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.

Machine 1 (sender) has the function:

string encode(vector<string> strs) {
  // ... your code
  return encoded_string;
}

Machine 2 (receiver) has the function:

vector<string> decode(string s) {
  //... your code
  return strs;
}

So Machine 1 does:

string encoded_string = encode(strs);

and Machine 2 does:

vector<string> strs2 = decode(encoded_string);

strs2 in Machine 2 should be the same as strs in Machine 1.

Implement the encode and decode methods.

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.

Companies:
Google, Square, Twitter

Related Topics:
String

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• Serialize and Deserialize Binary Tree (Hard)
• String Compression (Easy)
• Count Binary Substrings (Easy)

// OJ: https://leetcode.com/problems/encode-and-decode-strings/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Codec {
public:
    string encode(vector<string>& strs) {
        string ans;
        for (string &str : strs) {
            for (char c : str) {
                if (c == '$') ans.push_back(c);
                ans.push_back(c);
            }
            ans.push_back('$');
            ans.push_back('x');
        }
        return ans;
    }
    vector<string> decode(string s) {
        vector<string> ans;
        int i = 0, N = s.size();
        while (i < N) {
            string str;
            for (; i < N; ++i) {
                if (s[i] != '$') str.push_back(s[i]);
                else if (i + 1 < N && s[i + 1] == '$') str.push_back(s[i++]);
                else {
                    i += 2;
                    break;
                }
            }
            ans.push_back(str);
        }
        return ans;
    }
};