Discussion on Encryption Challenge

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2 weeks ago+ 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();
}

2 weeks ago+ 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();
    }

1 month ago+ 0 comments

def encryption(s: str) -> str:
    temp = s.replace(" ", "")
    cols = ceil(sqrt(len(temp)))
    batch = list(batched(temp, cols))
    result = [[word[c] for word in batch if len(word) > c] for c in range(cols)]
    return " ".join("".join(x) for x in result)

2 months ago+ 0 comments

Javascript

function encryption(s) {
    const noSpaces = s.replaceAll(' ', '');
    const strLen = noSpaces.length;   
    const row = Math.floor(Math.sqrt(strLen));
    const col = Math.ceil(Math.sqrt(strLen)); 
 
    const encryptedArr = [];
    let pos = 0;
    for(let i=0; i<strLen; i++) {
        const str = noSpaces.substr(i, 1);
        
        encryptedArr[pos] = encryptedArr[pos] ?? '';
        encryptedArr[pos] += str;
        
        pos = pos === col-1 ? 0 : pos+1;
    }
    
    return encryptedArr.join(' ');
}

2 months ago+ 0 comments

sample tes2: string length =8 rows =2 and cols =3 , floor and ceil rows *cols < L (8) readujust rows and cols such that rows *cols >= L and rows and rows*cols is minimu with multiple grid options Here we have two option 2*4 or 3*3, rows =2 and cols=4 is minimum area With this option it says wrong answer. The expected answer matches with 3*3 grid which not as per specs. mistake in problem

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