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In this challenge, you are required to shift characters in an inputted string n number of times and output the shifted string

Input

Input will first contain a string. In the next line, an integer, which denotes n will be present.

Output

  • If n is positive, shift the characters in the string to the right n times.
  • If n is negative, shift the characters in the string to the left n times.
  • If n is zero, don't shift the characters in the string.

After shifting (except when n is zero), print the shifted string.

Notes

  • The string will not be empty or null.
  • The string will not be longer than 100 characters and will only contain ASCII characters in range (space) to ~(tilde) (character codes 0x20 to 0x7E, inclusive). See ASCII table for reference.
  • The shift is cyclic.
  • The number n may be positive, negative, or zero.
  • n will always be greater than or equal to -1000 and lesser than or equal to 1000
  • You may take input via stdin or from command line arguments
  • The shifted string must be outputted in the stdout (or closest equivalent)
  • You may write a full program or a function which takes input and outputs the string in stdout or closest equivalent

Test Cases

1)

Hello world!
5             -->orld!Hello w

2)

Testing...
-3            -->ting...Tes

3)

~~~
1000          -->~~~

4)

12345
0             -->12345

5)

ABA
17            -->BAA

Scoring

This is , so the shortest submission (in bytes) wins.

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26 Answers 26

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5
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Pyth, 4 bytes

.>zQ

This is almost similar to my CJam 5 byte version, except that Pyth as a auto-eval input operator Q.

.>              # Cyclic right shift of 
  z             # Input first line as string
   Q            # Rest of the input as evaluated integer

Try it online here

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3
  • \$\begingroup\$ Exactly the same solution as this :-) \$\endgroup\$ Commented May 30, 2015 at 9:07
  • \$\begingroup\$ @CoolGuy Its pretty straight forward. Though, I did not see this in sandbox.. \$\endgroup\$ Commented May 30, 2015 at 9:11
  • \$\begingroup\$ Seems to no longer work for some reason. Here's a working alternative, also 4 bytes. \$\endgroup\$ Commented Aug 1, 2018 at 18:53
4
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Javascript (ES5), 55 52 bytes

p=prompt;with(p())p(slice(b=-p()%length)+slice(0,b))

Commented:

p = prompt; // store a copy of prompt function for reuse
with(p()) // extend scope chain with first input
    p( // print result
        slice(b = -p() % length) // take second input negated and modulo length
        +                        // and slice string by result
        slice(0, b) // concatenate with opposite slice
    )
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2
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CJam, 5 bytes

llim>

This is pretty straight forward.

l               e# Read the first line
 li             e# Read the second line and convert to integer
   m>           e# Shift rotate the first string by second integer places

Try it online here

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4
  • 1
    \$\begingroup\$ Would this fall under built-in functions? \$\endgroup\$ Commented May 30, 2015 at 18:12
  • \$\begingroup\$ @LegionMammal978 It is a built in function. But OP does not restrict the usage of built ins \$\endgroup\$ Commented May 30, 2015 at 18:22
  • 1
    \$\begingroup\$ Built-in functions are standard loopholes. \$\endgroup\$ Commented May 30, 2015 at 18:37
  • 5
    \$\begingroup\$ @LegionMammal978 you point to an answer which has almost 50-50 up/down votes. That is not a community decision. \$\endgroup\$ Commented May 30, 2015 at 18:45
2
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C, 93 bytes

main(a,v,n)char**v;{a=v[2]-v[1]-1;n=atoi(v[2]);a=a*(n>0)-n%a;printf("%s%.*s",v[1]+a,a,v[1]);}

More clear is the function-argument version that was modified to make the command line-argument version

f(s,n,c)char*s;{c=strlen(s);c=c*(n>0)-n%c;printf("%s%.*s",s+c,c,s);}

This one is only 68 bytes, which just goes to show how disadvantaged C is when dealing with command line arguments.

If the shift, n, is positive then strlen(s)-n%strlen(s) is the offset and if n is negative the offset is -n%strlen(s). The printf prints from the offset, c, to the end of the string, and then the final c characters from the beginning.

Examples:

$ ./rotstr "Hello world!" 5
orld!Hello w
$ ./rotstr "Testing..." -3
ting...Tes
$ ./rotstr "~~~" 1000
~~~
$ ./rotstr "12345" 0
12345
$ ./rotstr "ABA" 17
BAA
$ ./rotstr "Hello world!" -16
o world!Hell
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7
  • \$\begingroup\$ It doesn't work as expected for me. When v[2] is "1", the code simply outputs the string without any modifications. And only "~~~" and "12345" works. The rest of them gives wrong outputs. If they all rotated one more time, it would have been correcet. \$\endgroup\$ Commented May 30, 2015 at 11:50
  • \$\begingroup\$ I've tested it with both gcc and (with a slight modification main(a,v,n) -> n;main(a,v)) clang on linux and it works as expected. For gcc I'm using version 5.1.0 and compiling with gcc -o rotstr rotstr.c. What compiler are you using? \$\endgroup\$ Commented May 30, 2015 at 13:01
  • \$\begingroup\$ Tried making n global too. Same issue. I compiled using gcc file.c -o file. I'm using GCC 4.8.1 on windows. Is there any undefined behavior in your code? \$\endgroup\$ Commented May 30, 2015 at 13:04
  • \$\begingroup\$ Replacing v[2]-v[1]-1 with strlen(v[1]) might make a difference, that's the only place I can think of something subtle going on. Unfortunately I don't have access to a windows machine to test. \$\endgroup\$ Commented May 30, 2015 at 13:15
  • \$\begingroup\$ Yes. The code worked when I changed that. \$\endgroup\$ Commented May 30, 2015 at 13:16
2
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Python 3, 45 bytes

s=input();n=int(input());print(s[-n:]+s[:-n])

The core of the program is

s[-n:]+s[:-n]

All the rest is just clumsy work with I/O.

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2
  • 2
    \$\begingroup\$ This fail for the last ABA 17 test case, and would in general if |n| > length of string \$\endgroup\$ Commented May 30, 2015 at 16:29
  • \$\begingroup\$ if you use n=int(input())%len(s);, it would work for integers greater than the string length, but require 7 more characters \$\endgroup\$ Commented May 30, 2015 at 17:11
2
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K, 8 7 bytes

{|x!|y}

There is already a primitive "rotate" (!) which performs a generalization of this operation for lists. K strings are lists of characters, so it applies. The spec favors CJam and Pyth a bit, though, because K's rotate happens to go the opposite direction of what is desired. Wrapping ! in a function and negating the implicit argument x will do what we want:

  f:{(-x)!y}
{(-x)!y}
  f[5;"Hello world!"]
"orld!Hello w"
  f[-3;"Testing..."]
"ting...Tes"
  f[17;"ABA"]
"BAA"

A slightly shorter approach, suggested by kirbyfan64sos, is to do away with the parentheses and negation in favor of reversing the string (|) before and after the rotation.

If it weren't for this impedance mismatch, the solution would be simply

!

Called identically:

  f:!
!
  f[5;"Hello, World!"]
", World!Hello"
  f[-5;"Hello, World!"]
"orld!Hello, W"
  f[0;"Hello, World!"]
"Hello, World!"
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2
  • 1
    \$\begingroup\$ Would reversing the string with |, rotating that, and reversing it again yield the same result? If so, you can cut off a character. \$\endgroup\$ Commented May 30, 2015 at 22:24
  • \$\begingroup\$ Good point! That would work. \$\endgroup\$ Commented May 30, 2015 at 23:01
2
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Casio Basic, 27 bytes

StrRotate s,s,-n:Print s

As it turns out, there's a built-in for this on the Casio ClassPad! But it works in reverse, hence -n.

24 bytes for the code, 3 bytes to specify s,n as arguments.

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1
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Pip, 10 bytes

This could quite possibly be improved further. Still, for a language with no shift operator, 10 bytes ain't bad.

a@_M-b+,#a

Explanation:

            a, b are command-line args (implicit)
       ,#a  range(len(a))
    -b+     range(-b, len(a)-b)
a@_M        map(lambda x: a[x], range(-b, len(a)-b))
            Concatenate the list and print (implicit)

It works because string and list indexing in Pip is cyclical: "Hello"@9 == "Hello"@4 == "o".

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1
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rs, 180 chars

^(-\d+) (.*)/\1 \2\t
+^(-\d+) (.)(.*?)\t(.*)$/\1 \3\t\2\4
^(-\d+) \t/\1 
^(-?)(\d+)/\1 (_)^^(\2)
+_(_*) (.*)(.)$/\1 \3\2
^- /- \t
+^- (.*?)\t(.*?)(.)$/- \1\3\t\2
^-? +/
\t/

Live demo.

Most of this is reversing the string if the input number is negative. I took advantage of the fact that only some ASCII characters are valid input and used the tab to my advantage.

Note that I had to cheat a little: since rs is a single-line text modifier, I had to use <number> <text> as the input format.

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1
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Java, 167

enum S{;public static void main(String[]r){int n=-Integer.parseInt(r[1]),l=r[0].length();while(n<0)n+=l;n%=l;System.out.print(r[0].substring(n)+r[0].substring(0,n));}}

Takes the input through the command line.

funny enough, originally I had accidentally reversed how the string was supposed to be shifted. But fixing that mistake was shorter to just multiply n by -1 then to write the logic properly.

expanded:

enum Shift{
    ;
    public static void main(String[]args){
        int n=-Integer.parseInt(args[1]),length=args[0].length();
        while(n<0)n+=length;
        n%=length;
        System.out.print(args[0].substring(n)+args[0].substring(0,n));
    }
}
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6
  • \$\begingroup\$ Why do you have enum S{; ... }? \$\endgroup\$ Commented Jun 2, 2015 at 5:40
  • 1
    \$\begingroup\$ I opted to write the full program because 9 bytes wasn't really going to make a huge difference. Also it's a reminder when I look back to prefer enum S{;...} over class S{...} because (even though they take up the same number of bytes in this example) if I ever need to have an instance of the class, it takes one byte more with the enum version: enum S{X;...}. This helps if I want to declare a method or variable in the class without having to use the static keyword or explictly instantiating a new object of the class. \$\endgroup\$ Commented Jun 3, 2015 at 0:28
  • \$\begingroup\$ Wow! Nice. Never knew that enums can be used like this! \$\endgroup\$ Commented Jun 3, 2015 at 5:49
  • 1
    \$\begingroup\$ I know it's been close to two years since you've posted this, but you can golf a few things. Integer.parseInt can be new Integer (-5 bytes); and n%=l; can be removed if you change r[0].substring(n)+ to r[0].substring(n%=l)+ (-2 bytes). Also, you might want to specify this is Java 6, because in Java 7 or higher an enum with main-method isn't possible anymore. \$\endgroup\$ Commented May 22, 2017 at 7:29
  • \$\begingroup\$ too lazy to bother edit, but duely noted for the savings. \$\endgroup\$ Commented May 28, 2017 at 16:50
1
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Japt, 2 bytes

éV

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1
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Julia 0.6, 31 bytes

s|n=String(circshift([s...],n))

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1
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PHP>=7.1, 88 Bytes

for([,$s,$t]=$argv;$t;)$s=$t<0?substr($s,1).$s[!$t++]:$s[-1].substr($s,!$t--,-1);echo$s;

Testcases

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0
1
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Pascal (ISO standard 10206, “Extended Pascal”), 120 Bytes

program p(input,output);var s:string(100);n:integer;begin read(s,n);n:=(-n)mod length(s);writeLn(subStr(s,n+1),s:n)end.

Ungolfed:

program shiftCharactersInAString(input, output);
var
    line: string(100);
    n: integer;
begin
    read(line, n);
    { In Pascal, the result of the `mod` operation is non-negative. }
    n := (-n) mod length(line);
    { In Pascal, string indices are 1‑based. }
    writeLn(subStr(line, n + 1), line:n)
    { `line:n` is equivalent to `subStr(line, 1, n)` (if `n` ≤ `length(line)`). }
end.
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1
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05AB1E, 6 5 3 bytes

(._

Inputs in the order integer,string.

Try it online or verify all test cases.

Explanation:

(    # Negate the first (implicit) input-integer
 ._  # Rotate the second (implicit) input-string that many times towards the left
     # (which supports negative integers as rotating right instead)
     # (after which the result is output implicitly)

05AB1E (legacy), 6 5 bytes

sg+FÁ

Inputs in the order integer,string.

Try it online or verify all test cases.

Explanation:

s      # Swap so the two (implicit) inputs are in reversed order on the stack
 g     # Pop and push the length of the top input-string
  +    # Add it to the input-integer
   F   # Loop that many times
    Á  #  And rotate once towards the right during every iteration
       #  (using the last implicit input-string in the first iteration)

Since the legacy version of 05AB1E only has builtins for Rotate once towards the right/left, and not Rotate \$n\$ amount towards the right/left, I loop \$length + input\$ amount of times and rotate that many times towards the right.

For example:

  • "Testing..." and -3 will rotate \$10 + -3 = 7\$ times towards the right, resulting in ting...Tes.
  • "Hello world" and 5 will rotate \$11 + 5 = 16\$ times towards the right, resulting in worldHello.

The new version of 05AB1E does have a builtin for rotating \$n\$ amount of times towards the left, which is ._, saving 2 bytes.

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1
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Wolfram Language (Mathematica), 40 bytes

StringJoin@RotateRight[Characters@#,#2]&

StringJoin@RotateLeft[Characters@#,-#2]&

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1
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vemf, 1 byte

+

Left argument is string, right argument is offset.

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1
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Vyxal, 1 byte

ǔ

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Takes the number then the string. Prepend $ if you want the inputs the other way round.

Built-in solution for "rotate right". Rotation is modular so we don't have to worry about negative inputs.

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1
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Jelly, 3 bytes

N⁴ṙ

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Explanation

N⁴ṙ  # Main link. Takes an integer on the
     # left and a string on the right.
 ⁴   # Get the right argument (the string)
  ṙ  # And rotate it left by ↓ spaces
N    # The left argument negated
     # Implicit output
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1
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Python, 35 bytes

lambda s,n:s[(x:=-n%len(s)):]+s[:x]

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1
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Thunno J, \$ 8 \log_{256}(96) \approx \$ 6.58 bytes

LR_z0sAI

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No shift operator in Thunno so we have to do it ourselves.

Explanation

LR_z0sAI  # Implicit input        STACK: "Hello, World!", 5
LR        # Length range          STACK: 5, [0..12]
  _       # Swapped subtract      STACK: [-5..7]
   z0     # First input           STACK: [-5..7], "Hello, World!"
     sAI  # Swap and index        STACK: ['o', 'r', 'l', 'd', '!', 'H', 'e', 'l', 'l', 'o', ',', ' ', 'W']
          # J flag joins          STACK: "orld!Hello, W"
          # Implicit output
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0
\$\begingroup\$

Stax, 2 bytes

|)

Run and debug it

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0
\$\begingroup\$

Perl 5 + -palF, 26 bytes

$_=substr$_.$_,@F-<>%@F,@F

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0
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Ruby, 28 bytes

->s,n{s.chars.rotate(-n)*""}

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0
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HP‑41C series, 4 Bytes

Input string is located in alpha register and number to rotate on top of the stack (i. e. the X register). An extended functions module or HP‑41CX is required.

CHS     1 Byte   flip sign because AROT acts just the opposite as required
AROT    2 Bytes  rotate alpha register according to value in X register
AVIEW   1 Byte   display contents of alpha register
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0
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UiuaSBCS, 2 bytes

↻¯

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↻ rotate shifts cyclically to the right, so a ¯ negate has to be added to reverse the direction.

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