assumes all strings have a length that can be represented in 4 digits (maximum 9999 characters). If a string has 10,000 or more characters, the formatting f"{len(string):4}" will exceed the 4-character width, breaking the decoding logic.

Example of the problem:

codec = Codec()
long_string = "a" * 10000  # String with 10,000 characters
encoded = codec.encode([long_string])
# The length "10000" takes 5 characters, not 4
# This breaks the fixed-width assumption
decoded = codec.decode(encoded)  # Will fail or produce incorrect results

Solution: Use a more robust length encoding scheme:

def encode(self, strs: List[str]) -> str:
    encoded_parts = []
    for string in strs:
        # Use delimiter with escape sequence
        # Format: "length:string" where : is a delimiter
        length_str = str(len(string))
        # Include the length of the length string itself
        encoded_parts.append(f"{len(length_str)}:{length_str}{string}")
    return "".join(encoded_parts)

def decode(self, s: str) -> List[str]:
    decoded_strings = []
    i = 0
    while i < len(s):
        # Read length of length
        colon_pos = s.index(':', i)
        len_of_len = int(s[i:colon_pos])
        i = colon_pos + 1
        # Read actual length
        string_length = int(s[i:i + len_of_len])
        i += len_of_len
        # Read string
        decoded_strings.append(s[i:i + string_length])
        i += string_length
    return decoded_strings

2. Alternative: Using Non-Printable Delimiter

Another approach that avoids the length limitation:

def encode(self, strs: List[str]) -> str:
    # Use a non-printable character as delimiter
    # Format: "length\x00string" for each string
    result = []
    for string in strs:
        result.append(f"{len(string)}\x00{string}")
    return "".join(result)

def decode(self, s: str) -> List[str]:
    if not s:
        return []
  
    decoded_strings = []
    i = 0
    while i < len(s):
        # Find the null character delimiter
        null_pos = s.index('\x00', i)
        length = int(s[i:null_pos])
        i = null_pos + 1
        decoded_strings.append(s[i:i + length])
        i += length
    return decoded_strings

3. Most Robust: Variable-Length Encoding

The most flexible solution uses a format like "length#string":

def encode(self, strs: List[str]) -> str:
    return ''.join(f"{len(s)}#{s}" for s in strs)

def decode(self, s: str) -> List[str]:
    decoded = []
    i = 0
    while i < len(s):
        # Find the delimiter '#'
        delimiter_pos = s.index('#', i)
        length = int(s[i:delimiter_pos])
        i = delimiter_pos + 1
        decoded.append(s[i:i + length])
        i += length
    return decoded

This approach:

• Handles strings of any length
• Uses '#' as a delimiter between length and content
• The length prefix tells us exactly how many characters to read, so '#' appearing in the string content doesn't matter