Encode and Decode Strings Python Solution

LeetCode 271, Medium. Topics: Array, String, Design. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 271   # by problem number
lcpy gen -s encode_and_decode_strings   # by problem name

Problem

Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings.

Examples

Input: dummy_input = ["Hello","World"]
Output: "Hello,World"
Explanation: Machine 1:
Codec encoder = new Codec();
String msg = encoder.encode(strs);
Machine 1 ---msg---> Machine 2
Machine 2:
Codec decoder = new Codec();
String[] strs = decoder.decode(msg);

Constraints

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

Note:

The string may contain any possible characters out of 256 valid ASCII characters. Your algorithm should be generalized enough to work on any possible characters.
Do not use class member/global/static variables to store states. Your encode and decode algorithms should be stateless.
Do not rely on any library method such as eval or serialize methods. You should implement your own encode/decode algorithm.

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n)
    # Space: O(n)
    def encode(self, strs: list[str]) -> str:
        encoded = ""
        for s in strs:
            # Format: length + '#' + string
            encoded += str(len(s)) + "#" + s
        return encoded

    # Time: O(n)
    # Space: O(n)
    def decode(self, s: str) -> list[str]:
        decoded = []
        i = 0

        while i < len(s):
            # Find the delimiter '#'
            j = i
            while s[j] != "#":
                j += 1

            # Extract length
            length = int(s[i:j])

            # Extract string of that length
            decoded.append(s[j + 1 : j + 1 + length])

            # Move to next encoded string
            i = j + 1 + length

        return decoded

Complexity

Time	Space
O(n)	O(n)

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Encode and Decode Strings Python Solution

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