How to determine if a number is positive or negative

Question

I was asked in an interview  how to determine whether a number is positive or negative  The rules are that we should not use relational operators such as  lt   and  gt   built in java functions  like substring  indexOf  charAt  and startsWith   no regex  or API s   I did some homework on this and the code is given below  but it only works for integer type  But they asked me to write a generic code that works for float  double  and long       This might not be better way     S O P     number  gt  gt  31    amp  1     1    - ve number       ve number      any ideas from your side

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Write it using the conditional then take a look at the assembly code generated

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I don t know how exactly Java coerces numeric values  but the answer is pretty simple  if put in pseudocode  I leave the details to you    sign x      x    0    0    x x

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Combined generics with double API  Guess it s a bit of cheating  but at least we need to write only one method   static  lt T extends Number gt  boolean isNegative T number               return   number doubleValue     Double POSITIVE INFINITY     Double NEGATIVE INFINITY

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There is a function is the math library called signnum   http   www tutorialspoint com java lang math signum float htm http   www tutorialspoint com java lang math signum double htm

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if antennaHeight compareTo Double valueOf 0   gt  0    In the above code  antennaHeight compareTo Double valueOf 0    --- this  ll return int  comparing this with 0 gives the solution

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What about this   return   num       charAt 0      -

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What about the following   T sign T x        if x  0  return 0      return x Math abs x       Should work for every type T     Alternatively  one can define abs x  as Math sqrt x x   and if that is also cheating  implement your own square root function

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if    Double calcYourDouble    toString   contains  -            doThis    else doThat

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one more option I could think of   private static boolean isPositive Object numberObject    Long number   Long valueOf numberObject toString     return Math sqrt  number   number      number      private static boolean isPositive Object numberObject    Long number   Long valueOf numberObject toString     long signedLeftShifteredNumber   number  lt  lt  1     Signed left shift long unsignedRightShifterNumber   signedLeftShifteredNumber  gt  gt  gt  1     Unsigned right shift return unsignedRightShifterNumber    number

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If it is a valid answer  boolean IsNegative char   v  throws NullPointerException  ArrayIndexOutOfBoundException      return v 0    -

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Untested  but illustrating my idea   boolean IsNegative lt T gt  T v      return  v  amp    T -1

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You can do something like this     long   num   1E308   1E308   gt  gt  63     0     ve     -ve    The main idea here is that we cast to a long  and check the value of the most significant bit  As a double float between -1 and 0 will round to zero when cast to a long  we multiply by large doubles so that a negative float double will be less than -1  Two multiplications are required because of the existence of subnormals  it doesn t really need to be that big though

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I think there is a very simple solution   public boolean isPositive int float double long i       return    i-i   0   true   false       tell me if I m wrong

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If you are allowed to use      as seems to be the case  you can do something like that taking advantage of the fact that an exception will be raised if an array index is out of bounds  The code is for double  but you can cast any numeric type to a double  here the eventual loss of precision would not be important at all    I have added comments to explain the process  bring the value in  -2 0  -1 0  union  1 0  2 0   and a small test driver as well   class T       public static boolean positive double f              final boolean pos0      true          final boolean posn      false  true           if  f    0 0             return true          while  true                   If f is in  -1 0  1 0   multiply it by 2 and restart             try                  if  pos0  int  f                        f    2 0                     continue                                catch  Exception e                                If f is in  -2 0  -1 0  U  1 0  2 0   return the proper answer             try                  return posn  int    f 1 5  2                 catch  Exception e                                f is outside  -2 0  2 0   divide by 2 and restart             f    2 0                      static void check double f              System out println f     - gt      positive f             public static void main String args                for  double i   -10 0  i  lt   10 0  i               check i          check -1e24          check -1e-24          check 1e-24          check 1e24          The output is   -10 0 - gt  false -9 0 - gt  false -8 0 - gt  false -7 0 - gt  false -6 0 - gt  false -5 0 - gt  false -4 0 - gt  false -3 0 - gt  false -2 0 - gt  false -1 0 - gt  false 0 0 - gt  true 1 0 - gt  true 2 0 - gt  true 3 0 - gt  true 4 0 - gt  true 5 0 - gt  true 6 0 - gt  true 7 0 - gt  true 8 0 - gt  true 9 0 - gt  true 10 0 - gt  true -1 0E24 - gt  false -1 0E-24 - gt  false 1 0E-24 - gt  true 1 0E24 - gt  true

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The following is a terrible approach that would get you fired at any job       It depends on you getting a Stack Overflow Exception  or whatever Java calls it      And it would only work for positive numbers that don t deviate from 0 like crazy     Negative numbers are fine  since you would overflow to positive  and then get a stack overflow exception eventually  which would return false  or  yes  it is negative    Boolean isPositive lt T gt  T a      if a    0  return true    else         try             return isPositive a-1        catch StackOverflowException e              return false    It went way down there and eventually went kaboom

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This solution uses modulus  And yes  it also works for 0 5  tests are below  in the main method    public class Num        public static int sign long x            if  x    0L    x    1L  return  int  x          return x    Long MIN VALUE    x    x - 1L     x   -1   1             public static int sign double x            if  x    x  throw new IllegalArgumentException  NaN            if  x    0 d    x    1 d  return  int  x          if  x    Double POSITIVE INFINITY  return 1          if  x    Double NEGATIVE INFINITY  return -1          return x    x - 1 d     x   -1   1             public static int sign int x            return Num sign  long x              public static int sign float x            return Num sign  double x              public static void main String args               System out println Num sign Integer MAX VALUE       1         System out println Num sign 1       1         System out println Num sign 0       0         System out println Num sign -1       -1         System out println Num sign Integer MIN VALUE       -1          System out println Num sign Long MAX VALUE       1         System out println Num sign 1L       1         System out println Num sign 0L       0         System out println Num sign -1L       -1         System out println Num sign Long MIN VALUE       -1          System out println Num sign Double POSITIVE INFINITY       1         System out println Num sign Double MAX VALUE       1         System out println Num sign 0 5d       1         System out println Num sign 0 d       0         System out println Num sign -0 5d       -1         System out println Num sign Double MIN VALUE       -1         System out println Num sign Double NEGATIVE INFINITY       -1          System out println Num sign Float POSITIVE INFINITY       1         System out println Num sign Float MAX VALUE       1         System out println Num sign 0 5f       1         System out println Num sign 0 f       0         System out println Num sign -0 5f       -1         System out println Num sign Float MIN VALUE       -1         System out println Num sign Float NEGATIVE INFINITY       -1         System out println Num sign Float NaN       Throws an exception

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This solution uses no conditional operators  but relies on catching two excpetions   A division error equates to the number originally being  negative   Alternatively  the number will eventually fall off the planet and throw a StackOverFlow exception if it is positive    public static boolean isPositive  f                      int x             try                  x   1   int f   1                  return isPositive x 1                catch  StackOverFlow Error e                   return true                catch  Zero Division Error e                   return false

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This code covers all cases and types   public static boolean isNegative Number number        return  Double doubleToLongBits number doubleValue     amp  Long MIN VALUE     Long MIN VALUE      This method accepts any of the wrapper classes  Integer  Long  Float and Double  and thanks to auto-boxing any of the primitive numeric types  int  long  float and double  and simply checks it the high bit  which in all types is the sign bit  is set   It returns true when passed any of    any negative int Integer any negative long Long any negative float Float any negative double Double Double NEGATIVE INFINITY Float NEGATIVE INFINITY   and false otherwise

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The integer cases are easy  The double case is trickier  until you remember about infinities   Note  If you consider the double constants  part of the api   you can replace them with overflowing expressions like 1E308   2   int sign int i        if  i    0  return 0      if  i  gt  gt  31    0  return -1      return  1    int sign long i        if  i    0  return 0      if  i  gt  gt  63    0  return -1      return  1    int sign double f        if  f    f  throw new IllegalArgumentException  NaN        if  f    0  return 0      f    Double POSITIVE INFINITY      if  f    Double POSITIVE INFINITY  return  1      if  f    Double NEGATIVE INFINITY  return -1         this should never be reached  but I ve been wrong before        throw new IllegalArgumentException  Unfathomed double

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Two simple solutions  Works also for infinities and numbers -1  lt   r  lt   1 Will return  positive  for NaNs   String positiveOrNegative double number       return    int  number 0 0   gt  gt 31    0    positive     negative      String positiveOrNegative double number       return  number  0      int  number-1 0   gt  gt 31  0    positive     negative

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This will only works for everything except  0  2   boolean isPositive    n    n - 1     n    n    You can make a better solution like this  works except for  0  1    boolean isPositive     n    n - 0 5     n    0 5    n    You can get better precision by changing the 0 5 part with something like 2 m  m integer    boolean isPositive     n    n - 0 03125     n    0 03125    n

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static boolean isNegative double v      return new Double v  toString   startsWith  -

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Why not get the square root of the number  If its negative - java will throw an error and we will handle it            try               d   Math sqrt THE NUMBER                       catch   ArithmeticException e                 console putln  Number is negative

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Returns 0 if positive  nonzero if negative public long sign long value        return value  amp  0x8000000000000000L      Call like   long val1        double val2        float val3        int val4         sign  long  valN     Casting from double   float   integer to long should preserve the sign  if not the actual value

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It s easy to do this like  private static boolean isNeg T l            return  Math abs l-1  gt Math abs l

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Well  taking advantage of casting  since we don t care what the actual value is  perhaps the following would work  Bear in mind that the actual implementations do not violate the API rules  I ve edited this to make the method names a bit more obvious and in light of  chris  comment about the  -1  1  problem domain  Essentially  this problem does not appear to solvable without recourse to API methods within Float or Double that reference the native bit structure of the float and double primitives   As everybody else has said  Stupid interview question  Grr   public class SignDemo      public static boolean isNegative byte x        return    x  gt  gt  7    amp  1     1         public static boolean isNegative short x        return    x  gt  gt  15    amp  1     1         public static boolean isNegative int x        return    x  gt  gt  31    amp  1     1         public static boolean isNegative long x        return    x  gt  gt  63    amp  1     1         public static boolean isNegative float x        return isNegative  int x          public static boolean isNegative double x        return isNegative  long x          public static void main String   args             byte     System out printf  Byte  b n  isNegative  byte 1        System out printf  Byte  b n  isNegative  byte -1            short     System out printf  Short  b n  isNegative  short 1        System out printf  Short  b n  isNegative  short -1            int     System out printf  Int  b n  isNegative 1        System out printf  Int  b n  isNegative -1            long     System out printf  Long  b n  isNegative 1L        System out printf  Long  b n  isNegative -1L            float     System out printf  Float  b n  isNegative Float MAX VALUE        System out printf  Float  b n  isNegative Float NEGATIVE INFINITY            double     System out printf  Double  b n  isNegative Double MAX VALUE        System out printf  Double  b n  isNegative Double NEGATIVE INFINITY            interesting cases        This will fail because we can t get to the float bits without an API and        casting will round to zero     System out printf   -1 1   fail   b n  isNegative -0 5f

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It seems arbitrary to me because I don t know how you would get the number as any type  but what about checking Abs number     number   Maybe  amp  amp  number    0

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Integers are trivial  this you already know  The deep problem is how to deal with floating-point values  At that point  you ve got to know a bit more about how floating point values actually work   The key is Double doubleToLongBits    which lets you get at the IEEE representation of the number   The method s really a direct cast under the hood  with a bit of magic for dealing with NaN values   Once a double has been converted to a long  you can just use 0x8000000000000000L as a mask to select the sign bit  if zero  the value is positive  and if one  it s negative

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Try this without the code   x-SQRT x 2    2 x

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This one is roughly based on ItzWarty s answer  but it runs in logn time  Caveat  Only works for integers   Boolean isPositive int a      if a    -1  return false    if a    0  return false    if a    1  return true    return isPositive a 2

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You say     we should not use conditional operators   But this is a trick requirement  because    is also a conditional operator   There is also one built into      while  and for loops   So nearly everyone has failed to provide an answer meeting all the requirements   The only way to build a solution without a conditional operator is to use lookup table vs one of a few other people s solutions that can be boiled down to 0 1 or a character  before a conditional is met   Here are the answers that I think might work vs a lookup table    Nabb Steven Schlansker Dennis Cheung Gary Rowe

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if  v  lt  0  System out println  negative    else System out println  positive

[java] How to determine if a number is positive or negative?

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