Getting Numberformat exception with unknown source. How to fix this

Question

How to fix this issue?

java.lang.NumberFormatException: at java.lang.NumberFormatException.forInputString(Unknown Source)

I am doing some example problem and my code is working fine for the first string and digit. (Commented one)

But when change the new string and digit (Current one) I am getting this error :

java.lang.NumberFormatException: For input string: "299858953917872714814599237991174513476623756395992135212546127959342974628712329595771672911914471"
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Long.parseLong(Unknown Source)
    at java.lang.Long.parseLong(Unknown Source)
    at com.codejam.q1.problems.maxResult.removeDigit(maxResult.java:21)
    at com.codejam.q1.problems.maxResult.main(maxResult.java:10)

Here is my code. Anywhere I am missing something ?

public class maxResult {
    public static void main(String[] args) {
        //String str = "1231";
        String str = "2998589353917872714814599237991174513476623756395992135212546127959342974628712329595771672911914471";
        //char digit = '1';
        char digit = '3';
        System.out.println(removeDigit(str,digit));
    }
    
    public static String removeDigit(String number, char digit) {
        long result = 0;
        for(int i = 0; i<number.length(); i++) {
            char num = number.charAt(i);
            if(num == digit) {
                String myStr = number.substring(0, i) + number.substring(i + 1); 
                try{
                    long myNum = Long.parseLong(myStr);  
                    if(myNum > result) {
                        result = myNum;
                    }
                }
                catch (NumberFormatException ex){
                    ex.printStackTrace();
                }
            }
         }
        String s = String.valueOf(result);  
        return s;
    }
}

Even though I change int to long but no change in result.

If you are trying to remove every occurrence of the digit `3` from `str`, then you can simply do: `str.replaceAll("3", "");` Or did I misunderstand what you are trying to do? — Abra, May 01 '22 at 03:54
Alternatively, consider using class [java.math.BigInteger](https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/math/BigInteger.html) Refer to this SO question: [Is there an upper bound to BigInteger?](https://stackoverflow.com/questions/12693273/is-there-an-upper-bound-to-biginteger) — Abra, May 01 '22 at 04:15

Jason Barbour · Answer 1 · 2022-05-01T14:59:42.007

1

The number is too long for a long. Longs go from -9,223,372,036,854,775,808 to 9,223,372,036,854,775,808.

Try doing this :

public static String removeDigit(String number, char digit) {
    double temp = 0;
    String result="";
    for(int i = 0; i<number.length(); i++) {
        char num = number.charAt(i);
        if(num == digit) {
            String myStr = number.substring(0, i) + number.substring(i + 1); 
            try{
                double myNum = Double.parseDouble(myStr);  
                if(myNum > temp) {
                    temp = myNum;
                    result=myStr;
                }
            }
            catch (NumberFormatException ex){
                ex.printStackTrace();
            }
        }
    }
    return result;
}

edited May 01 '22 at 14:59

answered May 01 '22 at 03:45

Jason Barbour

11
2

what should I do here ? – David May 01 '22 at 03:47
I am converting string to long/int and then comaparing with result. I am returning the maximum value after removing the digit. – David May 01 '22 at 03:51
what are you trying to do when comparing the value? do you just want the biggest value of the long ?? – Jason Barbour May 01 '22 at 03:53
Do you want to remove only the first occurrence of that number then ? – Jason Barbour May 01 '22 at 03:53
In loop I am removing all the occurence of number one by one and then comparing with result and getting the maxium one. – David May 01 '22 at 03:56
So you want to remove only one number and keep the biggest answer? – Jason Barbour May 01 '22 at 03:58

score 1 · Answer 2 · answered May 01 '22 at 04:25

Your number is too big for a long value. The maximum long value is 9,223,372,036,854,775,807. You can use BigInteger, which essentially has no limit.

Using long

long result = 0;
// ...
long myNum = Long.parseLong(myStr);  
if(myNum > result) {
    result = myNum;
}
// ...
String s = String.valueOf(result);  
return s;

Using BigInteger

import java.math.BigInteger;
// ...
BigInteger result = BigInteger.ZERO;
// ...
BigInteger myNum = new BigInteger(myStr);
result = myNum.max(result);
// ...
return result.toString();

Sambhav Khandelwal · Answer 3 · 2022-05-04T08:10:19.757

1

The problem you get is that you are exceeding the limit of the int and the long. Let us see the limits of some number storing types and then use the best one:

Type	Size	Value	Exceeds
int	32 bit	-2,147,483,648 to 2,147,483,647	Yes
long	64 bit	-9,223,372,036,854,775,808 to 9,223,372,036,854,775,807	Yes
float	32 bit	3.40282347 x 1038 to 1.40239846 x 10-45	Yes
double	64 bit	1.7976931348623157 x 10308 to 4.9406564584124654 x 10-324	Yes
BigInteger	32 bit	2^64billion	No

Here, we find that BigInteger is the class we need to use. So, instead of using a long or int for it, use BigInteger. To know more about BigInteger, visit here.

Also to know how to use a big integers refer to the answer here

edited May 04 '22 at 08:10

answered May 01 '22 at 05:18

Sambhav Khandelwal

3,585
2
7
38

You evidently copied a source that used superscripted exponents but removed the superscripting, giving results which are quite wrong. Also BigInteger isn't infinite; its limit is about 2^64billion, a number which in decimal has about 19billion digits, which is _very_ big but finite. – dave_thompson_085 May 01 '22 at 06:03
Ok. @dave_thompson_085. I didnt know the vlue so i wrote infinite. U can edit and give the value there – Sambhav Khandelwal May 01 '22 at 09:04

score 0 · Answer 4 · answered May 01 '22 at 04:11

You can use BigDecimal instead of long.

public class Application {
        public static void main(String[] args) {
            //String str = "1231";
            String str = "2998589353917872714814599237991174513476623756395992135212546127959342974628712329595771672911914471";
            //char digit = '1';
            char digit = '3';
            System.out.println(removeDigit(str,digit));
        }

        public static BigDecimal removeDigit(String number, char digit) {
            BigDecimal result = BigDecimal.ZERO;
            for(int i = 0; i<number.length(); i++) {
                char num = number.charAt(i);
                if(num == digit) {
                    String myStr = number.substring(0, i) + number.substring(i + 1);
                    try{
                        BigDecimal myNum = new BigDecimal(myStr);
                        if(myNum.compareTo(result)>0) {
                            result = myNum;
                        }
                    }
                    catch (NumberFormatException ex){
                        ex.printStackTrace();
                    }
                }
            }
            return result;
        }
    }

Getting Numberformat exception with unknown source. How to fix this

4 Answers4