0s in every case without using 2d array (c++). It was a great hint! Thanks!

I did it like this and got all submissions 0.06. Great hint! :D https://www.hackerrank.com/challenges/encryption/submissions/code/16330491

I have the same approach with you, why everybody have to do it the hard way?

int main(){
    string s;
    cin >> s;
    auto n = s.size();
    int row = round(sqrt(n));
    int column;
    if (row >= sqrt(n)) column = row; else column = row + 1;
    for(int j=0;j<column;++j) {
        for(int i=j; i<n;i+=column)cout << s[i];
        cout << ' ';
    }
    return 0;
}

Sure:

#!/bin/python3

import sys
from math import ceil, sqrt


s = input().strip()
c = ceil(sqrt(len(s)))
print(' '.join(map(lambda x: s[x::c], range(c))))

Anyone see a way to turn this into a 1-liner without calculating c twice?

once you said it it actually seems to be stupid to solve it with arrays :D

int main() { string s;

cin >> s;

auto n = s.size();
int row, column;

row = = round(sqrt(n));

if( row >= sqrt(n) )
    column = row; else column = row + 1;

for( int j = 0; j < column; j++ )
{
    for( int i = j; i < n; i += column )
        cout << s[i];

    cout << ' ';
}

}

Sure, I did it this way.

0.02s or quicker in C#. Didn't even bother with the grid approach but it's got to be slower because of the extra iteration and memory writes. The additional space requirements of allocating a grid also impacts overall performance. And if we're talking about cryptography here, you want it to be as performant as possible.

Honestly, using a 2D array barely even crossed my mind. I spent too long making binary trees in C++ to be afraid of some simple index math.

0.1 seconds in Java on every test.

sometimes people overthink, I was going with the array approach then I saw this, thanks for the hint. here is my implementation in java

static void message(String string){
        int row = (int)Math.floor(Math.sqrt(string.length()));
        int column = (int)Math.ceil(Math.sqrt(string.length()));
        if(row * column < string.length()) row = column;
        for(int i = 0; i < column; i++){
            for(int j = i; j < string.length(); j = j + column ){
                System.out.print(string.charAt(j));
            }
            System.out.print(" ");
        }
    }

How can you the computation time?

std::string sentence;
getline(cin, sentence);

int length = sentence.length(),
    columns = ceil(sqrt(length)),
    rows = floor(sqrt(length));
if(rows*columns < length)
    rows = columns;

for(int i = 0; i < columns; ++i) {
    for(int j = 0; j < rows; ++j) {
        if(i + j * columns < length)
            cout << sentence[i + j * columns];
    }
    cout << ' ';
}

Thank you for the hint. O(s.Length)

string s = Console.ReadLine().Trim();
var rowLength = (int)Math.Sqrt(s.Length)+1;
if(Math.Sqrt(s.Length)-(int)Math.Sqrt(s.Length) == 0)
    rowLength = (int)Math.Sqrt(s.Length);

for(int index = 0; index < rowLength;index++){
    var verticalStrip = string.Empty;
    var k = index;   
    
    while( k < s.Length){
        verticalStrip += s[k];
        k += rowLength;
    }
    Console.Write(verticalStrip+ " ");
}

A solution to this question has been provided on the HackerRank solution repo! If you are interested in solutions to HackerRank problems as well as gaining some experience working on an GitHub open source project, check out HackerRank Solutions. We are working to make a repository of solutions to every HackerRank question.

Did it in this manner and got 0s in all test cases. https://www.hackerrank.com/challenges/encryption/forum/comments/280014

I did :)

s = list(input().strip())
L =  len(s)
row = math.floor(L**(1/2))
col = math.ceil(L**(1/2))

for i in range(col):
    x = 0     
    while True:
        try:
            print(s[i+x],end='')
            x+= col
        except:
            break        
    print(' ',end='')

I did it this way

int main(){
    char* s = (char *)malloc(10240 * sizeof(char));
    scanf("%s",s);
    int len = strlen(s);
    int columns = ceil(sqrt(len));
    for(int i = 0; i < columns; i++) {
        for(int j = i; j < len; j += columns) {
            printf("%c", s[j]);
        }
        printf(" ");
    }
    return 0;
}

all of the test cases take 0s

without 2d array

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int r,c;
    string s;
    cin>>s;
    int l=s.length();
    r=floor(sqrt(l));
    c=ceil(sqrt(l));
    if((r*c)<l)
        r++;
    for(int i=0;i<c;i++)
    {
        for(int j=i;j<l;j+=c)
        {
            cout<<s[j];
        }
        cout<<" ";
    }
    return 0;
}

How are checking the running time of your solution?

Hey that was a great tip!! Passed all test cases in 0s in Python.

Check it out https://www.hackerrank.com/challenges/encryption/submissions/code/50674389

yep, 2D array is very unnecessary.

include

include

include

include

include

using namespace std;

int main(){ string s; cin >> s; int count=0; int l=s.length(); int rows=floor(sqrt(l)); int col=ceil(sqrt(l)); if(rows*col

for(int i=0;i<rows&&m<l;i++){
    cout<<s[m];
    count++;
    m=m+col;}k++;n++;
    if(count<l)
cout<<" ";}


return 0;

}

how do you check the time per subimssion ?

Don't need to use a 2d array.

package main
import (
    "fmt"
    "math"
)

func main() {
    var text string
    fmt.Scan(&text)
    
    l := len(text)
    m := math.Sqrt(float64(l))
    lower := int(math.Floor(m))
    upper := int(math.Ceil(m))
    
    // upper * upper must be >= l, so it safe to do following. Or just simply do one lower += 1
    for lower * upper < l {
        lower += 1
    }
    
    index := 0
    for i := 0; i < upper; i++ {
        for j := 0; j < lower; j++ {
            index = i + j * upper
            if index >= l {
                break
            }
            fmt.Printf("%c", text[index])
        }
        fmt.Printf(" ")
    }
}

My code without using two dimensional array.

import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); int hop = (int)Math.ceil(Math.sqrt(s.length())); int i = 0, j = 1,count = 0; while ( j <= hop ){ while( i < s.length()){ System.out.print(s.charAt(i)); i+=hop; } System.out.print(" "); i = j; j++; } in.close(); } }

My Python3 solution. Do not have to use 2-d array.

#!/bin/python3

import sys
s = input().strip()
rows = int(len(s) ** 0.5)
if rows < len(s) ** 0.5:
    columns = rows + 1
else:
    columns = rows
    
for i in range(columns):
    print(s[i::columns], end = ' ')

Without any 2D array.

void encryption(string s) {
int n = s.length();
    int x = floor(sqrt(n));
    int y = ceil(sqrt(n));
    int k=0;
    if((x*y)<n)
        x++;
    for(int i=0; i<y; i++){
        for(int j=0; j<x; j++){
            k=i+(j*y);
            if(s[k]!='\0')
            cout<<s[k];
        }
        cout<<" ";
    }

}

This worked for me..

int rows = 0, cols = 0, i, j = 0; StringBuffer sb = new StringBuffer();

    //rows = (int)Math.floor(Math.pow(s.length(), 0.5));
    cols = (int)Math.ceil(Math.pow(s.length(), 0.5));

    for(i = 0; i < cols; i++)
    {
        for(j = i; j < s.length(); j += cols)
        {
            sb.append(s.charAt(j));
        }
        sb.append(" ");
    }

    return sb.toString();

solution without using 2d array https://www.hackerrank.com/challenges/encryption/forum/comments/483060

you mean like this '#include

using namespace std;

int main() {

string s;
getline(cin, s);
int length=s.length();
int ul=0,ll=0;
for(int i=1; i<=9; i++)
{
    if(i*i>length)
    {
        ul=i;
        ll=i-1;
        break;
    }
    else if(i*i==length)
    {
        ul=i;
        ll=i;
        break;
    }
}
//cout<<ul<<" "<<ll<<endl;
int i=0;
while(i<ul && i<s.length())
{
    int j=i;
    while(j<s.length())
    {
        cout<<s[j];
        j=j+ul;
        //cout<<j;
    }
    cout<<" ";
    i=i+1;
}

return 0;

}

How about Ruby?

def sqrt(l); Math.sqrt l end

def encryption(s)
  # strip off any whitespaces
  s.gsub! /\s+/, ''

  # length of the string without whitespaces
  len = s.length

  # let us set the (rows, cols) based on floor & ceil of the string length
  rows, cols = sqrt(len).floor, sqrt(len).ceil

  # if the product of rows and cols is less than string length, set both to same
  rows = cols = sqrt(len).ceil if rows * cols < len

  # each index of this array will hold chars (c..row), where for each c,
  # we get all the values until row
  encrypted = []
  (0...rows).each do |i|
    encrypted << s[((i % rows) * cols)..((i % rows) * cols) + cols - 1].chars
  end

  # we use the first entry and zip it with the rest of entries,
  # and join them to make a complete string
  encrypted[0].zip(*encrypted[1..-1]).map do |cipher|
    cipher.compact.join + ' '
  end.join
end

I solved it without arrays, too in Python. A very simple Python3 code aka Python at its best :)

import math

def encryption(s):
    s = s.replace(" ", "")
    sr = len(s) ** 0.5  # which is square root
    col = int(sr) if math.ceil(sr) == int(sr) else int(sr)+1

    word = ''
    for i in range(col):
        p = i
        while p < len(s):
            word = word + s[p]
            p += col
        word = word + ' '

    return word

Definitely did not use 2d array. Makes more sense it grab string at indexes when you know it's needed :+1:

Yes

for(int i = 0; i < w; i++) {
     for(int j = 0; j < h; j++) {
           if(j  w + i < z.length()) o += z.charAt(j * w + i);
					 else break;
     }
     o += ' ';
}

Edit: multiplication sign got removed for some reason.

i tried this and it passed all cases:

static String encryption(String s) { int len=s.length(); System.out.println(len); String s1 =""; for(int i = 0 ; irow) row+=1; System.out.println(col); for(int i=0 ; i s2+=" "; } return s2; }

k=len(s) w=[] r=int(math.sqrt(k)) f=r+1 for i in range(0,r+1): for j in range(i,len(s),f): w.append(s[j]) w.append(" ") w.pop() w="".join(w) return(w)

Pseudo-code to solve using iteration (without 2d array):

sn = replace spaces of s        
    L = length of  sn
    rows = math.floor(math.sqrt(L))
cols = math.ceil(math.sqrt(L))
if (rows * cols < L) then rows = cols
    accumulate array arr of size rows * cols
    let row = -1
    for i = 0 to L:
       if (i % cols == 0) then row++
         arr[((i % cols) * rows) + row] = sn[i]
    set result to an empty list
for i = 0 to length of arr:
       if (i % rows == 0 and i > 0) then result.append(" ")
         result.append(arr[i])
return string.join("", result)

Iterative is a little hard to get but here is my way: String e = new String(""); int r= (int)Math.sqrt(s.length()); if(r*r!=s.length()) r++; for(int i=0;i<r;i++){ int j=0; while(j*r+i<s.length()){ e = e+s.charAt(j*r+i); j++; } e=e+" "; } return e;

I haven't use 2d array in my solution

static String encryption(String s) {

char arr[] = s.toCharArray();
String ans = new String();
int length = 0;
if(Math.sqrt(arr.length) % 1 == 0){
 length = (int)(Math.sqrt(arr.length));
}else{
 length = (int)(Math.sqrt(arr.length)+1);
}

System.out.println(arr.length+"   "+length);

for(int i = 0; i < length ; i++)
{
    for(int j=i; j < arr.length ; j++){

        if(j < arr.length){

            ans+=arr[j];
            j+=(length-1);

        }
    }
   if(i != length){
       ans += ' ';
   }


}
return ans;





}

I did it with single array. Just calcualte the index and print.

It may be due to trailing spaces. Add a check for iter less than length, print only then. You might be using a direct cout.

Sir,I took care about trailing spaces, but irrespective of no. of trailing spaces result was same, ie.it passed few test cases but failed is some.

have you solved this ??? Iam getting the same problem

Have you resolved? Same problem for me. I also checked trailing space

same issue for me, did u solve it ?

same issue here

It is because you need to ensure that rows*cols >= Lemgth of string.

We need to keep rows <= cols so we keep rows as the floor of squareroot and cols as the ceiling.

now if rows*cols < L then you need to add another row to make things fit.

I'm having the same problem, did you solve it? What's stranger is that the first testcase doesn't mind whether or not I add a trailing space at the end, but the second two don't work even when they're identical (no trailing whitespace).

int main(){
    string s;
    cin >> s;
    
    // Remove spaces from string
    for (auto it = s.begin(); it != s.end();){
        if (*it == ' ') s.erase(it);
        else ++it;
    }
    
    // Caculate grid sizes
    int floor = sqrt(s.size());
    int ceiling = sqrt(s.size()) + 1;
    
    if (floor * ceiling < s.size())
        (floor < ceiling)? floor += 1 : ceiling += 1;
    
    // Create grid
    vector<string> encrypted(ceiling);
    for (int i = 0; i < floor; ++i) {
        for (int j = 0; j < ceiling; ++j) {
            encrypted[j] += s[j + i * ceiling];
        }
    }
    // Ugly output method to remove trailing spaces
    for (int i = 0; i < encrypted.size(); ++i) {
        std::cout << encrypted[i];
        if (i < encrypted.size()-1) std::cout << " ";
    }
    std::cout << std::endl;
    
    return 0;
}

I am facing same issue for test cases 1,3,6. I downloaded cases and output is correct but still test cases are failing. Anyone figured out cause?

After proceesing the input, my 2D matrix looked like this. When printing it column wise, ensure that cout is not executed when a emply space is encountered i.e '\0'

I am also facing the same issue. I even trimmed the trailing spaces. Still its showing wrong answer.

for(int i=0;i

Same problem :(

I am having the exact same problem.

This is my code:

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String message = in.nextLine();

        String ciphertext = encrypt(message);
        System.out.println(ciphertext);
    }

    private static char[][] buildGrid(String message, int rows, int columns) {
        char[][] grid = new char[rows][columns];
        char[] chars = message.toCharArray();
        int row;
        int column;
        for (int i = 0; i < chars.length; i++) {
            row = i / columns;
            column = i % columns;
            grid[row][column] = chars[i];
        }

        return grid;
    }

    private static String encrypt(String message) {
        int length = message.length();
        double lengthSqrt = Math.sqrt(length);
        int rows = (int) Math.floor(lengthSqrt);
        int columns = (int) Math.ceil(lengthSqrt);
        rows = (rows * columns < length) ? columns : rows;

        char[][] grid = buildGrid(message, rows, columns);
        
        StringJoiner sj = new StringJoiner(" ");
        StringBuilder sb = new StringBuilder();
        int charsTotal = 0;
        for (int column = 0; column < columns; column++) {
            sb.setLength(0);
            for (int row = 0; row < rows; row++) {
                if (charsTotal <= length) {
                    sb.append(grid[row][column]);
                    charsTotal++;
                }
            }
            sj.add(sb.toString());
        }

        return sj.toString();
    }
}

I had the same mistake and i solved it by checking whether it is a letter before printing it . You can identify your extra spaces by printing '#' or anything else instead of a space . Also , the row calculation is wrong when applied to second test case . I calculated row by Ceil(len of input / col)

Check if the character is null before printing.

could you please explain the for loop part? specialy sm[i::c] part

NICE AND SORT

Same idea but a little simpler:

import math
s = input()
w = int(math.ceil(len(s.strip())**.5))
print(" ".join([s[i::w] for i in range(w)]))

No need to calculate r here.

take care of the space(blank) and it works.

i am facing the same problem

Same problem here. My compiler outputs "fto ehg ee dd" for Sample #2 but Hackerrank outputs "fto ehg ee". Same thing with Test case #6, which is one of the test cases that Hackerrank graded "wrong". Here's my code:

https://www.hackerrank.com/challenges/encryption/submissions/code/11082541

I have the same problem. Spent a decent amount of time trying to find the bug, but I put it in Ideone and got the right answer.

http://ideone.com/dttF2J

I had the same problem. You are basically going out of bounds when reading the string ( which is a memory error ) The HackerRank compiler just stops executing the code when you go out of bounds. Suggestion : Check for boundary conditions

I'm guessing because you never initialized the default value in arr, so you don't know if its going to be a space character. So, if (arr[r][c] != ' ') is useless.

if (arr=='\0'); break

use this :----->

if(grid[j][i]==(char) 0){ break; }

Ignore the '\0' characters when displaying the output.

just add the condition if you are working in java char[i][j]!=0 since a emty cell in char array is 0 in java, here is my solution for that https://www.hackerrank.com/challenges/encryption/submissions/code/16072054

In this scenario if u used floor and ceil method then it gives row=2,col=3.. Actually to store that much data we need row=2,col=4.. So I done it and directly prints the characters without storing in the 2D array..

Same issue here, correct results are marked as failures:

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.next();
        int[] gridSize = findMinimumArea(s.length());
        char[][] grid = new char[gridSize[0]][gridSize[1]];
        
        int cursor = 0;
        for(int row = 0; row < gridSize[0]; row++) {
            for(int col = 0; col < gridSize[1]; col++) {
                if(cursor >= s.length()) {
                    break;
                } else {
                    grid[row][col] = s.charAt(cursor);
                    cursor++;
                }
            }
        }
        
        String ans = "";
        for(int col = 0; col < gridSize[1]; col++) {
            for(int row = 0; row < gridSize[0]; row++) {
                ans += grid[row][col];
            }
            ans += " ";
        }
        
        System.out.println(ans.trim());
        
    }
    
    // [rows, cols]
    private static int[] findMinimumArea(int n) {
        int min = (int)Math.round(Math.floor(Math.sqrt(n)));
        int max = (int)Math.ceil(Math.sqrt(n));
        for(int row = min; row <= max; row++) {
            for(int col = row; col <= max; col++) {
                if(row * col >= n) {
                    return new int[]{row,col};
                }
            }
        }
        throw new Error("shouldn't get here : " + n);
    }
}

Actually square is unique, there is no question of max/min.
Will change the statement