Circular shift string












3












$begingroup$


So,basically I am working on an algorithm for circular shifting a string upto a position.There are two parameters required here.




  1. String of text to shift

  2. Number of shifts required.


For example:



Given string: "Hello"
Shifts: 3
Final: "ollHe"


Ignore the limited validation here.



import java.lang.*;
import java.util.*;

public class Shift
{
public String shifting(String s,Integer sh)
{
Integer shifts=sh;
String string=s;
char shifted=new char[50];
Integer pos=0;
for(int i=string.length();i>shifts;i--)
{
System.out.println(""+i);
pos++;
shifted[pos]=string.charAt(i-1);
}
System.out.println("Shifting complete");
for(int j=0;j<shifts;j++)
{
System.out.println(""+j);
pos++;
shifted[pos]=string.charAt(j);
}
return new String(shifted);
}
public static void main(String args) {
System.out.println("Enter the string ");
Scanner sc=new Scanner(System.in);
String string=sc.nextLine();
System.out.println("So you have entered:"+string);
System.out.println("Enter the number of shifts:");
Integer shifts=sc.nextInt();
System.out.println("number of shifts is:"+shifts.toString());
System.out.println("Shifted string:"+new Shift().shifting(string,shifts));
}
}


`
Give your views on the code here.










share|improve this question











$endgroup$








  • 2




    $begingroup$
    assuming that the String is at most 50 characters is dangerous.
    $endgroup$
    – MrSmith42
    Dec 13 '16 at 12:57










  • $begingroup$
    You should add an example to your question.
    $endgroup$
    – MrSmith42
    Dec 13 '16 at 12:57






  • 1




    $begingroup$
    (The example still looks off in rev. 3: how did the o and ll swap places?) What about string.substring(shifts) + string.substring(0, shifts)?
    $endgroup$
    – greybeard
    Dec 13 '16 at 14:04










  • $begingroup$
    ("backticks" are for inline code snippets; there is a {}-button and a keyboard shortcut (<Ctrl>+k) in the edit-window to turn the selection into a code block (prefixing every line with four additional spaces if the selection doesn't already look like this (then it removes four spaces from the start of every line).))
    $endgroup$
    – greybeard
    Dec 13 '16 at 21:45










  • $begingroup$
    @greybeard..You are right.I was using the wrong way to do it.Hello->oHell->loHel->lloHe.Can you suggest the correct way?
    $endgroup$
    – Javasamurai
    Dec 14 '16 at 6:29


















3












$begingroup$


So,basically I am working on an algorithm for circular shifting a string upto a position.There are two parameters required here.




  1. String of text to shift

  2. Number of shifts required.


For example:



Given string: "Hello"
Shifts: 3
Final: "ollHe"


Ignore the limited validation here.



import java.lang.*;
import java.util.*;

public class Shift
{
public String shifting(String s,Integer sh)
{
Integer shifts=sh;
String string=s;
char shifted=new char[50];
Integer pos=0;
for(int i=string.length();i>shifts;i--)
{
System.out.println(""+i);
pos++;
shifted[pos]=string.charAt(i-1);
}
System.out.println("Shifting complete");
for(int j=0;j<shifts;j++)
{
System.out.println(""+j);
pos++;
shifted[pos]=string.charAt(j);
}
return new String(shifted);
}
public static void main(String args) {
System.out.println("Enter the string ");
Scanner sc=new Scanner(System.in);
String string=sc.nextLine();
System.out.println("So you have entered:"+string);
System.out.println("Enter the number of shifts:");
Integer shifts=sc.nextInt();
System.out.println("number of shifts is:"+shifts.toString());
System.out.println("Shifted string:"+new Shift().shifting(string,shifts));
}
}


`
Give your views on the code here.










share|improve this question











$endgroup$








  • 2




    $begingroup$
    assuming that the String is at most 50 characters is dangerous.
    $endgroup$
    – MrSmith42
    Dec 13 '16 at 12:57










  • $begingroup$
    You should add an example to your question.
    $endgroup$
    – MrSmith42
    Dec 13 '16 at 12:57






  • 1




    $begingroup$
    (The example still looks off in rev. 3: how did the o and ll swap places?) What about string.substring(shifts) + string.substring(0, shifts)?
    $endgroup$
    – greybeard
    Dec 13 '16 at 14:04










  • $begingroup$
    ("backticks" are for inline code snippets; there is a {}-button and a keyboard shortcut (<Ctrl>+k) in the edit-window to turn the selection into a code block (prefixing every line with four additional spaces if the selection doesn't already look like this (then it removes four spaces from the start of every line).))
    $endgroup$
    – greybeard
    Dec 13 '16 at 21:45










  • $begingroup$
    @greybeard..You are right.I was using the wrong way to do it.Hello->oHell->loHel->lloHe.Can you suggest the correct way?
    $endgroup$
    – Javasamurai
    Dec 14 '16 at 6:29
















3












3








3


2



$begingroup$


So,basically I am working on an algorithm for circular shifting a string upto a position.There are two parameters required here.




  1. String of text to shift

  2. Number of shifts required.


For example:



Given string: "Hello"
Shifts: 3
Final: "ollHe"


Ignore the limited validation here.



import java.lang.*;
import java.util.*;

public class Shift
{
public String shifting(String s,Integer sh)
{
Integer shifts=sh;
String string=s;
char shifted=new char[50];
Integer pos=0;
for(int i=string.length();i>shifts;i--)
{
System.out.println(""+i);
pos++;
shifted[pos]=string.charAt(i-1);
}
System.out.println("Shifting complete");
for(int j=0;j<shifts;j++)
{
System.out.println(""+j);
pos++;
shifted[pos]=string.charAt(j);
}
return new String(shifted);
}
public static void main(String args) {
System.out.println("Enter the string ");
Scanner sc=new Scanner(System.in);
String string=sc.nextLine();
System.out.println("So you have entered:"+string);
System.out.println("Enter the number of shifts:");
Integer shifts=sc.nextInt();
System.out.println("number of shifts is:"+shifts.toString());
System.out.println("Shifted string:"+new Shift().shifting(string,shifts));
}
}


`
Give your views on the code here.










share|improve this question











$endgroup$




So,basically I am working on an algorithm for circular shifting a string upto a position.There are two parameters required here.




  1. String of text to shift

  2. Number of shifts required.


For example:



Given string: "Hello"
Shifts: 3
Final: "ollHe"


Ignore the limited validation here.



import java.lang.*;
import java.util.*;

public class Shift
{
public String shifting(String s,Integer sh)
{
Integer shifts=sh;
String string=s;
char shifted=new char[50];
Integer pos=0;
for(int i=string.length();i>shifts;i--)
{
System.out.println(""+i);
pos++;
shifted[pos]=string.charAt(i-1);
}
System.out.println("Shifting complete");
for(int j=0;j<shifts;j++)
{
System.out.println(""+j);
pos++;
shifted[pos]=string.charAt(j);
}
return new String(shifted);
}
public static void main(String args) {
System.out.println("Enter the string ");
Scanner sc=new Scanner(System.in);
String string=sc.nextLine();
System.out.println("So you have entered:"+string);
System.out.println("Enter the number of shifts:");
Integer shifts=sc.nextInt();
System.out.println("number of shifts is:"+shifts.toString());
System.out.println("Shifted string:"+new Shift().shifting(string,shifts));
}
}


`
Give your views on the code here.







java algorithm stack






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 14 '16 at 6:31







Javasamurai

















asked Dec 13 '16 at 12:38









JavasamuraiJavasamurai

11915




11915








  • 2




    $begingroup$
    assuming that the String is at most 50 characters is dangerous.
    $endgroup$
    – MrSmith42
    Dec 13 '16 at 12:57










  • $begingroup$
    You should add an example to your question.
    $endgroup$
    – MrSmith42
    Dec 13 '16 at 12:57






  • 1




    $begingroup$
    (The example still looks off in rev. 3: how did the o and ll swap places?) What about string.substring(shifts) + string.substring(0, shifts)?
    $endgroup$
    – greybeard
    Dec 13 '16 at 14:04










  • $begingroup$
    ("backticks" are for inline code snippets; there is a {}-button and a keyboard shortcut (<Ctrl>+k) in the edit-window to turn the selection into a code block (prefixing every line with four additional spaces if the selection doesn't already look like this (then it removes four spaces from the start of every line).))
    $endgroup$
    – greybeard
    Dec 13 '16 at 21:45










  • $begingroup$
    @greybeard..You are right.I was using the wrong way to do it.Hello->oHell->loHel->lloHe.Can you suggest the correct way?
    $endgroup$
    – Javasamurai
    Dec 14 '16 at 6:29
















  • 2




    $begingroup$
    assuming that the String is at most 50 characters is dangerous.
    $endgroup$
    – MrSmith42
    Dec 13 '16 at 12:57










  • $begingroup$
    You should add an example to your question.
    $endgroup$
    – MrSmith42
    Dec 13 '16 at 12:57






  • 1




    $begingroup$
    (The example still looks off in rev. 3: how did the o and ll swap places?) What about string.substring(shifts) + string.substring(0, shifts)?
    $endgroup$
    – greybeard
    Dec 13 '16 at 14:04










  • $begingroup$
    ("backticks" are for inline code snippets; there is a {}-button and a keyboard shortcut (<Ctrl>+k) in the edit-window to turn the selection into a code block (prefixing every line with four additional spaces if the selection doesn't already look like this (then it removes four spaces from the start of every line).))
    $endgroup$
    – greybeard
    Dec 13 '16 at 21:45










  • $begingroup$
    @greybeard..You are right.I was using the wrong way to do it.Hello->oHell->loHel->lloHe.Can you suggest the correct way?
    $endgroup$
    – Javasamurai
    Dec 14 '16 at 6:29










2




2




$begingroup$
assuming that the String is at most 50 characters is dangerous.
$endgroup$
– MrSmith42
Dec 13 '16 at 12:57




$begingroup$
assuming that the String is at most 50 characters is dangerous.
$endgroup$
– MrSmith42
Dec 13 '16 at 12:57












$begingroup$
You should add an example to your question.
$endgroup$
– MrSmith42
Dec 13 '16 at 12:57




$begingroup$
You should add an example to your question.
$endgroup$
– MrSmith42
Dec 13 '16 at 12:57




1




1




$begingroup$
(The example still looks off in rev. 3: how did the o and ll swap places?) What about string.substring(shifts) + string.substring(0, shifts)?
$endgroup$
– greybeard
Dec 13 '16 at 14:04




$begingroup$
(The example still looks off in rev. 3: how did the o and ll swap places?) What about string.substring(shifts) + string.substring(0, shifts)?
$endgroup$
– greybeard
Dec 13 '16 at 14:04












$begingroup$
("backticks" are for inline code snippets; there is a {}-button and a keyboard shortcut (<Ctrl>+k) in the edit-window to turn the selection into a code block (prefixing every line with four additional spaces if the selection doesn't already look like this (then it removes four spaces from the start of every line).))
$endgroup$
– greybeard
Dec 13 '16 at 21:45




$begingroup$
("backticks" are for inline code snippets; there is a {}-button and a keyboard shortcut (<Ctrl>+k) in the edit-window to turn the selection into a code block (prefixing every line with four additional spaces if the selection doesn't already look like this (then it removes four spaces from the start of every line).))
$endgroup$
– greybeard
Dec 13 '16 at 21:45












$begingroup$
@greybeard..You are right.I was using the wrong way to do it.Hello->oHell->loHel->lloHe.Can you suggest the correct way?
$endgroup$
– Javasamurai
Dec 14 '16 at 6:29






$begingroup$
@greybeard..You are right.I was using the wrong way to do it.Hello->oHell->loHel->lloHe.Can you suggest the correct way?
$endgroup$
– Javasamurai
Dec 14 '16 at 6:29












4 Answers
4






active

oldest

votes


















3












$begingroup$

Do you have any constraints?




  • Is your example wrong? I would expected lloHe

  • If you are concerned of modifying your method parameters make them final. Copy it to local variables and don't change the variables is kind of useless.

  • Don't use short variables names (exception: the loop counter)

  • If you use Integer instead of int, you should handle null

  • Maybe a simple static utility method would be fine here, except you have some inheritance scenario in mind.

  • Maybe you should also indicate that you are shifting left for positive numbers

  • You should take the length of you string for the char.


Assuming your example is wrong, what about:



import java.util.Objects;
import javax.annotation.Nonnull;

public final class Shift
{
@Nonnull
public static String left( @Nonnull final String string, final int shift )
{
final int length = string.length();
if( length == 0 ) return "";
final int offset = ((shift % length) + length) % length; // get a positive offset

return string.substring( offset, length ) + string.substring( 0, offset );
}

public static void main( String... args )
{
assertEquals( "loHel", Shift.left( "Hello", -2 ) );
assertEquals( "oHell", Shift.left( "Hello", -1 ) );
assertEquals( "Hello", Shift.left( "Hello", 0 ) );
assertEquals( "elloH", Shift.left( "Hello", 1 ) );
assertEquals( "lloHe", Shift.left( "Hello", 2 ) );
assertEquals( "loHel", Shift.left( "Hello", 3 ) );
assertEquals( "oHell", Shift.left( "Hello", 4 ) );
assertEquals( "Hello", Shift.left( "Hello", 5 ) );
assertEquals( "elloH", Shift.left( "Hello", 6 ) );
assertEquals( "", Shift.left( "", 3 ) );
}

private static void assertEquals( String expected, String actual )
{
if( !Objects.equals( expected, actual ) ) throw new AssertionError( "Expected: >" + expected + "< was: >" + actual + "<" );
}
}





share|improve this answer











$endgroup$













  • $begingroup$
    Needs to take shift modulo string.length(), but otherwise this is the obvious way to do it.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:00










  • $begingroup$
    @PeterTaylor It's not that simple. I completed the validation and added some tests. Keep in mind that %in java is the remainder and may become negative.
    $endgroup$
    – mheinzerling
    Dec 14 '16 at 5:54










  • $begingroup$
    Yes, that's the reason I used the word "modulo" rather than the symbol %.
    $endgroup$
    – Peter Taylor
    Dec 14 '16 at 8:19



















0












$begingroup$

("Rotation" = "shift".)



You can perform any rotation in constant time by devising a String wrapper that maintains the string itself and the index into that string which is to be considered as the head character of a rotation. Now, in order to rotate $k$ characters, just add $k$ to that internal offset and take the modulo of string length.



If you want to get a String out of that wrapper, just call its toString(). There is, however, an optimization: if the state of the wrapper is not "dirty", it returns the previous cached rotation. Otherwise, a new string will be created and the state marked as not dirty.



I had this in mind:



import java.util.Objects;

public class StringRotation {

private final String string;
private int headIndex;
private boolean dirty;
private String lastCreatedString;

public StringRotation(String string) {
this.string = Objects.requireNonNull(string,
"The input string is null.");
this.dirty = true;
}

public StringRotation rotate(int rotationCount) {
headIndex = mod(headIndex + rotationCount, string.length());
dirty = true;
return this;
}

public char charAt(int index) {
checkIndexWithinBounds(index);
return string.charAt(mod(headIndex + index, string.length()));
}

@Override
public String toString() {
if (!dirty) {
return lastCreatedString;
}

int stringLength = string.length();
StringBuilder sb = new StringBuilder(stringLength);

for (int i = 0; i < stringLength; ++i) {
sb.append(charAt(i));
}

dirty = false;
lastCreatedString = sb.toString();
return lastCreatedString;
}

private void checkIndexWithinBounds(int index) {
int stringLength = string.length();

if (index < 0) {
throw new IndexOutOfBoundsException(
"The character index is negative: " + index);
}

if (index >= stringLength) {
throw new IndexOutOfBoundsException(
"The character index is too large: " + index +
". The length of this string is " + stringLength);
}
}

private static int mod(int a, int q) {
int ret = a % q;
return ret < 0 ? ret + q : ret;
}

public static void main(String args) {
for (int i = 0; i < 10; ++i) {
System.out.println(new StringRotation("hello").rotate(i));
}

System.out.println("---");

for (int i = 0; i > -10; --i) {
System.out.println(new StringRotation("hello").rotate(i));
}
}
}





share|improve this answer









$endgroup$













  • $begingroup$
    Maybe a little over-engineered without further requirements :)
    $endgroup$
    – mheinzerling
    Dec 13 '16 at 13:14






  • 1




    $begingroup$
    (Pity you can't extends java.lang.String.)
    $endgroup$
    – greybeard
    Dec 13 '16 at 13:59










  • $begingroup$
    @greybeard, but it is possible to implement CharSequence, and a wrapper approach should consider it.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 14:59










  • $begingroup$
    @PeterTaylor Perhaps. Yet CharSequence does not provide rotation methods, so using them would require StringRotation after all.
    $endgroup$
    – coderodde
    Dec 13 '16 at 15:04










  • $begingroup$
    Maybe I didn't express myself clearly enough. I was suggesting that a production version of StringRotation should implement CharSequence. (To be honest, if I went down the wrapper route then I would make the wrapper immutable with constructor StringRotation(CharSequence, int)).
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:41





















0












$begingroup$

In my opinion this is way better by a landslide:



import java.util.Scanner;

public class Example {

static String shifter(String stringToShift, int numberOfShifts){
for(int i = 0; i < numberOfShifts;i++)
{
// Store the character you want to shift
char temporaryChar = stringToShift.charAt(stringToShift.length() - 1);

// Store the character with the original String. Character first
stringToShift = temporaryChar + stringToShift;

// Now that we did that, the string is 1 character longer so we need to take a substring of it
stringToShift = stringToShift.substring(0, stringToShift.length()-1);
}

// After everything is done return the String
return stringToShift;
}

public static void main(String args) {
// Start getting input
Scanner input = new Scanner(System.in);

// Get String input
System.out.println("Enter the string ");
String answer = input.nextLine();
System.out.println("So you have entered: " + answer);

// Get int input
System.out.println("Enter the number of shifts:");
int numberOfShifts = input.nextInt();
System.out.println("number of shifts is: " + numberOfShifts);

// Shift and print out the answer
System.out.println("Shifted string: " + shifter(answer, numberOfShifts));

// Close input after you're done
input.close();

}

}


Pay attention at my style of coding and naming, it's better to have longer names so later you at least understand why the variable is there in the first place.






share|improve this answer











$endgroup$









  • 1




    $begingroup$
    This takes time Omega(mn) where m is the displacement of the shift and n is the length of the string. It's possible to do it in O(n).
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:03










  • $begingroup$
    I just wrote how to write better, more understandable code rather than efficient
    $endgroup$
    – Erlandas Aksomaitis
    Dec 13 '16 at 15:18










  • $begingroup$
    I disagree that it's better to replace a linear algorithm with a quadratic one unless you're massively reducing the constant factor. I'm not really convinced that this code is more understandable either. I can't understand the indexing in char temporaryChar = stringToShift.charAt(stringToShift.length()- i - 1);: it looks buggy and a quick test seems to support that.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:50










  • $begingroup$
    oh sorry didn't notice that, i'll fix that quick
    $endgroup$
    – Erlandas Aksomaitis
    Dec 13 '16 at 18:33










  • $begingroup$
    I have to agree with @PeterTaylor. When you start writing you own code instead of using the libraries, you should get faster and not slower, especial if it get harder to read your code in the end.
    $endgroup$
    – mheinzerling
    Dec 14 '16 at 6:07





















0












$begingroup$

fun getShiftedString(s: String, leftShifts: Int, rightShifts: Int): String {
val shift = leftShifts - rightShifts

return when (shift) {
0 -> s
else -> {
val strBuilder = StringBuilder(s)
s.indices.forEach { index ->
val newShift = if (shift > 0) s.length - shift else -shift
val newIndex = (index + newShift).absoluteValue % s.length
strBuilder[newIndex] = s[index]
}
strBuilder.toString()
}
}


}





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    4 Answers
    4






    active

    oldest

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    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    3












    $begingroup$

    Do you have any constraints?




    • Is your example wrong? I would expected lloHe

    • If you are concerned of modifying your method parameters make them final. Copy it to local variables and don't change the variables is kind of useless.

    • Don't use short variables names (exception: the loop counter)

    • If you use Integer instead of int, you should handle null

    • Maybe a simple static utility method would be fine here, except you have some inheritance scenario in mind.

    • Maybe you should also indicate that you are shifting left for positive numbers

    • You should take the length of you string for the char.


    Assuming your example is wrong, what about:



    import java.util.Objects;
    import javax.annotation.Nonnull;

    public final class Shift
    {
    @Nonnull
    public static String left( @Nonnull final String string, final int shift )
    {
    final int length = string.length();
    if( length == 0 ) return "";
    final int offset = ((shift % length) + length) % length; // get a positive offset

    return string.substring( offset, length ) + string.substring( 0, offset );
    }

    public static void main( String... args )
    {
    assertEquals( "loHel", Shift.left( "Hello", -2 ) );
    assertEquals( "oHell", Shift.left( "Hello", -1 ) );
    assertEquals( "Hello", Shift.left( "Hello", 0 ) );
    assertEquals( "elloH", Shift.left( "Hello", 1 ) );
    assertEquals( "lloHe", Shift.left( "Hello", 2 ) );
    assertEquals( "loHel", Shift.left( "Hello", 3 ) );
    assertEquals( "oHell", Shift.left( "Hello", 4 ) );
    assertEquals( "Hello", Shift.left( "Hello", 5 ) );
    assertEquals( "elloH", Shift.left( "Hello", 6 ) );
    assertEquals( "", Shift.left( "", 3 ) );
    }

    private static void assertEquals( String expected, String actual )
    {
    if( !Objects.equals( expected, actual ) ) throw new AssertionError( "Expected: >" + expected + "< was: >" + actual + "<" );
    }
    }





    share|improve this answer











    $endgroup$













    • $begingroup$
      Needs to take shift modulo string.length(), but otherwise this is the obvious way to do it.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:00










    • $begingroup$
      @PeterTaylor It's not that simple. I completed the validation and added some tests. Keep in mind that %in java is the remainder and may become negative.
      $endgroup$
      – mheinzerling
      Dec 14 '16 at 5:54










    • $begingroup$
      Yes, that's the reason I used the word "modulo" rather than the symbol %.
      $endgroup$
      – Peter Taylor
      Dec 14 '16 at 8:19
















    3












    $begingroup$

    Do you have any constraints?




    • Is your example wrong? I would expected lloHe

    • If you are concerned of modifying your method parameters make them final. Copy it to local variables and don't change the variables is kind of useless.

    • Don't use short variables names (exception: the loop counter)

    • If you use Integer instead of int, you should handle null

    • Maybe a simple static utility method would be fine here, except you have some inheritance scenario in mind.

    • Maybe you should also indicate that you are shifting left for positive numbers

    • You should take the length of you string for the char.


    Assuming your example is wrong, what about:



    import java.util.Objects;
    import javax.annotation.Nonnull;

    public final class Shift
    {
    @Nonnull
    public static String left( @Nonnull final String string, final int shift )
    {
    final int length = string.length();
    if( length == 0 ) return "";
    final int offset = ((shift % length) + length) % length; // get a positive offset

    return string.substring( offset, length ) + string.substring( 0, offset );
    }

    public static void main( String... args )
    {
    assertEquals( "loHel", Shift.left( "Hello", -2 ) );
    assertEquals( "oHell", Shift.left( "Hello", -1 ) );
    assertEquals( "Hello", Shift.left( "Hello", 0 ) );
    assertEquals( "elloH", Shift.left( "Hello", 1 ) );
    assertEquals( "lloHe", Shift.left( "Hello", 2 ) );
    assertEquals( "loHel", Shift.left( "Hello", 3 ) );
    assertEquals( "oHell", Shift.left( "Hello", 4 ) );
    assertEquals( "Hello", Shift.left( "Hello", 5 ) );
    assertEquals( "elloH", Shift.left( "Hello", 6 ) );
    assertEquals( "", Shift.left( "", 3 ) );
    }

    private static void assertEquals( String expected, String actual )
    {
    if( !Objects.equals( expected, actual ) ) throw new AssertionError( "Expected: >" + expected + "< was: >" + actual + "<" );
    }
    }





    share|improve this answer











    $endgroup$













    • $begingroup$
      Needs to take shift modulo string.length(), but otherwise this is the obvious way to do it.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:00










    • $begingroup$
      @PeterTaylor It's not that simple. I completed the validation and added some tests. Keep in mind that %in java is the remainder and may become negative.
      $endgroup$
      – mheinzerling
      Dec 14 '16 at 5:54










    • $begingroup$
      Yes, that's the reason I used the word "modulo" rather than the symbol %.
      $endgroup$
      – Peter Taylor
      Dec 14 '16 at 8:19














    3












    3








    3





    $begingroup$

    Do you have any constraints?




    • Is your example wrong? I would expected lloHe

    • If you are concerned of modifying your method parameters make them final. Copy it to local variables and don't change the variables is kind of useless.

    • Don't use short variables names (exception: the loop counter)

    • If you use Integer instead of int, you should handle null

    • Maybe a simple static utility method would be fine here, except you have some inheritance scenario in mind.

    • Maybe you should also indicate that you are shifting left for positive numbers

    • You should take the length of you string for the char.


    Assuming your example is wrong, what about:



    import java.util.Objects;
    import javax.annotation.Nonnull;

    public final class Shift
    {
    @Nonnull
    public static String left( @Nonnull final String string, final int shift )
    {
    final int length = string.length();
    if( length == 0 ) return "";
    final int offset = ((shift % length) + length) % length; // get a positive offset

    return string.substring( offset, length ) + string.substring( 0, offset );
    }

    public static void main( String... args )
    {
    assertEquals( "loHel", Shift.left( "Hello", -2 ) );
    assertEquals( "oHell", Shift.left( "Hello", -1 ) );
    assertEquals( "Hello", Shift.left( "Hello", 0 ) );
    assertEquals( "elloH", Shift.left( "Hello", 1 ) );
    assertEquals( "lloHe", Shift.left( "Hello", 2 ) );
    assertEquals( "loHel", Shift.left( "Hello", 3 ) );
    assertEquals( "oHell", Shift.left( "Hello", 4 ) );
    assertEquals( "Hello", Shift.left( "Hello", 5 ) );
    assertEquals( "elloH", Shift.left( "Hello", 6 ) );
    assertEquals( "", Shift.left( "", 3 ) );
    }

    private static void assertEquals( String expected, String actual )
    {
    if( !Objects.equals( expected, actual ) ) throw new AssertionError( "Expected: >" + expected + "< was: >" + actual + "<" );
    }
    }





    share|improve this answer











    $endgroup$



    Do you have any constraints?




    • Is your example wrong? I would expected lloHe

    • If you are concerned of modifying your method parameters make them final. Copy it to local variables and don't change the variables is kind of useless.

    • Don't use short variables names (exception: the loop counter)

    • If you use Integer instead of int, you should handle null

    • Maybe a simple static utility method would be fine here, except you have some inheritance scenario in mind.

    • Maybe you should also indicate that you are shifting left for positive numbers

    • You should take the length of you string for the char.


    Assuming your example is wrong, what about:



    import java.util.Objects;
    import javax.annotation.Nonnull;

    public final class Shift
    {
    @Nonnull
    public static String left( @Nonnull final String string, final int shift )
    {
    final int length = string.length();
    if( length == 0 ) return "";
    final int offset = ((shift % length) + length) % length; // get a positive offset

    return string.substring( offset, length ) + string.substring( 0, offset );
    }

    public static void main( String... args )
    {
    assertEquals( "loHel", Shift.left( "Hello", -2 ) );
    assertEquals( "oHell", Shift.left( "Hello", -1 ) );
    assertEquals( "Hello", Shift.left( "Hello", 0 ) );
    assertEquals( "elloH", Shift.left( "Hello", 1 ) );
    assertEquals( "lloHe", Shift.left( "Hello", 2 ) );
    assertEquals( "loHel", Shift.left( "Hello", 3 ) );
    assertEquals( "oHell", Shift.left( "Hello", 4 ) );
    assertEquals( "Hello", Shift.left( "Hello", 5 ) );
    assertEquals( "elloH", Shift.left( "Hello", 6 ) );
    assertEquals( "", Shift.left( "", 3 ) );
    }

    private static void assertEquals( String expected, String actual )
    {
    if( !Objects.equals( expected, actual ) ) throw new AssertionError( "Expected: >" + expected + "< was: >" + actual + "<" );
    }
    }






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 14 '16 at 6:10

























    answered Dec 13 '16 at 12:59









    mheinzerlingmheinzerling

    2,644917




    2,644917












    • $begingroup$
      Needs to take shift modulo string.length(), but otherwise this is the obvious way to do it.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:00










    • $begingroup$
      @PeterTaylor It's not that simple. I completed the validation and added some tests. Keep in mind that %in java is the remainder and may become negative.
      $endgroup$
      – mheinzerling
      Dec 14 '16 at 5:54










    • $begingroup$
      Yes, that's the reason I used the word "modulo" rather than the symbol %.
      $endgroup$
      – Peter Taylor
      Dec 14 '16 at 8:19


















    • $begingroup$
      Needs to take shift modulo string.length(), but otherwise this is the obvious way to do it.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:00










    • $begingroup$
      @PeterTaylor It's not that simple. I completed the validation and added some tests. Keep in mind that %in java is the remainder and may become negative.
      $endgroup$
      – mheinzerling
      Dec 14 '16 at 5:54










    • $begingroup$
      Yes, that's the reason I used the word "modulo" rather than the symbol %.
      $endgroup$
      – Peter Taylor
      Dec 14 '16 at 8:19
















    $begingroup$
    Needs to take shift modulo string.length(), but otherwise this is the obvious way to do it.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:00




    $begingroup$
    Needs to take shift modulo string.length(), but otherwise this is the obvious way to do it.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:00












    $begingroup$
    @PeterTaylor It's not that simple. I completed the validation and added some tests. Keep in mind that %in java is the remainder and may become negative.
    $endgroup$
    – mheinzerling
    Dec 14 '16 at 5:54




    $begingroup$
    @PeterTaylor It's not that simple. I completed the validation and added some tests. Keep in mind that %in java is the remainder and may become negative.
    $endgroup$
    – mheinzerling
    Dec 14 '16 at 5:54












    $begingroup$
    Yes, that's the reason I used the word "modulo" rather than the symbol %.
    $endgroup$
    – Peter Taylor
    Dec 14 '16 at 8:19




    $begingroup$
    Yes, that's the reason I used the word "modulo" rather than the symbol %.
    $endgroup$
    – Peter Taylor
    Dec 14 '16 at 8:19













    0












    $begingroup$

    ("Rotation" = "shift".)



    You can perform any rotation in constant time by devising a String wrapper that maintains the string itself and the index into that string which is to be considered as the head character of a rotation. Now, in order to rotate $k$ characters, just add $k$ to that internal offset and take the modulo of string length.



    If you want to get a String out of that wrapper, just call its toString(). There is, however, an optimization: if the state of the wrapper is not "dirty", it returns the previous cached rotation. Otherwise, a new string will be created and the state marked as not dirty.



    I had this in mind:



    import java.util.Objects;

    public class StringRotation {

    private final String string;
    private int headIndex;
    private boolean dirty;
    private String lastCreatedString;

    public StringRotation(String string) {
    this.string = Objects.requireNonNull(string,
    "The input string is null.");
    this.dirty = true;
    }

    public StringRotation rotate(int rotationCount) {
    headIndex = mod(headIndex + rotationCount, string.length());
    dirty = true;
    return this;
    }

    public char charAt(int index) {
    checkIndexWithinBounds(index);
    return string.charAt(mod(headIndex + index, string.length()));
    }

    @Override
    public String toString() {
    if (!dirty) {
    return lastCreatedString;
    }

    int stringLength = string.length();
    StringBuilder sb = new StringBuilder(stringLength);

    for (int i = 0; i < stringLength; ++i) {
    sb.append(charAt(i));
    }

    dirty = false;
    lastCreatedString = sb.toString();
    return lastCreatedString;
    }

    private void checkIndexWithinBounds(int index) {
    int stringLength = string.length();

    if (index < 0) {
    throw new IndexOutOfBoundsException(
    "The character index is negative: " + index);
    }

    if (index >= stringLength) {
    throw new IndexOutOfBoundsException(
    "The character index is too large: " + index +
    ". The length of this string is " + stringLength);
    }
    }

    private static int mod(int a, int q) {
    int ret = a % q;
    return ret < 0 ? ret + q : ret;
    }

    public static void main(String args) {
    for (int i = 0; i < 10; ++i) {
    System.out.println(new StringRotation("hello").rotate(i));
    }

    System.out.println("---");

    for (int i = 0; i > -10; --i) {
    System.out.println(new StringRotation("hello").rotate(i));
    }
    }
    }





    share|improve this answer









    $endgroup$













    • $begingroup$
      Maybe a little over-engineered without further requirements :)
      $endgroup$
      – mheinzerling
      Dec 13 '16 at 13:14






    • 1




      $begingroup$
      (Pity you can't extends java.lang.String.)
      $endgroup$
      – greybeard
      Dec 13 '16 at 13:59










    • $begingroup$
      @greybeard, but it is possible to implement CharSequence, and a wrapper approach should consider it.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 14:59










    • $begingroup$
      @PeterTaylor Perhaps. Yet CharSequence does not provide rotation methods, so using them would require StringRotation after all.
      $endgroup$
      – coderodde
      Dec 13 '16 at 15:04










    • $begingroup$
      Maybe I didn't express myself clearly enough. I was suggesting that a production version of StringRotation should implement CharSequence. (To be honest, if I went down the wrapper route then I would make the wrapper immutable with constructor StringRotation(CharSequence, int)).
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:41


















    0












    $begingroup$

    ("Rotation" = "shift".)



    You can perform any rotation in constant time by devising a String wrapper that maintains the string itself and the index into that string which is to be considered as the head character of a rotation. Now, in order to rotate $k$ characters, just add $k$ to that internal offset and take the modulo of string length.



    If you want to get a String out of that wrapper, just call its toString(). There is, however, an optimization: if the state of the wrapper is not "dirty", it returns the previous cached rotation. Otherwise, a new string will be created and the state marked as not dirty.



    I had this in mind:



    import java.util.Objects;

    public class StringRotation {

    private final String string;
    private int headIndex;
    private boolean dirty;
    private String lastCreatedString;

    public StringRotation(String string) {
    this.string = Objects.requireNonNull(string,
    "The input string is null.");
    this.dirty = true;
    }

    public StringRotation rotate(int rotationCount) {
    headIndex = mod(headIndex + rotationCount, string.length());
    dirty = true;
    return this;
    }

    public char charAt(int index) {
    checkIndexWithinBounds(index);
    return string.charAt(mod(headIndex + index, string.length()));
    }

    @Override
    public String toString() {
    if (!dirty) {
    return lastCreatedString;
    }

    int stringLength = string.length();
    StringBuilder sb = new StringBuilder(stringLength);

    for (int i = 0; i < stringLength; ++i) {
    sb.append(charAt(i));
    }

    dirty = false;
    lastCreatedString = sb.toString();
    return lastCreatedString;
    }

    private void checkIndexWithinBounds(int index) {
    int stringLength = string.length();

    if (index < 0) {
    throw new IndexOutOfBoundsException(
    "The character index is negative: " + index);
    }

    if (index >= stringLength) {
    throw new IndexOutOfBoundsException(
    "The character index is too large: " + index +
    ". The length of this string is " + stringLength);
    }
    }

    private static int mod(int a, int q) {
    int ret = a % q;
    return ret < 0 ? ret + q : ret;
    }

    public static void main(String args) {
    for (int i = 0; i < 10; ++i) {
    System.out.println(new StringRotation("hello").rotate(i));
    }

    System.out.println("---");

    for (int i = 0; i > -10; --i) {
    System.out.println(new StringRotation("hello").rotate(i));
    }
    }
    }





    share|improve this answer









    $endgroup$













    • $begingroup$
      Maybe a little over-engineered without further requirements :)
      $endgroup$
      – mheinzerling
      Dec 13 '16 at 13:14






    • 1




      $begingroup$
      (Pity you can't extends java.lang.String.)
      $endgroup$
      – greybeard
      Dec 13 '16 at 13:59










    • $begingroup$
      @greybeard, but it is possible to implement CharSequence, and a wrapper approach should consider it.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 14:59










    • $begingroup$
      @PeterTaylor Perhaps. Yet CharSequence does not provide rotation methods, so using them would require StringRotation after all.
      $endgroup$
      – coderodde
      Dec 13 '16 at 15:04










    • $begingroup$
      Maybe I didn't express myself clearly enough. I was suggesting that a production version of StringRotation should implement CharSequence. (To be honest, if I went down the wrapper route then I would make the wrapper immutable with constructor StringRotation(CharSequence, int)).
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:41
















    0












    0








    0





    $begingroup$

    ("Rotation" = "shift".)



    You can perform any rotation in constant time by devising a String wrapper that maintains the string itself and the index into that string which is to be considered as the head character of a rotation. Now, in order to rotate $k$ characters, just add $k$ to that internal offset and take the modulo of string length.



    If you want to get a String out of that wrapper, just call its toString(). There is, however, an optimization: if the state of the wrapper is not "dirty", it returns the previous cached rotation. Otherwise, a new string will be created and the state marked as not dirty.



    I had this in mind:



    import java.util.Objects;

    public class StringRotation {

    private final String string;
    private int headIndex;
    private boolean dirty;
    private String lastCreatedString;

    public StringRotation(String string) {
    this.string = Objects.requireNonNull(string,
    "The input string is null.");
    this.dirty = true;
    }

    public StringRotation rotate(int rotationCount) {
    headIndex = mod(headIndex + rotationCount, string.length());
    dirty = true;
    return this;
    }

    public char charAt(int index) {
    checkIndexWithinBounds(index);
    return string.charAt(mod(headIndex + index, string.length()));
    }

    @Override
    public String toString() {
    if (!dirty) {
    return lastCreatedString;
    }

    int stringLength = string.length();
    StringBuilder sb = new StringBuilder(stringLength);

    for (int i = 0; i < stringLength; ++i) {
    sb.append(charAt(i));
    }

    dirty = false;
    lastCreatedString = sb.toString();
    return lastCreatedString;
    }

    private void checkIndexWithinBounds(int index) {
    int stringLength = string.length();

    if (index < 0) {
    throw new IndexOutOfBoundsException(
    "The character index is negative: " + index);
    }

    if (index >= stringLength) {
    throw new IndexOutOfBoundsException(
    "The character index is too large: " + index +
    ". The length of this string is " + stringLength);
    }
    }

    private static int mod(int a, int q) {
    int ret = a % q;
    return ret < 0 ? ret + q : ret;
    }

    public static void main(String args) {
    for (int i = 0; i < 10; ++i) {
    System.out.println(new StringRotation("hello").rotate(i));
    }

    System.out.println("---");

    for (int i = 0; i > -10; --i) {
    System.out.println(new StringRotation("hello").rotate(i));
    }
    }
    }





    share|improve this answer









    $endgroup$



    ("Rotation" = "shift".)



    You can perform any rotation in constant time by devising a String wrapper that maintains the string itself and the index into that string which is to be considered as the head character of a rotation. Now, in order to rotate $k$ characters, just add $k$ to that internal offset and take the modulo of string length.



    If you want to get a String out of that wrapper, just call its toString(). There is, however, an optimization: if the state of the wrapper is not "dirty", it returns the previous cached rotation. Otherwise, a new string will be created and the state marked as not dirty.



    I had this in mind:



    import java.util.Objects;

    public class StringRotation {

    private final String string;
    private int headIndex;
    private boolean dirty;
    private String lastCreatedString;

    public StringRotation(String string) {
    this.string = Objects.requireNonNull(string,
    "The input string is null.");
    this.dirty = true;
    }

    public StringRotation rotate(int rotationCount) {
    headIndex = mod(headIndex + rotationCount, string.length());
    dirty = true;
    return this;
    }

    public char charAt(int index) {
    checkIndexWithinBounds(index);
    return string.charAt(mod(headIndex + index, string.length()));
    }

    @Override
    public String toString() {
    if (!dirty) {
    return lastCreatedString;
    }

    int stringLength = string.length();
    StringBuilder sb = new StringBuilder(stringLength);

    for (int i = 0; i < stringLength; ++i) {
    sb.append(charAt(i));
    }

    dirty = false;
    lastCreatedString = sb.toString();
    return lastCreatedString;
    }

    private void checkIndexWithinBounds(int index) {
    int stringLength = string.length();

    if (index < 0) {
    throw new IndexOutOfBoundsException(
    "The character index is negative: " + index);
    }

    if (index >= stringLength) {
    throw new IndexOutOfBoundsException(
    "The character index is too large: " + index +
    ". The length of this string is " + stringLength);
    }
    }

    private static int mod(int a, int q) {
    int ret = a % q;
    return ret < 0 ? ret + q : ret;
    }

    public static void main(String args) {
    for (int i = 0; i < 10; ++i) {
    System.out.println(new StringRotation("hello").rotate(i));
    }

    System.out.println("---");

    for (int i = 0; i > -10; --i) {
    System.out.println(new StringRotation("hello").rotate(i));
    }
    }
    }






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Dec 13 '16 at 13:07









    coderoddecoderodde

    15.7k538127




    15.7k538127












    • $begingroup$
      Maybe a little over-engineered without further requirements :)
      $endgroup$
      – mheinzerling
      Dec 13 '16 at 13:14






    • 1




      $begingroup$
      (Pity you can't extends java.lang.String.)
      $endgroup$
      – greybeard
      Dec 13 '16 at 13:59










    • $begingroup$
      @greybeard, but it is possible to implement CharSequence, and a wrapper approach should consider it.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 14:59










    • $begingroup$
      @PeterTaylor Perhaps. Yet CharSequence does not provide rotation methods, so using them would require StringRotation after all.
      $endgroup$
      – coderodde
      Dec 13 '16 at 15:04










    • $begingroup$
      Maybe I didn't express myself clearly enough. I was suggesting that a production version of StringRotation should implement CharSequence. (To be honest, if I went down the wrapper route then I would make the wrapper immutable with constructor StringRotation(CharSequence, int)).
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:41




















    • $begingroup$
      Maybe a little over-engineered without further requirements :)
      $endgroup$
      – mheinzerling
      Dec 13 '16 at 13:14






    • 1




      $begingroup$
      (Pity you can't extends java.lang.String.)
      $endgroup$
      – greybeard
      Dec 13 '16 at 13:59










    • $begingroup$
      @greybeard, but it is possible to implement CharSequence, and a wrapper approach should consider it.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 14:59










    • $begingroup$
      @PeterTaylor Perhaps. Yet CharSequence does not provide rotation methods, so using them would require StringRotation after all.
      $endgroup$
      – coderodde
      Dec 13 '16 at 15:04










    • $begingroup$
      Maybe I didn't express myself clearly enough. I was suggesting that a production version of StringRotation should implement CharSequence. (To be honest, if I went down the wrapper route then I would make the wrapper immutable with constructor StringRotation(CharSequence, int)).
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:41


















    $begingroup$
    Maybe a little over-engineered without further requirements :)
    $endgroup$
    – mheinzerling
    Dec 13 '16 at 13:14




    $begingroup$
    Maybe a little over-engineered without further requirements :)
    $endgroup$
    – mheinzerling
    Dec 13 '16 at 13:14




    1




    1




    $begingroup$
    (Pity you can't extends java.lang.String.)
    $endgroup$
    – greybeard
    Dec 13 '16 at 13:59




    $begingroup$
    (Pity you can't extends java.lang.String.)
    $endgroup$
    – greybeard
    Dec 13 '16 at 13:59












    $begingroup$
    @greybeard, but it is possible to implement CharSequence, and a wrapper approach should consider it.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 14:59




    $begingroup$
    @greybeard, but it is possible to implement CharSequence, and a wrapper approach should consider it.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 14:59












    $begingroup$
    @PeterTaylor Perhaps. Yet CharSequence does not provide rotation methods, so using them would require StringRotation after all.
    $endgroup$
    – coderodde
    Dec 13 '16 at 15:04




    $begingroup$
    @PeterTaylor Perhaps. Yet CharSequence does not provide rotation methods, so using them would require StringRotation after all.
    $endgroup$
    – coderodde
    Dec 13 '16 at 15:04












    $begingroup$
    Maybe I didn't express myself clearly enough. I was suggesting that a production version of StringRotation should implement CharSequence. (To be honest, if I went down the wrapper route then I would make the wrapper immutable with constructor StringRotation(CharSequence, int)).
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:41






    $begingroup$
    Maybe I didn't express myself clearly enough. I was suggesting that a production version of StringRotation should implement CharSequence. (To be honest, if I went down the wrapper route then I would make the wrapper immutable with constructor StringRotation(CharSequence, int)).
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:41













    0












    $begingroup$

    In my opinion this is way better by a landslide:



    import java.util.Scanner;

    public class Example {

    static String shifter(String stringToShift, int numberOfShifts){
    for(int i = 0; i < numberOfShifts;i++)
    {
    // Store the character you want to shift
    char temporaryChar = stringToShift.charAt(stringToShift.length() - 1);

    // Store the character with the original String. Character first
    stringToShift = temporaryChar + stringToShift;

    // Now that we did that, the string is 1 character longer so we need to take a substring of it
    stringToShift = stringToShift.substring(0, stringToShift.length()-1);
    }

    // After everything is done return the String
    return stringToShift;
    }

    public static void main(String args) {
    // Start getting input
    Scanner input = new Scanner(System.in);

    // Get String input
    System.out.println("Enter the string ");
    String answer = input.nextLine();
    System.out.println("So you have entered: " + answer);

    // Get int input
    System.out.println("Enter the number of shifts:");
    int numberOfShifts = input.nextInt();
    System.out.println("number of shifts is: " + numberOfShifts);

    // Shift and print out the answer
    System.out.println("Shifted string: " + shifter(answer, numberOfShifts));

    // Close input after you're done
    input.close();

    }

    }


    Pay attention at my style of coding and naming, it's better to have longer names so later you at least understand why the variable is there in the first place.






    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      This takes time Omega(mn) where m is the displacement of the shift and n is the length of the string. It's possible to do it in O(n).
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:03










    • $begingroup$
      I just wrote how to write better, more understandable code rather than efficient
      $endgroup$
      – Erlandas Aksomaitis
      Dec 13 '16 at 15:18










    • $begingroup$
      I disagree that it's better to replace a linear algorithm with a quadratic one unless you're massively reducing the constant factor. I'm not really convinced that this code is more understandable either. I can't understand the indexing in char temporaryChar = stringToShift.charAt(stringToShift.length()- i - 1);: it looks buggy and a quick test seems to support that.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:50










    • $begingroup$
      oh sorry didn't notice that, i'll fix that quick
      $endgroup$
      – Erlandas Aksomaitis
      Dec 13 '16 at 18:33










    • $begingroup$
      I have to agree with @PeterTaylor. When you start writing you own code instead of using the libraries, you should get faster and not slower, especial if it get harder to read your code in the end.
      $endgroup$
      – mheinzerling
      Dec 14 '16 at 6:07


















    0












    $begingroup$

    In my opinion this is way better by a landslide:



    import java.util.Scanner;

    public class Example {

    static String shifter(String stringToShift, int numberOfShifts){
    for(int i = 0; i < numberOfShifts;i++)
    {
    // Store the character you want to shift
    char temporaryChar = stringToShift.charAt(stringToShift.length() - 1);

    // Store the character with the original String. Character first
    stringToShift = temporaryChar + stringToShift;

    // Now that we did that, the string is 1 character longer so we need to take a substring of it
    stringToShift = stringToShift.substring(0, stringToShift.length()-1);
    }

    // After everything is done return the String
    return stringToShift;
    }

    public static void main(String args) {
    // Start getting input
    Scanner input = new Scanner(System.in);

    // Get String input
    System.out.println("Enter the string ");
    String answer = input.nextLine();
    System.out.println("So you have entered: " + answer);

    // Get int input
    System.out.println("Enter the number of shifts:");
    int numberOfShifts = input.nextInt();
    System.out.println("number of shifts is: " + numberOfShifts);

    // Shift and print out the answer
    System.out.println("Shifted string: " + shifter(answer, numberOfShifts));

    // Close input after you're done
    input.close();

    }

    }


    Pay attention at my style of coding and naming, it's better to have longer names so later you at least understand why the variable is there in the first place.






    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      This takes time Omega(mn) where m is the displacement of the shift and n is the length of the string. It's possible to do it in O(n).
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:03










    • $begingroup$
      I just wrote how to write better, more understandable code rather than efficient
      $endgroup$
      – Erlandas Aksomaitis
      Dec 13 '16 at 15:18










    • $begingroup$
      I disagree that it's better to replace a linear algorithm with a quadratic one unless you're massively reducing the constant factor. I'm not really convinced that this code is more understandable either. I can't understand the indexing in char temporaryChar = stringToShift.charAt(stringToShift.length()- i - 1);: it looks buggy and a quick test seems to support that.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:50










    • $begingroup$
      oh sorry didn't notice that, i'll fix that quick
      $endgroup$
      – Erlandas Aksomaitis
      Dec 13 '16 at 18:33










    • $begingroup$
      I have to agree with @PeterTaylor. When you start writing you own code instead of using the libraries, you should get faster and not slower, especial if it get harder to read your code in the end.
      $endgroup$
      – mheinzerling
      Dec 14 '16 at 6:07
















    0












    0








    0





    $begingroup$

    In my opinion this is way better by a landslide:



    import java.util.Scanner;

    public class Example {

    static String shifter(String stringToShift, int numberOfShifts){
    for(int i = 0; i < numberOfShifts;i++)
    {
    // Store the character you want to shift
    char temporaryChar = stringToShift.charAt(stringToShift.length() - 1);

    // Store the character with the original String. Character first
    stringToShift = temporaryChar + stringToShift;

    // Now that we did that, the string is 1 character longer so we need to take a substring of it
    stringToShift = stringToShift.substring(0, stringToShift.length()-1);
    }

    // After everything is done return the String
    return stringToShift;
    }

    public static void main(String args) {
    // Start getting input
    Scanner input = new Scanner(System.in);

    // Get String input
    System.out.println("Enter the string ");
    String answer = input.nextLine();
    System.out.println("So you have entered: " + answer);

    // Get int input
    System.out.println("Enter the number of shifts:");
    int numberOfShifts = input.nextInt();
    System.out.println("number of shifts is: " + numberOfShifts);

    // Shift and print out the answer
    System.out.println("Shifted string: " + shifter(answer, numberOfShifts));

    // Close input after you're done
    input.close();

    }

    }


    Pay attention at my style of coding and naming, it's better to have longer names so later you at least understand why the variable is there in the first place.






    share|improve this answer











    $endgroup$



    In my opinion this is way better by a landslide:



    import java.util.Scanner;

    public class Example {

    static String shifter(String stringToShift, int numberOfShifts){
    for(int i = 0; i < numberOfShifts;i++)
    {
    // Store the character you want to shift
    char temporaryChar = stringToShift.charAt(stringToShift.length() - 1);

    // Store the character with the original String. Character first
    stringToShift = temporaryChar + stringToShift;

    // Now that we did that, the string is 1 character longer so we need to take a substring of it
    stringToShift = stringToShift.substring(0, stringToShift.length()-1);
    }

    // After everything is done return the String
    return stringToShift;
    }

    public static void main(String args) {
    // Start getting input
    Scanner input = new Scanner(System.in);

    // Get String input
    System.out.println("Enter the string ");
    String answer = input.nextLine();
    System.out.println("So you have entered: " + answer);

    // Get int input
    System.out.println("Enter the number of shifts:");
    int numberOfShifts = input.nextInt();
    System.out.println("number of shifts is: " + numberOfShifts);

    // Shift and print out the answer
    System.out.println("Shifted string: " + shifter(answer, numberOfShifts));

    // Close input after you're done
    input.close();

    }

    }


    Pay attention at my style of coding and naming, it's better to have longer names so later you at least understand why the variable is there in the first place.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 13 '16 at 18:55

























    answered Dec 13 '16 at 13:49









    Erlandas AksomaitisErlandas Aksomaitis

    174




    174








    • 1




      $begingroup$
      This takes time Omega(mn) where m is the displacement of the shift and n is the length of the string. It's possible to do it in O(n).
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:03










    • $begingroup$
      I just wrote how to write better, more understandable code rather than efficient
      $endgroup$
      – Erlandas Aksomaitis
      Dec 13 '16 at 15:18










    • $begingroup$
      I disagree that it's better to replace a linear algorithm with a quadratic one unless you're massively reducing the constant factor. I'm not really convinced that this code is more understandable either. I can't understand the indexing in char temporaryChar = stringToShift.charAt(stringToShift.length()- i - 1);: it looks buggy and a quick test seems to support that.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:50










    • $begingroup$
      oh sorry didn't notice that, i'll fix that quick
      $endgroup$
      – Erlandas Aksomaitis
      Dec 13 '16 at 18:33










    • $begingroup$
      I have to agree with @PeterTaylor. When you start writing you own code instead of using the libraries, you should get faster and not slower, especial if it get harder to read your code in the end.
      $endgroup$
      – mheinzerling
      Dec 14 '16 at 6:07
















    • 1




      $begingroup$
      This takes time Omega(mn) where m is the displacement of the shift and n is the length of the string. It's possible to do it in O(n).
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:03










    • $begingroup$
      I just wrote how to write better, more understandable code rather than efficient
      $endgroup$
      – Erlandas Aksomaitis
      Dec 13 '16 at 15:18










    • $begingroup$
      I disagree that it's better to replace a linear algorithm with a quadratic one unless you're massively reducing the constant factor. I'm not really convinced that this code is more understandable either. I can't understand the indexing in char temporaryChar = stringToShift.charAt(stringToShift.length()- i - 1);: it looks buggy and a quick test seems to support that.
      $endgroup$
      – Peter Taylor
      Dec 13 '16 at 15:50










    • $begingroup$
      oh sorry didn't notice that, i'll fix that quick
      $endgroup$
      – Erlandas Aksomaitis
      Dec 13 '16 at 18:33










    • $begingroup$
      I have to agree with @PeterTaylor. When you start writing you own code instead of using the libraries, you should get faster and not slower, especial if it get harder to read your code in the end.
      $endgroup$
      – mheinzerling
      Dec 14 '16 at 6:07










    1




    1




    $begingroup$
    This takes time Omega(mn) where m is the displacement of the shift and n is the length of the string. It's possible to do it in O(n).
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:03




    $begingroup$
    This takes time Omega(mn) where m is the displacement of the shift and n is the length of the string. It's possible to do it in O(n).
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:03












    $begingroup$
    I just wrote how to write better, more understandable code rather than efficient
    $endgroup$
    – Erlandas Aksomaitis
    Dec 13 '16 at 15:18




    $begingroup$
    I just wrote how to write better, more understandable code rather than efficient
    $endgroup$
    – Erlandas Aksomaitis
    Dec 13 '16 at 15:18












    $begingroup$
    I disagree that it's better to replace a linear algorithm with a quadratic one unless you're massively reducing the constant factor. I'm not really convinced that this code is more understandable either. I can't understand the indexing in char temporaryChar = stringToShift.charAt(stringToShift.length()- i - 1);: it looks buggy and a quick test seems to support that.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:50




    $begingroup$
    I disagree that it's better to replace a linear algorithm with a quadratic one unless you're massively reducing the constant factor. I'm not really convinced that this code is more understandable either. I can't understand the indexing in char temporaryChar = stringToShift.charAt(stringToShift.length()- i - 1);: it looks buggy and a quick test seems to support that.
    $endgroup$
    – Peter Taylor
    Dec 13 '16 at 15:50












    $begingroup$
    oh sorry didn't notice that, i'll fix that quick
    $endgroup$
    – Erlandas Aksomaitis
    Dec 13 '16 at 18:33




    $begingroup$
    oh sorry didn't notice that, i'll fix that quick
    $endgroup$
    – Erlandas Aksomaitis
    Dec 13 '16 at 18:33












    $begingroup$
    I have to agree with @PeterTaylor. When you start writing you own code instead of using the libraries, you should get faster and not slower, especial if it get harder to read your code in the end.
    $endgroup$
    – mheinzerling
    Dec 14 '16 at 6:07






    $begingroup$
    I have to agree with @PeterTaylor. When you start writing you own code instead of using the libraries, you should get faster and not slower, especial if it get harder to read your code in the end.
    $endgroup$
    – mheinzerling
    Dec 14 '16 at 6:07













    0












    $begingroup$

    fun getShiftedString(s: String, leftShifts: Int, rightShifts: Int): String {
    val shift = leftShifts - rightShifts

    return when (shift) {
    0 -> s
    else -> {
    val strBuilder = StringBuilder(s)
    s.indices.forEach { index ->
    val newShift = if (shift > 0) s.length - shift else -shift
    val newIndex = (index + newShift).absoluteValue % s.length
    strBuilder[newIndex] = s[index]
    }
    strBuilder.toString()
    }
    }


    }





    share








    New contributor




    leagmain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$


















      0












      $begingroup$

      fun getShiftedString(s: String, leftShifts: Int, rightShifts: Int): String {
      val shift = leftShifts - rightShifts

      return when (shift) {
      0 -> s
      else -> {
      val strBuilder = StringBuilder(s)
      s.indices.forEach { index ->
      val newShift = if (shift > 0) s.length - shift else -shift
      val newIndex = (index + newShift).absoluteValue % s.length
      strBuilder[newIndex] = s[index]
      }
      strBuilder.toString()
      }
      }


      }





      share








      New contributor




      leagmain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$
















        0












        0








        0





        $begingroup$

        fun getShiftedString(s: String, leftShifts: Int, rightShifts: Int): String {
        val shift = leftShifts - rightShifts

        return when (shift) {
        0 -> s
        else -> {
        val strBuilder = StringBuilder(s)
        s.indices.forEach { index ->
        val newShift = if (shift > 0) s.length - shift else -shift
        val newIndex = (index + newShift).absoluteValue % s.length
        strBuilder[newIndex] = s[index]
        }
        strBuilder.toString()
        }
        }


        }





        share








        New contributor




        leagmain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$



        fun getShiftedString(s: String, leftShifts: Int, rightShifts: Int): String {
        val shift = leftShifts - rightShifts

        return when (shift) {
        0 -> s
        else -> {
        val strBuilder = StringBuilder(s)
        s.indices.forEach { index ->
        val newShift = if (shift > 0) s.length - shift else -shift
        val newIndex = (index + newShift).absoluteValue % s.length
        strBuilder[newIndex] = s[index]
        }
        strBuilder.toString()
        }
        }


        }






        share








        New contributor




        leagmain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.








        share


        share






        New contributor




        leagmain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered 9 mins ago









        leagmainleagmain

        1




        1




        New contributor




        leagmain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        leagmain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        leagmain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























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