Please help me to detect the error

Question

This code works fine for all decimals except 0.(3).
I wasted my 1 hr but unable to figure out where I went wrong and why numberformatexception is coming for this particular input, but didn't got any solution.
LukArToDo help me to find the mistake.

https://code.sololearn.com/cV86SWCV8ySp/?ref=app

Answer 1

Buddy, put after line 42 and before line 43:
r = Double.parseDouble(String.format("%."+(a[0].length()+a[1].length())+"f", r));

Answer 2

I even think it would be enough to put a[1] .length() as number of decimal places in String.format() expression, but with (a[0].length()+a[1].length()) we will get better precision.
So, you can test solution with :
r = Double.parseDouble(String.format("%."+ a[1].length()+"f", r));

Answer 3

RAHUL JAISWAL
You're welcome 😉

Answer 4

Double precision gives a precision of up to 15 decimal places, and it is out the int range (number with 15 digits).
The enigma is in which cases a double precision gives an "unusual" result.
For example:
5.6 + 5.8 = 11.399999999999999
or
100.266-100.0=0.26600000000000534 
or in your case :
0.33 *10 = 3.3000000000003
etc...

Therefore, we will format the obtained number "r" into the number with the desired decimal places using String.format()

In short, the String.format("%."+(a[0].length()+a[1].length())+"f", r)  syntax tells Java to return your variable (r) with (a[0].length()+a[1].length()) decimal places (which is a length of our input) in decimal representation of a floating-point number (f) from the start of the format specifier (%).
So, if your input is 0.(3) -> length/or number of decimal places in String.format() expression is 5
So, if r=3.000000000004 above String.format() will return String "3.00000"
After Double.parseDouble("3.00000") you will get r=3.0

Answer 5

Line 45 parseInt is what is crashing because the string overflows an integer number. Your double math on line 42 ends up with a tiny precision error getting the .0000000000000004

Answer 6

It also does it for 0.(6)
0.(9) doesn't result in an error, but also doesn't result in the correct answer.

Edit, I guess it's about as close as you can get to correct though. 😉

Answer 7

Yes, that's true. lol, I didn't really look into what your program was doing yet at that point. I hadn't even seen that it was an epsilon issue yet. Just saw the error and that it was beyond the limits of an int. long day = tired. lol

Answer 8

The error is originating from this part of your call to your lowest method;
Integer.parseInt(a[0]+a[1])
In the event that you enter 0.(3) as your input a[0] is "3" and a[1] is "0000000000000004". The 2 Strings are concatenated to create "30000000000000004". When this String is parsed for an int it results in an error. The max size for an int value in Java is 2147483647.

You may want to try using a long type instead, but your loop in the lowest method will most likely time out in the playground.

Answer 9

It seems a simple fix for the issue would be to programmatically change a repeating number that begins immediately after the decimal I.E. 0.(3) to something like 0.3(3) as this seems to work fine.

Answer 10

It happens for 0.(6) also.

Answer 11

Probably not the best solution, but this seems to work. At least for 0.(3) and 0.(6). Otherwise, you may try doing some epsilon comparison checking.

int lowest_a = 0;
try {
lowest_a = Integer.parseInt(a[0]+a[1]);
} catch (NumberFormatException e) {
a[1] = "0";
lowest_a = Integer.parseInt(a[0]+a[1]);
}
lowest (lowest_a ,(k-1)*(int)Math.pow(10,a[1].length()));

Answer 12

ChaoticDawg Thanks frnd it works

Answer 13

It can be due to floating point precision error ,but why it is working fine with 0.(2) , 0.(7) , etc

Answer 14

ChaoticDawg Yes I know that , but I want to know why this precision error comes only in case of 0.(3)

Answer 15

ChaoticDawg But even if I use long , it will not work as the actual value will change , giving incorrect outputs.

Answer 16

LukArToDo Thanks , it is best method , can you explain it.

Answer 17

LukArToDo Thanks alot for the explanation

Answer 18

Your code is poorly written.

Bad variable names, no comments, not splitting up code into functions with good function names.

You would find bugs far easier, if you did those things. I've started adding decent variable names, and it already makes more sense:
https://code.sololearn.com/ch0TJXj6N2jD/?ref=app

Answer 19

John Wells Yes that is a precious error but that is coming when I am multiplying 0.33 *10 = 3.3000000000003 , further Subtracting it with 0.3, gives 3.000000000004

Answer 20

ChaoticDawg John Wells right, so what is the best way to solve this, other than converting 0.(3) to 0.3(3).