Problem Statement

Design an algorithm to implement these two functions: one that can effectively convert the list of strings to a single string (encoding) and another that can revert this single string back to the original list (decoding).

Constraints:

• You cannot use any serialization methods such as eval.
• The goal is to ensure that after encoding and then decoding, the output is identical to the input.

Machine 1 (Sender):

• encode(List<String> strs): Takes a list of strings and converts it to a single encoded string.

Machine 2 (Receiver):

• decode(String s): Takes the encoded string and reconstructs the original list of strings.

Ensure that after sending the encoded string from Machine 1 to Machine 2, the list of strings obtained on Machine 2 is exactly the same as the one sent from Machine 1.

Examples

1. Example 1:

   • Input: strs = ["dog", "cat", "bird"]
   • Expected Output: ["dog", "cat", "bird"]
   • Justification: The list of strings contains three simple words. The encoded version should accurately store each word, and the decode function should reconstruct the exact original list.

2. Example 2:

   • Input: strs = ["hello,world", "foo!bar", ""]
   • Expected Output: ["hello,world", "foo!bar", ""]
   • Justification: This test case includes strings with special characters like commas and exclamation marks, as well as an empty string. The encoding must handle special characters and empty strings properly without losing any data.

3. Example 3:

   • Input: strs = ["", "", "empty"]
   • Expected Output: ["", "", "empty"]
   • Justification: This test case includes multiple empty strings and a non-empty string. The encoding should differentiate between multiple empty strings and preserve the original order and count of the strings.

Constraints:

• 1 <= strs.length <= 200
• 0 <= strs[i].length <= 200
• strs[i] contains any possible characters out of 256 valid ASCII characters.

Solution

To solve this problem, we encode a list of strings into a single string by storing each string's length followed by a special separator (#) and then the string itself. This approach helps us keep track of each string's length, making it easy to decode them later. The length acts as a marker that tells us where each string starts and ends, allowing the decode function to reconstruct the original list of strings accurately.

During decoding, we read the encoded string from the beginning, finding the separator (#) to determine the length of each encoded string segment. With the length information, we can extract each string correctly and add it back to the list. This method is efficient because it avoids complex parsing and uses a simple marker-based strategy to keep the encoded format clear and easy to decode.

Step-by-Step Algorithm

Encoding:

1. Initialize encoded_string: Create an empty StringBuilder named encoded_string to store the final encoded string.
2. Iterate through the input list strs: Loop through each string s in the list strs.
   • Calculate and Append Length: For each string s, calculate its length using s.length().
   • Append Separator and String: Append the length, followed by the separator #, and then the string s itself to encoded_string.
   • Purpose: The length and separator # help to differentiate between strings during decoding.
3. Return Encoded String: Convert the StringBuilder to a string using toString() and return it. This string represents the entire list in an encoded format.

Decoding:

1. Initialize List strs: Create an empty list strs to store the decoded strings.
2. Initialize Pointer i: Start with i = 0, pointing to the beginning of the encoded string s.
3. Iterate through Encoded String s: Use a while loop to process the entire encoded string until i reaches the end.
   • Find the Next Separator #: Locate the next # separator using s.indexOf('#', i).
   • Extract Length: Extract the substring from i to the position of # to get the length of the next string. Convert this length from a string to an integer using Integer.parseInt.
   • Extract String: Use the extracted length to get the actual string segment from the encoded string, starting just after # for length characters.
   • Add to List: Append the extracted string to the list strs.
   • Update Pointer i: Move the pointer i to the position after the extracted string to continue processing.
4. Return Decoded List: After processing all strings, return the list strs.

Algorithm Walkthrough

Input: ["hello,world", "foo!bar", ""]

1. Encoding:
   • Input List: ["hello,world", "foo!bar", ""]
   • Initialize encoded_string as an empty StringBuilder.
   • For the first string "hello,world":
     • Length is 11. Append 11#hello,world to encoded_string.
   • For the second string "foo!bar":
     • Length is 7. Append 7#foo!bar to encoded_string.
   • For the third string "":
     • Length is 0. Append 0# to encoded_string.
   • Encoded Result: encoded_string becomes "11#hello,world7#foo!bar0#".

2. Decoding:
   • Encoded String: "11#hello,world7#foo!bar0#"
   • Initialize strs as an empty list and set i = 0.
   • First Decoding Step:
     • Find # at position 2. Extract length 11 from s.substring(0, 2).
     • Extract "hello,world" using s.substring(3, 14).
     • Add "hello,world" to strs.
     • Update i to 14 (end of the extracted string).
   • Second Decoding Step:
     • Find # at position 15. Extract length 7 from s.substring(14, 15).
     • Extract "foo!bar" using s.substring(16, 23).
     • Add "foo!bar" to strs.
     • Update i to 23.
   • Third Decoding Step:
     • Find # at position 24. Extract length 0 from s.substring(23, 24).
     • Extract an empty string "".
     • Add "" to strs.
     • Update i to 25.
   • Final Decoded List: ["hello,world", "foo!bar", ""].

Code

Complexity Analysis

Time Complexity

Encode Function:

• The encode function goes through each string in the list once. For each string, it calculates the length and appends it with a separator (#) to the result.
• The total time taken depends on the total number of characters across all strings, which is N.

Time Complexity of encode: O(N), where N is the total number of characters in all strings.

Decode Function:

• The decode function scans through the entire encoded string once. It finds each separator (#) and extracts the original strings based on their lengths.
• The time taken is also proportional to the total length of the encoded string, which is N.

Time Complexity of decode: O(N), where N is the total length of the encoded string.

Space Complexity

1. Encoding:

   • The space complexity of the encode method is O(N), where N is the length of the final encoded string stored in the StringBuilder.

2. Decoding:

   • The space complexity of the decode method is O(M) for storing the result list, where M is the number of strings obtained after the split operation. Since M can be proportional to N (the size of the input string), the overall space complexity is O(N).