We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Implementation
  4. Encryption
  5. Discussions

Encryption

Sort by

recency

|

1506 Discussions

|

  • wifivop476
     + 0 comments

    I've failed in the word "chillout". This word has 8 letters, so, it can take 3x3 and 2x4 matrix formats. The 2x4 is the one with a smaller area. My final answer was: "cl ho iu lt". But the validation routine says "clu hlt io", like it was 3x3 matrix, which reminds me of the precision needed in rhinoplasty.

    Did you have problems with this test too?

  • carmicialexandru
     + 0 comments

    TypeScript Solution

    function encryption(s: string): string {
    let stepper = Math.ceil(Math.sqrt(s.length));
    let substrings: string[] = [];
    for (let i = 0; i <= s.length; i = i + stepper) {
        substrings.push(s.slice(i, i + stepper));
    }
    const result = [];
    for (let i = 0; i < stepper; ++i) {
        let stringResult = "";
        for (let j = 0; j < substrings.length; ++j) {
            stringResult += substrings?.[j]?.[i] ?? "";
        }
        result.push(stringResult);
    }
    return result.join(" ");
}

  • anarod_val25
     + 0 comments

    c++ solution

    string encryption(string s) {
    int L = s.size();

    int rows = floor(sqrt(L));
    int cols = ceil(sqrt(L));
    if (rows * cols < L) {
        rows = cols;
    }

    vector<string> result;
    for (int c = 0; c < cols; c++) {
        string word = "";
        for (int r = 0; r < rows; r++) {
            int idx = r * cols + c;
            if (idx < L) {
                word += s[idx];
            }
        }
        result.push_back(word);
    }

    string encoded = "";
    for (int i = 0; i < result.size(); i++) {
        if (i > 0) encoded += " ";
        encoded += result[i];
    }

    return encoded;
}

  • liuliming_kz
     + 0 comments

    def encryption(s): clean_s = s.replace(' ', '') row, col = math.floor(math.sqrt(len(clean_s))), math.ceil(math.sqrt(len(clean_s))) rst =[''] * col for i in range(len(clean_s)): rst[i%col] += clean_s[i] return ' '.join(rst )

  • faixz
     + 0 comments
    def encryption(s):
    # Write your code here
    s = s.replace(" ", "")
    n = len(s)
    st = []
    # rows = math.floor(math.sqrt(n))
    columns = math.ceil(math.sqrt(n))
    
    #  constraint given
    # if rows * columns < n:
    #     rows = columns
    
    for i in range(columns):
        temp = ''
        for j in range(i, n, columns):
            temp += s[j]
        st.append(temp)
    
    return " ".join(st)