Java's NumberFormatException

[Anand Francis]

Guys this a java program in which i've encountered Java's NUMBERFORMATEXC EPTION error can you please help me out with it.......thanks in advance :)
Question: Find Kaprekar number between 1-5000.
A kaprekar number example:
Example 1:
9 = 81(square of entered number) = 8+1(sum of digits of square)=9(origi nal number)

Example 2:
45 = 2025(square of entered number) = 20+25(sum of digits of square)=45(orig inal number)

Example 3:
297 = 88209(square of entered number) = 88+209(sum of digits of square)=297(ori ginal number)


[import java.io.*;
public class Kaprekar
{
BufferedReader inp=new BufferedReader( new InputStreamRead er(System.in));

int number,sqr,sum, count;
int lim1,lim2;
int len1,len2,len3;
String str1;
String sum1,sum2;

void enter_limits()t hrows IOException
{
System.out.prin t("Enter the 1st limit:");
lim1=Integer.pa rseInt(inp.read Line());
System.out.prin t("Enter the 2nd limit:");
lim2=Integer.pa rseInt(inp.read Line());
System.out.prin tln("Limit is from "+lim1+" -- "+lim2);
}

void calc()
{
for(int i=lim1;i<=lim2; i++)
{
count=0;
sqr=i*i;
str1=Integer.to String(sqr);
len1=str1.lengt h();

if(i%2==0)
{
len2=len1/2;
sum1=str1.subst ring(0,len2);
sum2=str1.subst ring(len2,len1) ;
int s1=Integer.pars eInt(sum1);//NumberFormatxce ption
int s2=Integer.pars eInt(sum2);//NumberFormatxce ption
sum=s1+s2;
if(sum==i)
{
System.out.prin t(i+" is a Kaprekar number.");
count++;
}
}
else if(i%2!=0&&i>0)
{
len3=len1-1;
len2=len3/2;
sum1=str1.subst ring(0,len2);
sum2=str1.subst ring(len2,len1) ;
int s1=Integer.pars eInt(sum1);//NumberFormatxce ption
int s2=Integer.pars eInt(sum2);//NumberFormatxce ption
sum=s1+s2;
if(sum==i)
{
System.out.prin t(i+" is a Kaprekar number.");
count++;
}
}
}
}

public static void main()throws IOException
{
Kaprekar k=new Kaprekar();
k.enter_limits( );
k.calc();
}
}]

Output:

java.lang.Numbe rFormatExceptio n: For input string: ""
at java.lang.Numbe rFormatExceptio n.forInputStrin g(NumberFormatE xception.java:4 8)
at java.lang.Integ er.parseInt(Int eger.java:468)
at java.lang.Integ er.valueOf(Inte ger.java:553)
at Kaprekar.calc(K aprekar.java:50 )
at Kaprekar.main(K aprekar.java:66 )

[r035198x]

When you split a single digit number you end up with one number and the empty string. For your purposes an empty string should be interpreted as zero so add

[CODE=java]if(sum1.equals( "")) {
sum1 = "0";
}
if(sum2.equals( "")) {
sum2 = "0";
}[/CODE]

Also, your algorithm will probably miss some numbers because you are not searching through all possible substrings.

[r035198x]

Actually you may find it easier to treat the one digit cases as special cases so they don't mess up with your algorithm structure that much:
[CODE=java]
public class Kaprekar {
public static void kaprekar() {
int lim1 = 1;
int lim2 = 1000;
for (int i = lim1; i <= lim2; i++) {
if (i == 1) {
System.out.prin tln(i + " is a Kaprekar number.");
}
int ilength = Integer.toStrin g(i).length();
if (ilength == 1) {
continue;
}
int sqr = i * i;
String str1 = Integer.toStrin g(sqr);
int len1 = str1.length();
String sum1 = str1.substring( 0, len1 - ilength);
String sum2 = str1.substring( len1 - ilength, len1);
int s1 = Integer.parseIn t(sum1);
int s2 = Integer.parseIn t(sum2);
int sum = s1 + s2;
if (sum == i) {
System.out.prin tln(i + " is a Kaprekar number.");
}
}
}

public static void main(String[] args) {
kaprekar();
}
}


[/CODE]