Rotate a number

Question

Given a positive number n, rotate its base-10 digits m positions rightward. That is, output the result of m steps of moving the last digit to the start. The rotation count m will be a non-negative integer.

You should remove leading zeroes in the final result, but not in any of the intermediate steps. For example, for the test case 100,2 => 1, we first rotate to 010, then to 001, then finally drop the leading zeroes to get 1.

Tests

n,m => Output

123,1 => 312
123,2 => 231
123,3 => 123
123,4 => 312
1,637 => 1
10,1 => 1
100,2 => 1
10,2 => 10 
110,2 => 101
123,0 => 123
9998,2 => 9899

Answer 1

Japt -N, 2 bytes

Takes m as a string and V=n as an integer or string, outputs an integer. Prepend s or ì for +1 byte if we have to take both as integers.

éV

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Answer 2

R, 51 bytes

function(n,m,p=10^nchar(n))sum(n*p^(0:m))%/%10^m%%p

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Numeric solution (that fails for combinations of n & m that cause it to exceed R's numeric range): chains the digits of n, m times (so: 123 => 123123123123 for m=4) and then calculates DIV 10^m (so: 12312312 for m=4) MOD 10^digits(n) (so: 312).

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R, 61 53 bytes

Edit: -8 bytes thanks to Giuseppe

function(n,m,N=nchar(n),M=10^(m%%N))n%%M*10^N/M+n%/%M

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Text-based function that Rotates by combining the two parts of the number together, so does not go out of numeric range: puts the last (m MOD digits(n)) digits of n first, followed by the other digits of n.

Answer 3

Python 3, 61 57 bytes

i=input
n=i()
k=int(i())%len(n)
print(int(n[-k:]+n[:-k]))

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Uses string slicing to move the last k digits at the beginning and converts it to an integer to remove the leading zeroes.

-4 bytes thanks to Lyxal

Answer 4

05AB1E, 4 bytes

(._ï

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Explanation

(._ï
(     : get negative of m
 ._   : rotate n left negative m times
   ï  : remove leading zeros

Answer 5

Perl 5 + -pl, 26 bytes

eval'$_=chop.$_;'x<>;$_|=0

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Answer 6

Python 3, 39 bytes

lambda n,m:int(((n*m)[-m:]+n)[:len(n)])

Try it online! Or see the test-suite.

How?

Rotating n right by m is the same as rotating n right by m modulo length n (m%len(n)), which is the concatenation of the last m%len(n) digits with the first len(n)-m%len(n) digits.

A simple slice would give us

lambda n,m:int(n[-m%len(n):]+n[:-m%len(n)])

for 43 bytes. To remove the need for the repeated -m% we can instead concatenate the last m%len(n) digits with all the digits of n and then take the first len(n) digits. This is

lambda n,m:int((n[-m%len(n):]+n)[:len(n)])

for 42 bytes. The n[-m%len(n):] can then be replaced with taking the rightmost m digits of m ns concatenated together, (n*m)[-m:] giving us the 39 byte solution.

Answer 7

[Taxi], 1698 1678 bytes

^{No need to wrap single-byte plan names in quote marks. 0.6% byte reduction!}

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to Chop Suey.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Addition Alley.1 is waiting at Starchild Numerology.Go to Starchild Numerology:n 1 l 1 l 1 l 2 l. Pickup a passenger going to Addition Alley.Go to Addition Alley:w 1 r 3 r 1 r 1 r.Pickup a passenger going to The Underground.Go to Chop Suey:n 1 r 2 r.[1]Switch to plan 2 if no one is waiting.Pickup a passenger going to Narrow Path Park.Go to Narrow Path Park:n 1 l 1 r 1 l.Go to Chop Suey:e 1 r 1 l 1 r.Switch to plan 1.[2]Go to Narrow Path Park:n 1 l 1 r 1 l.Switch to plan 3 if no one is waiting.Pickup a passenger going to Chop Suey.Go to Chop Suey:e 1 r 1 l 1 r.Switch to plan 2.[3]Go to Chop Suey:e 1 r 1 l 1 r.[a]Go to The Underground:s 1 r 1 l.Switch to plan b if no one is waiting.Pickup a passenger going to The Underground.Go to Fueler Up:s.Go to Chop Suey:n 3 r 1 l.Pickup a passenger going to Chop Suey.Switch to plan a.[b]Go to Chop Suey:n 2 r 1 l.[4]Switch to plan 5 if no one is waiting.Pickup a passenger going to Narrow Path Park.Go to Narrow Path Park:n 1 l 1 r 1 l.Go to Chop Suey:e 1 r 1 l 1 r.Switch to plan 4.[5]Go to Narrow Path Park:n 1 l 1 r 1 l.[c]Switch to plan d if no one is waiting.Pickup a passenger going to KonKat's.Go to KonKat's:e 1 r.Pickup a passenger going to KonKat's.Go to Narrow Path Park:n 2 l.Switch to plan c.[d]Go to KonKat's:e 1 r.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s.Pickup a passenger going to The Babelfishery.Go to KonKat's:n.Go to The Babelfishery:s.Pickup a passenger going to Post Office.Go to Post Office:n 1 l 1 r.

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I chose to get fired rather than sacrifice the bytes required to return to the garage at the end. I have checked both very long inputs and very long rotations and the net gain is positive so you never run out of gas.

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Formatted for legibility and with comments:

[ Pick up the inputs, add 1 to the second, and chop the first into pieces. ]
Go to Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to Chop Suey.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l 1 r.
Pickup a passenger going to Addition Alley.
1 is waiting at Starchild Numerology.
Go to Starchild Numerology:n 1 l 1 l 1 l 2 l. 
Pickup a passenger going to Addition Alley.
Go to Addition Alley:w 1 r 3 r 1 r 1 r.
Pickup a passenger going to The Underground.
Go to Chop Suey:n 1 r 2 r.

[ Reverse the order the charaters are stored in so we can right-shift instead of left-shift. ]
[1]
Switch to plan 2 if no one is waiting.
Pickup a passenger going to Narrow Path Park.
Go to Narrow Path Park:n 1 l 1 r 1 l.
Go to Chop Suey:e 1 r 1 l 1 r.
Switch to plan 1.
[2]
Go to Narrow Path Park:n 1 l 1 r 1 l.
Switch to plan 3 if no one is waiting.
Pickup a passenger going to Chop Suey.
Go to Chop Suey:e 1 r 1 l 1 r.
Switch to plan 2.
[3]
Go to Chop Suey:e 1 r 1 l 1 r.

[ Loop the required times, rotating the passengers at Chop Suey each time. ]
[a]
Go to The Underground:s 1 r 1 l.
Switch to plan b if no one is waiting.
Pickup a passenger going to The Underground.
Go to Fueler Up:s.
Go to Chop Suey:n 3 r 1 l.
Pickup a passenger going to Chop Suey.
Switch to plan a.
[b]
Go to Chop Suey:n 2 r 1 l.

[ Reverse the character order again. ]
[4]
Switch to plan 5 if no one is waiting.
Pickup a passenger going to Narrow Path Park.
Go to Narrow Path Park:n 1 l 1 r 1 l.
Go to Chop Suey:e 1 r 1 l 1 r.
Switch to plan 4.
[5]
Go to Narrow Path Park:n 1 l 1 r 1 l.

[ Concatenate the passengers at Narrow Path Park. ]
[c]
Switch to plan d if no one is waiting.
Pickup a passenger going to KonKat's.
Go to KonKat's:e 1 r.
Pickup a passenger going to KonKat's.
Go to Narrow Path Park:n 2 l.
Switch to plan c.

[ Convert to a number to remove leading zeros and then back to a string so the Post Office can handle it. ]
[d]
Go to KonKat's:e 1 r.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s.
Pickup a passenger going to The Babelfishery.
Go to KonKat's:n.
Go to The Babelfishery:s.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 l 1 r.

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Answer 8

MATL, 3 bytes

YSU

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Takes n as a string and m as an integer.

Explanation

YS   % Shift first input second input number of times
  U  % Convert to integer to remove leading 0s

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MATL, 5 bytes

ViYSU

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This answer takes both the inputs as integers.

Answer 9

Charcoal, 9 bytes

ＩＩ⭆θ§θ⁻κη

Try it online! Link is to verbose version of code. Explanation:

   θ        Input `n` as a string
  ⭆         Map over characters and join
       κ    Current index
      ⁻     Subtract
        η   Input `m`
    §       Cyclically indexed into
     θ      Input `n` as a string
 Ｉ          Cast to integer
Ｉ           Cast to string
            Implicitly print

Conveniently if you try to Subtract an integer and a string then the string gets cast to integer.

Answer 10

Python 3, 47 bytes

f=lambda n,m:m and f(n[-1]+n[:-1],m-1)or int(n)

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Inputs \$n\$ as a string and \$m\$ as an integer.
Returns rotated \$n\$ as an integer.

Answer 11

APL+WIN, 8 7 bytes

Prompts for n as integer and m as string:

⍎(-⎕)⌽⎕

Try it online! Courtesy of Dyalog Classic

Answer 12

JavaScript (ES6), 36 bytes

Expects (m)(n), where n is a string and m is either a string or an integer.

m=>g=n=>m--?g(n%10+n.slice(0,-1)):+n

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Answer 13

C (gcc)-lm, ^{65 \$\cdots\$ 56} 55 bytes

Saved a byte thanks to ceilingcat!!!

e;f(n,m){for(e=log10(n);m--;)n=n%10*exp10(e)+n/10;m=n;}

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Inputs integers \$n\$ and \$m\$.
Base-10 digitally rotates \$n\$ right \$m\$-times and returns it.

Answer 14

Pyth, 4 bytes

v.>z

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Explanation

v.>zQ
    Q  : first line of input evaluated 
   z   : second line of input as string
 .>    : cyclically rotate second line right by number in first line
v      : evaluate to remove leading 0s

Answer 15

Keg, -hr, 11 bytes

÷(¿|")⑷⅍⑸⅀ℤ

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Explained

÷(¿|")⑷⅍⑸⅀ℤ
÷               # Split m into individual numbers
 (¿|")          # n times, shift the stack right
      ⑷⅍⑸      # turn each character into a string
           ⅀ℤ   # sum stack and convert to integer. `-hr` prints it as integer

Answer 16

Java (JDK), 66 bytes

(n,x)->new Long((""+n+n).substring(x=(n=(""+n).length())-x%n,x+n))

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Answer 17

J, 11 bytes

(".@|.":)~-

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How it works

Uses @Bubbler's tacit trick for (F x) G (H y) = (G~F)~H.

(".@|.":)~-
          - negate y to shift right
(       )~  flip arguments, so ((-y) ".@|. (":x))
      ":    convert x to string
    |.      shift that by negated y
 ".@        and convert back to number

Answer 18

Io, 89 bytes

Uses a reduce trick to assign different lines of STDIN to variables.

File standardInput readLines reduce(a,b,a splitAt(-b asNumber)reverse join)asNumber print

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Io, 56 bytes

method(a,b,doString(a splitAt(-b asNumber)reverse join))

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Answer 19

Ruby -nl, 34 bytes

->m{($_*-~m*2)[~~/$/*m,~/$/].to_i}

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Takes \$n\$ from STDIN and \$m\$ as an argument. Concatenates \$n\$ \$2(m+1)\$ times, then from this string takes the substring of length \$d\$ (where \$d\$ is the number of digits in \$n\$) that begins \$m(d+1)\$ characters from the end. In the code, $_ is \$n\$ and ~/$/ gives \$d\$.

Example

For \$n=123\$, \$m=2\$:

1. Concatenate \$n\$ \$2(m+1)=6\$ times: 123123123123123123
2. Count back from the end \$m(d+1)=8\$ characters: 123123123123123123
3. Take substring of length \$d=3\$: 123123123123123123

Answer 20

Python 3.8 (pre-release), ⁴² 40 bytes

lambda x,r:int(x[(a:=-r%len(x)):]+x[:a])

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Answer 21

Jelly, (4?) 5 bytes

^{4 if we may accept a list of digits (remove the leading D).}

DṙN}Ḍ

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How?

DṙN}Ḍ - Link: integer, n; integer, m
D     - convert to base ten
   }  - use m as the input of:
  N   -   negate
 ṙ    - rotate (n) left by (-m)
    Ḍ - convert from base ten

Answer 22

CJam, 10 7 6 bytes

Saved 3 bytes by remembering that you can preform most array operations on strings.

-1 byte from @my pronoun is monicareinstate noting that m> takes arguments in either order.

rr~m>~

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Explanation:

rr       Read two string inputs
  ~      Parse m to number
   m>    Rotate n string right m times
     ~   Parse n to number to remove leading zeros
         (implicit) output

Old version, 7 bytes:

q~\sm>~

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Explanation:

q~        Take input as a string, evaluate to two numbers
  \       Swap order
   s      Convert n to string
    m>    Rotate n string right m times
      ~   Parse n to number to remove leading zeros
          (implicit) output

Answer 23

APL (Dyalog Extended), 4 bytes (SBCS)

Anonymous tacit infix function. Takes string n as right argument and number m as left argument.

⍎-⍛⌽

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⍎ execute the result of

-⍛ negating the left argument, then using that to

⌽ cyclically rotate the right argument

Answer 24

(non-competing) L=tn, 10

cα;θ2$cD=0

Explanation (in order of execution):

cα;         -  Turn first number into a list of digits
    2$c     -  Fetch second number as a number 
   θ        -  Shift list from first step by number from second step
       D=0  -  Drop while equals to zero

It prints result as a string.

Note: The language is in progress and I did add stuff to code golf this. So it is not a competing submission.

Answer 25

Ruby, 44 40 bytes

->a,b{a.to_s.chars.rotate(-b).join.to_i}

-4 from Dingus.

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Ruby, 19 bytes

->a,b{a.rotate(-b)}

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taking a list of digits

Answer 26

Wolfram Language (Mathematica), 43 bytes

FromDigits@RotateRight[IntegerDigits@#,#2]&

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Answer 27

Retina 0.8.2, 29 bytes

,.+
$*_
+`(.*)(\d)_
$2$1
^0+

Try it online! Link includes test cases. Takes input as n,m. Explanation:

,.+
$*_

Convert m to unary.

+`(.*)(\d)_
$2$1

Rotate n m times. This is O(m³) because of the way the regex backtracks trying to find a second match. Right-to-left matching, anchoring the match at the start, or rewriting the code to take input as m,n would reduce the time complexity (at a cost of a byte of course).

^0+

Delete leading zeros.

Answer 28

Scala, 61 bytes

(n,m)=>{val s=n+""size;val(a,b)=n+""splitAt s-m%s;b++a toInt}

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Answer 29

PHP, 45 43 bytes

Saved 2 bytes, realized we can shorten the variable names.

<?=(int)(substr($s,-$n).substr($s,0,-$n))?>

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Explanation:

<?= ?>       Shorthand for <?php echo ;?>
  (int)      Typecast string to int, removes 0s from prefix
   substr()  substr(string,start,[length]), returns part of string, 
             if range go out of bounds, starts again from the opposite end.
             Basically returns part of from a 'circular' string.    
  

Answer 30

JavaScript (V8), 47 bytes

(n,m,k=(e=n+'').length)=>+(e+e).substr(k-m%k,k)

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