Encode and Decode Strings
Problem Statement: The task is to design two methods, encode
and decode
, to encode a list of strings into a single string and decode the single string back into the original list of strings. The encoding algorithm should be reversible.
1. Clarify the problem:
- Are the input strings guaranteed to be non-empty?
- How do we handle an empty list of strings as input?
- Is there a limit on the length of individual strings or the total number of strings?
2. Analyze the problem:
- Input: List of strings
- Output: Encoded string (for
encode
), decoded list of strings (fordecode
) - Constraints:
- The input list can contain empty strings.
- The encoding algorithm should be reversible.
- There may be no limit on the length of individual strings or the total number of strings.
3. Design an algorithm:
- Encoding (
encode
):- Concatenate all the input strings, separated by a delimiter (e.g., "#").
- Prepend each string with its length, followed by a delimiter (e.g., "3#hello#5#world").
- Decoding (
decode
):- Split the encoded string by the delimiter to get individual strings.
- Parse the length of each string and extract the corresponding substring.
- Return the list of decoded strings.
4. Explain your approach:
- For encoding, we will concatenate the strings and prepend each string with its length.
- For decoding, we will split the encoded string, parse the lengths, and extract the substrings to reconstruct the original list of strings.
5. Write clean and readable code:
python
class Codec:
def encode(self, strs):
encoded = ""
delimiter = "#"
for s in strs:
encoded += str(len(s)) + delimiter + s
return encoded
def decode(self, s):
decoded = []
delimiter = "#"
i = 0
while i < len(s):
delimiter_index = s.find(delimiter, i)
length = int(s[i:delimiter_index])
decoded.append(s[delimiter_index + 1 : delimiter_index + 1 + length])
i = delimiter_index + 1 + length
return decoded
6. Test your code: Let's test the code with some example and additional test cases:
python
# Example test case
codec = Codec()
strs = ["hello", "world", "leetcode"]
encoded = codec.encode(strs)
decoded = codec.decode(encoded)
# Expected output:
# encoded = "5#hello5#world8#leetcode"
# decoded = ["hello", "world", "leetcode"]
# Additional test cases
strs = []
encoded = codec.encode(strs)
decoded = codec.decode(encoded)
# Expected output:
# encoded = ""
# decoded = []
strs = [""]
encoded = codec.encode(strs)
decoded = codec.decode(encoded)
# Expected output:
# encoded = "0#"
# decoded = [""]
strs = ["a", "", "b"]
encoded = codec.encode(strs)
decoded = codec.decode(encoded)
# Expected output:
# encoded = "1#a0#1#b"
# decoded = ["a", "", "b"]
7. Optimize if necessary: The provided solution has a straightforward approach and doesn't have any major optimizations to be made. The time complexity of both encoding and decoding is O(n), where n is the total length of the input strings.
8. Handle error cases:
In the given solution, we assume that the input will be valid and within the specified constraints. We do not handle cases where the list of strings is None
or the encoded string is empty.
9. Discuss complexity analysis: The time complexity of the encoding and decoding operations is O(n), where n is the total length of the input strings. This is because we iterate through each string in the input list once. The space complexity is O(n) as well since we store the encoded or decoded string as a single string.