Explanation on Integer MAX VALUE and Integer MIN VALUE to find min and max value in an array

Question

I don t seem to understand how Integer MAX VALUE and Integer MIN VALUE help in finding the min and max value in an array   I understand how this method  pseudocode below  works when finding the min and max values   max   A 0   min   A 0  for each i in A   if A i   gt  max then max   A i    if A i   lt  min then min   A i     But as for this method  I don t understand the purpose of Integer MAX VALUE and Integer MIN VALUE   import java util Scanner   class MyClass        public static void main String   args             int   numbers     declaring the data type of numbers         numbers   new int 3     assigning the number of values numbers will contain         int smallest   Integer MAX VALUE  largest   Integer MIN VALUE           Scanner input   new Scanner System in            System out println  Please enter 3 numbers             for int counter   0  counter lt numbers length counter                  numbers counter    input nextInt                       for int i   0  i lt numbers length  i                  if numbers i  lt smallest                  smallest   numbers i               else if numbers i  gt largest                  largest   numbers i                      System out println  Largest is   largest           System out println  Smallest is   smallest               System out println Integer MAX VALUE  gives 2147483647 System out println Integer MIN VALUE  gives -2147483648   So what purpose do Integer MIN VALUE and Integer MIN VALUE serve in the comparisons

User · Answer

Instead of initializing the variables with arbitrary values (for example int smallest = 9999, largest = 0) it is safer to initialize the variables with the largest and smallest values representable by that number type (that is int smallest = Integer.MAX_VALUE, largest = Integer.MIN_VALUE).

Since your integer array cannot contain a value larger than Integer.MAX_VALUE and smaller than Integer.MIN_VALUE your code works across all edge cases.

User · Answer

By initializing the min max values to their extreme opposite  you avoid any edge cases of values in the input  Either one of min max is in fact one of those values  in the case where the input consists of only one of those values   or the correct min max will be found   It should be noted that primitive types must have a value  If you used Objects  ie Integer   you could initialize value to null and handle that special case for the first comparison  but that creates extra  needless  code  However  by using these values  the loop code doesn t need to worry about the edge case of the first comparison   Another alternative is to set both initial values to the first value of the input array  never a problem - see below  and iterate from the 2nd element onward  since this is the only correct state of min max after one iteration  You could iterate from the 1st element too - it would make no difference  other than doing one extra  needless  iteration over the first element   The only sane way of dealing with inout of size zero is simple  throw an IllegalArgumentException  because min max is undefined in this case

User · Answer

but as for this method  I don t understand the purpose of Integer MAX VALUE and Integer MIN VALUE    By starting out with smallest set to Integer MAX VALUE and largest set to Integer MIN VALUE  they don t have to worry later about the special case where smallest and largest don t have a value yet  If the data I m looking through has a 10 as the first value  then numbers i  lt smallest will be true  because 10 is  lt  Integer MAX VALUE  and we ll update smallest to be 10  Similarly  numbers i  gt largest will be true because 10 is  gt  Integer MIN VALUE and we ll update largest  And so on   Of course  when doing this  you must ensure that you have at least one value in the data you re looking at  Otherwise  you end up with apocryphal numbers in smallest and largest     Note the point Onome Sotu makes in the comments         if the first item in the array is larger than the rest  then the largest item will always be Integer MIN VALUE because of the else-if statement    Which is true  here s a simpler example demonstrating the problem  live copy    public class Example       public static void main String   args  throws Exception           int   values    5  1  2           int smallest   Integer MAX VALUE          int largest    Integer MIN VALUE          for  int value   values                if  value  lt  smallest                    smallest   value                else if  value  gt  largest                    largest   value                                  System out println smallest          largest      1  2 -- WRONG           To fix it  either    Don t use else  or Start with smallest and largest equal to the first element  and then loop the remaining elements  keeping the else if    Here s an example of that second one  live copy    public class Example       public static void main String   args  throws Exception           int   values    5  1  2           int smallest   values 0           int largest    values 0           for  int n   1  n  lt  values length    n                int value   values n               if  value  lt  smallest                    smallest   value                else if  value  gt  largest                    largest   value                                  System out println smallest          largest      1  5

[java] Explanation on Integer.MAX_VALUE and Integer.MIN_VALUE to find min and max value in an array

Examples related to java

Examples related to arrays

Examples related to numbers