phouthasak Does anyone feel like the instructions to this challenege is kind of vague?
Shafaet Please suggest how can we improve this challenge.
kucao class Solution{ public static void main(String []argh) {
Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++) { try { long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=-128 && x<=127)System.out.println("* byte"); //Complete the code if(x >= -Math.pow(2, 15) && x <= Math.pow(2, 15) - 1) System.out.println("* short"); if(x >= -Math.pow(2, 31) && x <= Math.pow(2, 31) - 1) System.out.println("* int"); if(x >= -Math.pow(2, 63) && x <= Math.pow(2, 63) - 1) System.out.println("* long"); } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } }}
Kinara108 Hey Kucao, I had the same solution as yours, with just this minor change, and I got it wrong. Not able to understand why. Can someone explain ?
if(x >= (short)-Math.pow(2, 15) && x <=(short) Math.pow(2, 15) - 1) System.out.println("* short"); if(x >= (int)-Math.pow(2, 31) && x <=(int) Math.pow(2, 31) - 1) System.out.println("* int"); if(x >= (long)-Math.pow(2, 63) && x <=(long) Math.pow(2, 63) - 1) System.out.println("* long");
ialmustafa casting will truncate value
kanhaiyabgs but here it every casting is in range its range we are dealing with int types so where comes the question of casting
zjesse1314 That messed me up. I casted and was not able to get the as long value. Once I remove it worked.
amitrajitbose9 We do not need to cast explicitly. We are just required to check the applicable ranges.
if(x>=-128 && x<=127) System.out.println("* byte"); if(x>=-(Math.pow(2,15)) && x<=(Math.pow(2,15)-1)) System.out.println("* short"); if(x>=-(Math.pow(2,31)) && x<=(Math.pow(2,31)-1)) System.out.println("* int"); if(x>=-(Math.pow(2,63)) && x<=(Math.pow(2,63)-1)) System.out.println("* long");
barathvicky i tried using appying the range directly but i couldnt able to clear two test cases
anshul_sri002 which?
barathvicky u added pow function but i tried with entering the range values like for byte -128 to 127 but i am not able to clear the 2 testcases among 4
balavemulakonda but then..i thnk this wont work?
shwetham369 Thanks...
ritabandas2000 Casting is necessary because Math.pow function returns value in double
Tutankhamun (int) Math.pow(2, 31) casting would be done before the subtraction I believe.
Do you need to cast?
also, you really don't need to check for long. Any value not throwing an exception will fit.mkvarma use x<= (short) (Math.pow(2,15)-1.0)
pimk85 Simply write your own 2^x function and use that instead of Math.pow which has an upper limit:
public static long power(int x){ long r = 1; for (int k=0;k<x;k++){ r = r * 2; } return r; }
np_assignment Nice
aniketwarey [deleted]aniketwarey I think for loop is troubling everyone here (including me ). For loop limit should be the first int we are taking . Its just not any random number, its actually the number of inputs which different test cases gonna have.
hrkhandique same here
deekshithsagar73 same
shankarpentyala use (short)(Math.pow(2,15)-1) instead of (short) Math.pow(2,15) -1 because:
The value of Math.pow(2,15) = 32768.0 ,when you type cast it to short its value becomes -32768 and when you subtract it from 1 the value is -32769 .So,first subtract 1 from the Math.pow(2,15) then type cast to short.The same logic for the int and long.
jakkampudimurali Agreed. That makes sense.
vardhangaharwar not running
srikanthdikalla if (x >= Short.MIN_VALUE && x <= Short.MAX_VALUE) System.out.println("* short"); if (x >= Integer.MIN_VALUE && x <= Integer.MAX_VALUE) System.out.println("* int"); if (x >= Long.MIN_VALUE && x <= Long.MAX_VALUE) System.out.println("* long");
deekshithsagar73 no no impossible
papnai_preeti We can reply a redundant condition check through the below mentioned code:
try{ long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=Byte.MIN_VALUE && x<=Byte.MAX_VALUE)System.out.println("* byte\n* short\n* int\n* long"); else if(x>=Short.MIN_VALUE && x<=Short.MAX_VALUE)System.out.println("* short\n* int\n* long"); else if(x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE)System.out.println("* int\n* long"); else System.out.println("* long"); }catch(Exception ex){ System.out.println(sc.next()+" can't be fitted anywhere."); }
manoj_c_balakri1 Why are we including -1 at the end. can anyone explain me?
deekshithsagar73 noo
Rishabhgupta9695 because of range
barathvicky can you explain it clearly please
luigo - 1 is the bit for the negative sign "-"
ahjain2706 any of the data type has a specific range. take byte! its range is from -128 to 127; if you raise - 2 to the power of 7 i.e. -128. (this number is in the required range!). but if you raise 2 to the power of 7 i.e. 128. (this number is not in the required range(127)). So, all you need is subtract 1 from 128 and hence the code :
// sample code. if(x> -Math.pow(2,7) && x < Math.pow(2,7) - 1){ System.out.println("* byte"); System.out.println("* short"); System.out.println("* int"); System.out.println("* long"); }
singhsaurabh_ki1 bacause, like byte range is 8 so last in 0 to 7 meanse 8-1=7 hope you got it
ghazi261 It's to do with the range that data type can have.
For example short can have minimum -32,768 to maximum of 32,767.
Same case for int & long, which is ONE less number for it's maximum.
It's easier to write (2^31 - 1) than it is to write 2147483648 - 1 = 2147483647
skhayalian long 64 –9,223,372,036,854,775,808 to 9 ,223,372,036,854,775,807
int 32 –2,147,483,648 to 2,147,483,647
short 16 –32,768 to 32,767
byte 8 –128 to 127
if you recognize the positive number of the primitive variable is alway less with 1, check the Byte the minimum is -128 and the positive is not 128, its 128 -1 = 127, so we add -1 so we can have the right positive range and have the right value. wish i was helpfullghazi261 Thanks for your reply. I'm aware of the data type ranges. Simply just inserting the ranges within an if statements isn't enough and doesn't pass you the test cases 3 & 4.
Nevertheless, I've managed to solve the challenge! :D
anshul_sri002 Because in range of every databytes from negative to positive(postive is (-1) less than to negative ) as in byte -128 to +127.
Manish_Sharma_01 If first if condition work then no other if will work.
sashanksingh1128 (short)-Math.pow(2, 15) What is the use of minus sign '-' there???
Matriux short range of value lie to (-2 ^15 to 2^15-1)
anks2906 if(x >= -Math.pow(2, 63) && x <= Math.pow(2, 63) - 1) this particular condition checking isnt required at all.
According to problem statement the long data type can fit in all possibe input except when it doesnt. So the initialization and assignment part
long x=sc.nextLong(); along with the try and catch exception checking and handling blocks take care of that.
smritirastogi247 hey,kinara.....please explain me the needs for type casting every Math.pow().....??
ekanshgupta1997 yes here's need of this function
cse_15bcs7223 hey dear Smiriti there is no need math.pow() fuction as u can see there is long in all cases except can't fitted else condition so use simply this code import java.util.; import java.io.;
class Solution{ public static void main(String []argh) {
Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++) { try { long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=-128 && x<=127)System.out.println("* byte"); if(x>=-32768 && x<=32767)System.out.println("* short"); if(x>=-2147483648 && x<=2147483647)System.out.println("* int"); System.out.println("* long"); //Complete the code } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } }}
vora_raj71 I just have one silly question. Say we are doing for short then I wanted to know why we are doing Math.pow(2,15) and not Math.pow(2,16) considering we want 2 to the power of 16 and not 15.
h1705410 [deleted]
rahulshetty96 hey kucao,why did you converted the int number to long?what is the need of doing that?please reply.
vijay456in int is greater than short and byte only. If we take an int variable, that can also be used to store even byte value and also short value bcoze int is bigger than short and byte. like that if we take a long, that can be used to compare with any of byte, short, int.
deekshithsagar73 reply
jgj_io I solved it by using Datatype.MAX_VALUE; DataType.MIN_VALUE; ex. Long.MAX_VALUE;
Which might not be the point of the exercise, but I think it's really robust code.
cletusw This really is the best way to accomplish this task.
jagadishmanjuna1 what method should i use
deekshithsagar73 all in one
deekshithsagar73 no no its waste
shubhamrathii [deleted]shubhamganguly11 I did it the same way! ran only the sample test others were wrong
sashanksingh1128 I tired to do the same but got wrong answer for test case 2 and 3.
long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x >=(short)Short.MIN_VALUE && x <=(short)Short.MAX_VALUE) System.out.println("* short"); if(x >=(int)Integer.MIN_VALUE && x <=(int)Integer.MAX_VALUE) System.out.println("* int"); if(x >=(long)Long.MIN_VALUE && x <=(long)Long.MAX_VALUE) System.out.println("* long");
Matriux if(x>=Byte.MIN_VALUE && x<=Byte.MAX_VALUE) System.out.println("* byte"); if(x>=Short.MIN_VALUE && x<=Short.MAX_VALUE) System.out.println("* short"); if(x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE) System.out.println("* int"); if(x>=Long.MIN_VALUE && x<=Long.MAX_VALUE) System.out.println("* long"); }
Matriux if(x>=Byte.MIN_VALUE && x<=Byte.MAX_VALUE) System.out.println("* byte"); if(x>=Short.MIN_VALUE && x<=Short.MAX_VALUE) System.out.println("* short"); if(x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE) System.out.println("* int"); if(x>=Long.MIN_VALUE && x<=Long.MAX_VALUE) System.out.println("* long"); } using Wrapper Class we can solve it.
Akarshit_Joshi thanks this one is helpful
snath402 It worked
shankyftw Can you tell me what does the -1 do in the if statement?
rs2357 If you are still curious about this now, it accounts for 0.
deekshithsagar73 am i
deekshithsagar73 it does what it doos
karthiban The range of byte is (2^7) to (2^7)-1. ie -128 to 127
Fixer [deleted]
nagendrapatel86 Hey Shafaet, I had some masking techique that can not using any library but am not implemented it,
A&0=0
Vimsow The return you get from Math.pow is in doubleand you're comparing it with x i.e.long. Can we do it? Don't you think there's flaw here? If no! Could you please explain why? Using MIN and MAX is the best option I suppose.
dharani8696 can you please tell me why we want to use -1 ??
gurpriya06 because one of the value is zero. to adjust that !
sravaniteja short range is from -32768 to 32767 but not from -32768 to 32767 so we use '-1' in Math.pow(2,15)
palash_rawat What's the difference between -32768 to 32767 and -32768 to 32767
deekshithsagar73 the difference is in th question
atulr830 [deleted]
innocent_badmash hey Kucao, With the help of this code i creared only Test case #0. for other test case it is wrong. Can you please help me.
dzhang124 I had the same solution. But i wasnt able to run it. System said it is a worng answer. Can someone please explain? Thanks!
deekshithsagar73 explain
rjranjan11 thanks
deekshithsagar73 welcome
duvvadaharikris1 i have written the same code but it shows error in last two testcases
Janardan77 me too
deekshithsagar73 u did it wrong
neil_cuajotor why is
if(x>=-32768 && x<=32768)
Different from:
if(x >= -Math.pow(2, 15) && x <= Math.pow(2, 15) - 1)
I tried the first one since its the value given by java documentation but it doesn't work on the last two test cases. can anyone please tell me..
gklogodazhki //if(x >= -Math.pow(2, 15) && x <= Math.pow(2, 15) - 1) //if(x >= - 32768 && x <= 32768 - 1) if(x >= -32768 && x <= 32767)
rishabh0_9 I also have same problem .. and till now didn't understand .. so please any'ne can explain me
raphaelsoto717 instead of if(x>=-32768 && x<=32768) try if(x >= -32768 && x <= 32767) I think it's because 0 counts and a bit so the range ends at 32767 instead of 32768
deekshithsagar73 yes
sravaniteja short range is from -32768 to 32767 but not from -32768 to 32767
LoveenDwivedi9 i think its related to priority x>=-128 && x<=127 we havent write it as x>=-128 && x<=128-1 the reason is <= operator has greater priority then -(minus)
surajahuja69 plz explain this solution
deekshithsagar73 noo
patraism why sc.next() is used in catch??? why cant we use x instead ?
deekshithsagar73 i dont know
adi_gupta95 Thanks.. It helped!
deekshithsagar73 fake
nddai96 what about test case -100000000000000000??? it's not work???
HarsimratKohli try this :- long num2 = (long)Math.pow((double)2,(double)63); long x = sc.nextLong(); ...... ...... if(x>=-m2 && x<=m2-1)System.out.println("* long");
deekshithsagar73 yes
charchitmadan can you please tell what sc.next() does?
marcusalexander .next method reads the next token of input
Arpit23 please explain how sc.next() working outside the try{}. System.out.println(sc.next()+" can't be fitted anywhere.");
Krishankant [deleted]
adhi_khamitkar i entered the actual values in test cases. It threw error for the value 9223372036854775807 saying that it can't store that number for long.
Avery_King Add an L at the end -- 9223372036854775807L -- specifies a comparison with the long type
Shraddha_gami @kucao why you had writeen "can't be fitted anywhere" in catch block which exception will be thrown for this?
patnala07 can you please explain me this code ?
adiaholic i'm unable to submit this above code as well don't understand why
nirariz hye.can someone explain what is the meaning of -Math.pow(2,15) and when we need to use -Math.pow?and also the function of -Math.pow.tq
rsb0017 why compiler showing error out of range if(x>=-9223372036854775808 && x<= 9223372036854775807) as 9223372036854775807 is the range of long
vikash_nitmz For long value you must put 'L'after the value.(use 9223372036854775807L)
toufiqj27 if(x >= -Math.pow(2, 15) && x <= Math.pow(2, 15) - 1) System.out.println("* short"); if(x >= -Math.pow(2, 31) && x <= Math.pow(2, 31) - 1) System.out.println("* int"); if(x >= -Math.pow(2, 63) && x <= Math.pow(2, 63) - 1) System.out.println("* long"); please can you explain these lines ??...deekshithsagar73 yes
chandmumu why we use try?
dryabi01 Kucao, and all
Kucao's solution is NOT exactly correct. What if the input is number = 10. Number 10 can be fit into ANY primitive type. So, every single "if" in Kucao's code will be duplicated in the output.
Better to use if-else if construct. Like this
long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=-128 && x<=127)System.out.println("* byte\n* short\n* int\n* long"); else if( x>=((-1)*(Math.pow(2,15))) && x<=((Math.pow(2,15)-1)) )System.out.println("* short\n* int\n* long"); else if( x>=((-1)*(Math.pow(2,31))) && x<=((Math.pow(2,31)-1)) )System.out.println("* int\n* long"); else if( x>=((-1)*(Math.pow(2,63))) && x<=((Math.pow(2,63)-1)) )System.out.println("* long");
sgadade7 Not working for Cases #2 and #3
vikash_nitmz you write exact value for 2^63 & 2^31,and it will work.
misoli_fash U don't need -1 at the end of each code result still is same
juangiacosabrk In your solution Math.pow(2,63) is actually going to return an Int, it will return the largest int it can be calculated (2^31)-1. You have to specifically point out which numbers are going to be considered longs by appending an L at the end of the number (Like this: 600851475143L). This is because in java long literals end with an L. Source
I have almost the same solution as you, with the exception that I have every max and min dataType value stored in a different variable (like minInt, maxInt, minShort and so on).
Even though, I cant get it right on the test scenarios once the solution is submitted. Its quite frustrating.
MichaelWittmann Instead of wasting your time and resources doing mathematical calculations, you should just use final static values that are kept in classes. For example classes Byte, Short, Int, Long have final static variables for their MIN_VALUE and MAX_VALUE. So you can write your program like this to make it much more efficient. Don't forget, we are computer engineers. Engineer is a person who solves problem and solved problems efficiently. Solving a problem alone is not engineering.
if (x >= Byte.MIN_VALUE && x <= Byte.MAX_VALUE) { System.out.println("* byte"); } if (x >= Short.MIN_VALUE && x <= Short.MAX_VALUE) { System.out.println("* short"); } if (x >= Integer.MIN_VALUE && x <= Integer.MAX_VALUE) { System.out.println("* int"); } if (x >= Long.MIN_VALUE && x <= Long.MAX_VALUE) { System.out.println("* long"); }
vuppugovind3 thank you MichaelWittmann this worked, great.
arhankundu99 in the following line of code: "System.out.println(sc.next()+" can't be fitted anywhere.");",instead of "sc.next",Why can't we write "sc.nextLong"?
sven15 Some of the other comments already mentioned it, but basically the nextLong() token threw an error, so reading the same token with next() allows us to repeat the token as part of the error message.
prernasharma3011 This solution is giving 2 testcases wrong. What is the problem? Can anyone please explain ?
Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++) { try { long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=-128 && x<=127) System.out.println("* byte"); //Complete the code if(x>=-32768 && x<=32767) System.out.println("* short"); if(x>=Math.pow(-2,31) && x<Math.pow(2,31)) System.out.println("* int"); if(x>=Math.pow(-2,63) && x<Math.pow(2,63)) System.out.println("* long"); } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } sc.close();}
1RV15EC035 All test cases have been passed
There is only one wrong in your code , you just make the last test case as if(x>=Math.pow(-2,63) && x<= Math.pow(2,63)).
You basically needed to add an equal there, else it wont work.
Hope this helps
HK1427 WHY WE USE SC.NEXT() IN CATCH PLS EXPLAIN!
adil_H7 if you entered the catch block, an exception has been raised.
most surely, the exception is due to the statement
long x=sc.nextLong();Hence, you cannot use nextLong() to obtain that large a number.
A different way to obtain this number is to use next(). This function returns the next token as a String (in other words, as is). Since every input is a valid String, this function will not fail and will surely give you the next token (the token you failed to obtain using nextLong()).
abhishek_goel287 Cool
ragulkumarv333 can someone explain this code.please
getsadzeg I think it is very unnecessary to check if it fits into long. We have x variable in long datatype in the beginning, and also if it fits byte, short and int all the way, it'll definitely fit in long. So you can just print that it fits into long, without checking.
P.S. Why doesn't HackerRank have code formatting?..
eshandangwal9991 Can we use range instead of Math.pow()?How Math.pow() working?
SinghGurkirat we should not compare long value with double
Jenko I think that is better to use the static max value instead of computing them every run. So in my if conditions i put the Long.MAX_VALUE and Long.MIN_VALUE and so on for each primitive type.
deepshaswat Hi,
As per the Java site for Java 8 for int: 2^32 and for long 2^64 are the accepted value but while submission, the ans for TC 2,3 and 4 do not accept the values of 2^32 and 2^64.
reformingpanda unsigned representations are only accessible through the objects Integer and Long, not through the primitive data types int and long that are referenced in this problem.
jmuzikant Shafaet: I ve tryied to avoid the Math functions with this code but only It passed the first test. In my eclipse after a big number it stop to read from the input:
long two_exp_31 = two_exp_15*two_exp_15*2; long two_exp_63_minus_one = two_exp_31*two_exp_31-1+two_exp_31*two_exp_31; if(x>=-(two_exp_15-1) && x<=(two_exp_15-1))System.out.println("* short"); if(x>=-(two_exp_31-1) && x<=(two_exp_31-1))System.out.println("* int"); if(x>=-(two_exp_63_minus_one) && x<=(two_exp_63_minus_one)) System.out.println("* long");
Could you give me an idea about the rest of the test cases?Thanks
deekshithsagar73 yes
lap_minguez using import java.lang.*;
The rest of the code is this
if(x>=Short.MIN_VALUE && x<=Short.MAX_VALUE)System.out.println("* short"); if(x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE)System.out.println("* int"); if(x>=Long.MIN_VALUE && x<=Long.MAX_VALUE)System.out.println("* long");
glebocki_przemy1 Thanks for the tip using the MIN_VALUE and MAX_VALUE I had no Idea that there were those in primitive data type classes.
anup_gawande Hi Thanks for your solution. I just wanted to add something in it. All classes in the java.lang package are imported by default. Thus you do not need to import java.lang.*;
nicolas_saavedr1 tnks for the tip, i used those method and every test they ran.
zerovpp thank you,very godd!
deekshithsagar73 good*
pinakibnrj8 thanks a lot, learnd to use these functions
angel_aarti_29 Thanks man .... i was frustraed as my code was,nt submitting.
nadhikary this is aweful, why can't the code read as value instead off the static variables? no need to write code if it cannot read direct value.
rvtwr123 if (number >= Byte.MIN_VALUE && number <= Byte.MAX_VALUE) System.out.println("* byte\n* short\n* int\n* long"); else if (number >= Short.MIN_VALUE && number <= Short.MAX_VALUE) System.out.println("* short\n* int\n* long"); else if (number >= Integer.MIN_VALUE && number <= Integer.MAX_VALUE) System.out.println("* int\n* long"); else if (number >= Long.MIN_VALUE && number <= Long.MAX_VALUE) System.out.println("* long");
Evision the MAX_VALUE and MIN_VALUE would..
yusufm_salh you can teach the datatypes sizes in another way,I am a begginer in java ,but yet advanced in c# ,I had no idea how to think of this problem,kindly try to improve the "Cannot fit statment".
SevMardi Please, I had the same problem as above. I wasn't able to understand the question/challange easily. I would suggest the following.
Problem
Task
Explanation
Example
Input
output < you have this as well>
Doing this, I think, should provide a good understanding of what needs to happen.
And please, try to keep it plain and clean English.
Thank you!
keithchongwy I agree. We went from explaining the data types to input format and output format. I havent got around the idea of UX, but now I see why its important XD
deekshithsagar73 welcome
mubarizmirzayev1 Instead of writing unnecessary comments to everybody, you can solve the given task.
deekshithsagar73 instead of posting unnecessary comments to everybody,you can solve the given task
mubarizmirzayev1 Already, did. Genious guy
jorp78 Scanner sc = new Scanner(System.in); int t = sc.nextInt(); for (int i = 0; i < t; i++) { try { long x = sc.nextLong(); System.out.println(x + " can be fitted in:"); if (x >= Byte.MIN_VALUE && x <= Byte.MAX_VALUE) { System.out.println("* byte"); } if (x >= Short.MIN_VALUE && x <= Short.MAX_VALUE) { System.out.println("* short"); } if (x >= Integer.MIN_VALUE && x <= Integer.MAX_VALUE) { System.out.println("* int"); } if (x >= Long.MIN_VALUE && x <= Long.MAX_VALUE) { System.out.println("* long"); } } catch (Exception e) { System.out.println(sc.next() + " can't be fitted anywhere."); } }ksaigopi222 Hi, Why you used sc.next() in the Exception block ? Can you please Explain me ?
eulen_mesquita [deleted]andreasschmidth long x = sc.nextLong() throws the Exceptions, so the Token (the number) will never get parsed. So he has to parse the number via sc.next() as a String, so its available in the error message.
deekshithsagar73 noo
CSECG67 thanku so much sir. i understand now how to solve the problems like these.
IrfanShaikh I'm also do this but why my test case are failed...? can you help me please..
stelabr2 Ahh that's how you access the MAX_VALUE and MIN_VALUE constants. Thanks. Just a note, you could remove
if (x >= Long.MIN_VALUE && x <= Long.MAX_VALUE).xhas to be alongby definition so we don't have to check it.
rovi8930 [deleted]
sslohara_giri import java.util.; import java.io.;
class Solution{ public static void main(String []argh) {
Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++) { try { long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=-128 && x<=127)System.out.println("* byte"); if(x>=-32768 && x<=32767)System.out.println("* short"); if(x>=-Math.pow(2, 31) && x<=Math.pow(2,31)-1)System.out.println("* int"); System.out.println("* long"); } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } }}
petrikb66 if(x>=-(1<<15) && x<=(1<<15)-1)System.out.printf("* short\n"); if(x>=-(1<<31) && x<=(1<<31)-1)System.out.printf("* int\n"); if(x>=-(1L<<63) && x<=(1L<<63)-1)System.out.printf("* long\n");alexgordongbs Wow! I can see a hand of the C-programmer here... :-)
charchitmadan why i have to use this statement if(x>= -Math.pow(2, 63) && x=-9,223,372,036,854,775,808 && x<=9,223,372,036,854,775,807) why this statement gives me error which says very large number
kavyasree42 since it is a long literal use L at the end of that number. (x=-9223372036854775808L && x<=9223372036854775807L)
charchitmadan [deleted]vcuong0909 import java.util.; import java.io.;
class Solution{ public static void main(String []argh) {
Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++) { try { long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=-128 && x<=127)System.out.println("* byte"); if(x>=Short.MIN_VALUE && x<= Short.MAX_VALUE) System.out.println("* short"); if(x>=Integer.MIN_VALUE && x<= Integer.MAX_VALUE) System.out.println("* int"); if(x>=Long.MIN_VALUE && x<= Long.MAX_VALUE) System.out.println("* long"); //Complete the code } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } }}
GoelAnuj // try this one. clean code and now need of pow function
import java.util.; import java.io.; import java.math.*;
class Solution{ public static void main(String []argh) {
Scanner sc = new Scanner(System.in); int t = sc.nextInt(); for(int i=0; i<t; i++) { try { long num = sc.nextLong(); System.out.println(num+" can be fitted in:"); if( num >= -128 && num <= 127)System.out.println("* byte"); if( num >= Short.MIN_VALUE && num <= Short.MAX_VALUE) System.out.println("* short"); if( num >= Integer.MIN_VALUE && num <= Integer.MAX_VALUE) System.out.println("* int"); if( num >= Long.MIN_VALUE && num <= Long.MAX_VALUE) System.out.println("* long"); } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } }}
kpavankumar6231 i dont know what is the difference between 1.(x < Math.pow(2,63)) and 2.(x <= Math.pow(2,63)-1).i think both are same but if i write first statement in code two test cases fail but second one all test cases pass. plz any one elaborate it, thank you.
arjunaju_r_k [deleted]
dJOKERdUMMY [deleted]
vikas_chaubey12 it can be solved just took a little more effort, this code is working fine.
import java.util.; import java.io.; import java.math.BigInteger;
class Solution{
public static void main(String []argh) { long l1 = 9223372036854775807l; long l2 = -9223372036854775808l; long i1 = 2147483647l; long i2 = -2147483648l; long s1 =32767l; long s2 = -32768l; long b1 = 127l; long b2 = -128l; BigInteger la =BigInteger.valueOf(l1); BigInteger lb =BigInteger.valueOf(l2); BigInteger ia =BigInteger.valueOf(i1); BigInteger ib =BigInteger.valueOf(i2); BigInteger sa = BigInteger.valueOf(s1); BigInteger sb =BigInteger.valueOf(s2); BigInteger ba =BigInteger.valueOf(b1); BigInteger bb =BigInteger.valueOf(b2); Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++) { BigInteger k = sc.nextBigInteger(); int a1,a2,b5,b6,c1,c2,d1,d2; a1 = k.compareTo(la); a2 = k.compareTo(lb); b5= k.compareTo(ia); b6 = k.compareTo(ib); c1 = k.compareTo(sa); c2 = k.compareTo(sb); d1 = k.compareTo(ba); d2 = k.compareTo(bb); if(a1==1 || a2 == -1){ System.out.println(k + " can't be fitted anywhere."); } else{ System.out.println(k +" can be fitted in:"); } if ((d1== -1||d1==0) && (d2 == 1||d2==0)){ System.out.println("* byte"); } if ((c1== -1||c1==0) && (c2 == 1||c2==0)){ System.out.println("* short"); } if ((b5== -1||b5==0) && (b6 == 1||b6==0)){ System.out.println("* int"); } if ((a1== -1||a1==0) && (a2 == 1||a2==0)){ System.out.println("* long"); } } }}
rajujnvgupta @vikas_chaubey12 plz explian the following contion. how it work ? if (a1==1 || a2 == -1) { }
ghazi261 not op, but @rajujnvgupta
|| is the Logical OR operator. -> returns a boolean which is true if and only if both of its operands are true. b = 3 > 2 || 5 < 7; // b is true b = 2 > 3 || 5 < 7; // b is still true b = 2 > 3 || 5 > 7; // now b is false
You should check out Java Operator and their Precedence
RamalakshmiG [deleted]rubi2690 thanks
deekshithsagar73 welcome
gowridevi214 yes...
aqs5998 yes it's extemely vague. It does not cover what bits are to begin with. Also it does not state on how to solve the problem to begin with.
avi89 A simple solution that doesn't involve the Java Math Class is to use the MIN and MAX fields on the Byte, Short, Integer, and Long wrapper classes.
try { long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=Byte.MIN_VALUE && x<=Byte.MAX_VALUE){ System.out.println("* byte"); } if (x>=Short.MIN_VALUE && x<=Short.MAX_VALUE){ System.out.println("* short"); } if (x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE){ System.out.println("* int"); } if (x>=Long.MIN_VALUE && x<=Long.MAX_VALUE){ System.out.println("* long"); } }
lahouari and to avoid further tests then write the following :
public static void main(String[] args) { Scanner sc = new Scanner(System.in); int T = sc.nextInt(); for (int i = 0; i < T; i++) { try { long n = sc.nextLong(); System.out.println(n + " can be fitted in:"); if (n >= Byte.MIN_VALUE && n <= Byte.MIN_VALUE) System.out.println("* byte\n* short\n* int\n* long"); else if (Short.MIN_VALUE && n <= Short.MIN_VALUE) System.out.println("* short\n* int\n* long"); else if (Integer.MIN_VALUE && n <= Integer.MIN_VALUE) System.out.println("* int\n* long"); else System.out.println("* long"); } catch (Exception e) { System.out.println(sc.next() + " can't be fitted anywhere."); } } sc.close(); }since byte < short < int < long
keithchongwy This works! Thank you so much!
deekshithsagar73 welcome
bpg168 Thanks, this is nice
deekshithsagar73 welcome
shanaka111 true
deekshithsagar73 yeah it is not compiling
CryptCode This should work without any problem and you don't have to remember range of any Datatypes. Happy Coding!
import java.util.*; import java.io.*; class Solution{ public static void main(String []argh){ Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++){ try{ long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=Byte.MIN_VALUE && x<=Byte.MAX_VALUE) System.out.println("* byte"); if(x>=Short.MIN_VALUE && x<=Short.MAX_VALUE) System.out.println("* short"); if(x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE) System.out.println("* int"); if(x>=Long.MIN_VALUE && x<=Long.MAX_VALUE) System.out.println("* long"); } catch(Exception e){ System.out.println(sc.next()+" can't be fitted anywhere."); } } } }
feyashah14 why cant we write actual values of range? As in for short why cant we write 32767
hitesh_chopra My Solution:
public static void main(String[] args) { Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++) { try { long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x == (byte)x)System.out.println("* byte"); if(x == (short)x)System.out.println("* short"); if(x == (int)x)System.out.println("* int"); if(x == (long)x)System.out.println("* long"); } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } }
ashishdalal_yes IF INTEGER TOO LONG ,ERROR COMES FOR ANYONE:
USE 'L' AT ENDING OF LONG.
HERE'S HOW IT IS: if(x>=-9223372036854775808L && x<=9223372036854775807L )System.out.println("* long");
AND UPVOTE IF IT HELPS YOU OUT.
bluskript import java.util.; import java.io.;
class Solution{ public static void main(String []argh) {
Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i=0;i<t;i++) { try { long x=sc.nextLong(); System.out.println(x+" can be fitted in:"); if(x>=-128 && x<=127)System.out.println("* byte"); if(x==(short)x)System.out.println("* short"); if(x==(int)x)System.out.println("* int"); if(x==(long)x)System.out.println("* long"); //Complete the code } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } }}
tamas_benno suggesting to use the fixed numeric min and max value for size check / if(x>=-128 && x<=127) / is not a good pattern.
I think the wrapper Classes MIN_VALUE, MAX_VALUE is more safe for this job.
if(x>= Byte.MIN_VALUE && x<=Byte.MAX_VALUE) ... etc
metinyalcinkaya import java.util.*; import java.io.*; class Solution{ public static void main(String []argh) { Scanner sc = new Scanner(System.in); int t=sc.nextInt(); for(int i = 0;i < t;i++) { try { long x = sc.nextLong(); System.out.println(x + " can be fitted in:"); if(x>=Byte.MIN_VALUE && x<=Byte.MAX_VALUE){ System.out.println("* byte"); } if(x>=Short.MIN_VALUE && x<=Short.MAX_VALUE) { System.out.println("* short"); } if(x>=Integer.MIN_VALUE && x<=Integer.MAX_VALUE) { System.out.println("* int"); } if(x>=Long.MIN_VALUE && x<=Long.MAX_VALUE) { System.out.println("* long"); } } catch(Exception e) { System.out.println(sc.next()+" can't be fitted anywhere."); } } } }
brettl86 Best Practice
lovermann Nice job!
prem_irudayaraj why is the code not taking input 5 into consideration ?? if we see the input and output in the question the number 5 is ignored
Pyxylation This is totally late. But, its not being ignored. Its the amount of inputs that are about to be passed through. So, its used with the loop above the try catch block.
umakanth @Shafaet_Ashraf The Sample Input given doesn't have the no of Test cases. And the problem statement doesn't speak about boolean data type.
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