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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Encryption
  5. Discussions

Encryption

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1498 Discussions

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  • daniel_marques_1
     + 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.

    Did you have problemns with this test too?

  • yashdeora98294
     + 0 comments

    Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-encryption-problem-solution.html

  • giorgibregvadze1
     + 0 comments

    Javascript

    function encryption(s) {
    const rows = Math.floor(Math.sqrt(s.length));
    const columns = Math.ceil(Math.sqrt(s.length));
    const realNumberOfRows = s.length > rows * columns? rows + 1:rows;
    const wordsArray = [];
    const encryptedArray = [];
    for(let i = 0; i < realNumberOfRows; i++){wordsArray.push(s.split("").splice(i*columns,columns));
    }
    for(let i = 0; i < columns; i++){
        const newWord = [];
        wordsArray.forEach(item=>newWord.push(item[i]));
        encryptedArray.push(newWord.join(''));
    }  
    return encryptedArray.join(' ');
}

  • oluwashetemipe
     + 0 comments

    public static String encryption(String s) { // Write your code here int size = s.length();

    int x = (int)Math.floor(Math.sqrt(size));
int y = (int)Math.ceil(Math.sqrt(size));
if(x * y < size){
   x = y;
}
char[][] t = new char[x][y];
int index = 0;
for(int i = 0; i < x; i++ ){
    for(int j = 0; j < y; j++){
        if(index == size){
            break;
        }
        t[i][j] = s.charAt(index);
        index++;

    }
}
int io = 0;
StringBuilder enc = new StringBuilder();
for(int i = 0; i < y; i++){
    if( i != 0){
        enc.append(" ");
    }
    for(int j = 0; j < x; j++){
        if(t[j][i] == 0){
            continue;
        }
        enc.append(t[j][i]);
        io++;  
        if(io == size){
            break;
        }
    }
}

return enc.toString();
}

  • emhhacker
     + 0 comments

    My Java solution with o(n * m) time complexity and o(n) space complexity:

    public static String encryption(String s) {
        // remove the spaces from the string
        s.replaceAll(" ", "");
        
        // get the len of the string
        int length = s.length();
        
        // set x equal to floor of sqrt of len
        int x = (int) Math.floor(Math.sqrt(length));
        
        // set y equal to ceil of sqrt of len
        int y = (int) Math.ceil(Math.sqrt(length));
        if(x * y < length) x += 1;
        
        // print each char of the string x by y
        StringBuilder encryptedString = new StringBuilder();
        for(int i = 0; i < y; i++){
            for(int j = 0; j < x; j++){
                int idx = j * y + i;
                if(idx < length)
                    encryptedString.append(s.charAt(idx));
            }
            encryptedString.append(' '); //line break
        }
        return encryptedString.toString();
    }