271 Encode and Decode Strings

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);
strs2in Machine 2 should be the same asstrsin Machine 1.
Implement theencodeanddecodemethods.
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 asevalor serialize methods. You should implement your own encode/decode algorithm.
A somewhat trival, but accepted solution
    // Encodes a list of strings to a single string.
    string encode(vector<string>& strs) {
        string encoded = "";
        for (string s : strs) {
            for (char c : s) {
                encoded += c + 1;
            }
            encoded += ' ';
        }
        return encoded;
    }
​
    // Decodes a single string to a list of strings.
    vector<string> decode(string s) {
        vector<string> decoded;
        string temp = "";
        for (char c : s) {
            if (c == ' ') {
                decoded.push_back(temp);
                temp.clear();
            }
            else temp += c - 1;
        }
        return decoded;
    }
Congrats on the find
vector<int> encode(string &s, vector<int> &key) 
{
    if (key.size() < s.length()) {
        cout << "invalid key length" << endl;
        exit(0);
    }
​
    vector<int> encoded;
    int iter = 0;
​
    for (char c : s) {
        if (iter % 2 == 0) 
            encoded.push_back((int)c * key[iter++]);
        else encoded.push_back((int)c + key[iter++]);
    }
    return encoded;
}
​
string decode(vector<int> &encoded, vector<int> &key)
{
    string decoded = "";
    int iter = 0;
​
    for (int i : encoded) {
        if (iter % 2 == 0)
            decoded += (char)(i / key[iter++]);
        else decoded += (char)(i - key[iter++]);
    }
​
    return decoded;
}
​
string concate_input(int argc, char **argv)
{    
    string final_con = "";
    for (int i = 1; i < argc; i++) {
        final_con += argv[i];
    }
    return final_con;
}
​
int main(int argc, char **argv)
{
    if (argc < 2) {
        cout << "expected one or more arguments" << endl;
        exit(0);
    }
​
    // support 20 characters
    vector<int> key =  { 
        42,    282,    -264,    -287,    12,
        280,    145,    192,    274,    -56,
        -43,    -279,    -297,    125,    10,
        176,    279,    -234,    -243,    -184 };
​
    string msg = concate_input(argc, argv);
    cout << msg << endl;
    vector<int> encoded = encode(msg, key);
​
    cout << "encoded: ";
    for (int i: encoded) cout << i << ' ';
    cout << endl;
​
    string decoded = decode(encoded, key);
    cout << "decoded: " << decoded << endl;
}
270 Closest Binary Search Tree Value
277 Find the Celebrity
Last modified 4yr ago