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(" ")
    }
}

Why use 2D array when it can be done in single iteration...

void encryption(string s) {
    int row=ceil(sqrt(s.length()));
    int column=floor(sqrt(s.length()));
    for(int i=0;i<row;i++)
    {
        for(int j=i;j<s.length();j=j+row)
        cout<<s[j];
            
        cout<<" ";
    }
}

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 = ' ')

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.

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);
    }
}

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

NICE AND SORT

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

I would like to help you, but I can't because I don't know what you are doing.

I just resubmitted my solution and it is still accepted, so the system seems to be working correctly.

i too have the same problem ... please let me know if you find some mistake

same problem here

I am facing the same problem.

Actually the problem in my case is that the string serialization was ok but it was producing the wrong equals since I was adding the null character '\u0000' to the result.

You mean to say this:

1>point int to the instructions ..............................., where is floor function and is ceil function.