Develop Reference

Java String Concatenation Error "String index out of range: -1"

First off I know there are a lot of questions about "String index out of range", I have looked through them but can't find anyone having the same problem as me.
I have to write a simple program:
Given a string, return a "rotated left n" version where the first n chars are moved to the end.
leftN("Hello",2) → "lloHe"
leftN("java",0) → "java"
leftN("Hi,1") → "iH"
So I wrote the following:
package string;
public class LeftN {
public static String leftN(String str, int n) {
if (str.length() > 1 && n > 0) {
String a = str.substring(n);
String b = str.substring(0, n);
return a + b;
} else {
return str;
}
}
}
Question:
When I return just a or just b I get a valid output (if I add the output of a and b on paper I am getting the rotated left n version of the string). However, when I return the concatenation of a + b I get the String index out of range: -1 error, what can it be that is causing this?
Now I know that this error has to do with referencing a value that is out of bounds for the string and understand how this works when creating a substring. What is really confusing me is how adding two seemingly valid strings can give me this error?
Note: I have a test class provided that I am testing it against to see if it's giving the right output but I am not sure if I am allowed to post it online so that is why I am not providing it.

I can reproduce your exception by passing an n greater than the length of str:
leftN("abc", 4);
results in:
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: -1
at java.base/java.lang.String.substring(String.java:1850)
at Ideone.leftN(Main.java:12)
at Ideone.main(Main.java:22)
You need to handle the case of n being too large (or, indeed, negative). For example:
Throw an IllegalArgumentException
Use n % str.length() instead.

Removing a substring from a string, repeatedly

Problem:
Remove the substring t from a string s, repeatedly and print the number of steps involved to do the same.
Explanation/Working:
For Example: t = ab, s = aabb. In the first step, we check if t is
contained within s. Here, t is contained in the middle i.e. a(ab)b.
So, we will remove it and the resultant will be ab and increment the
count value by 1. We again check if t is contained within s. Now, t is
equal to s i.e. (ab). So, we remove that from s and increment the
count. So, since t is no more contained in s, we stop and print the
count value, which is 2 in this case.
So, here's what I have tried:
Code 1:
static int maxMoves(String s, String t) {
int count = 0,i;
while(true)
{
if(s.contains(t))
{
i = s.indexOf(t);
s = s.substring(0,i) + s.substring(i + t.length());
}
else break;
++count;
}
return count;
}
I am just able to pass 9/14 test cases on Hackerrank, due to some reason (I am getting "Wrong Answer" for rest of the cases). After a while, I found out that there is something called replace() method in Java. So, I tried using that by replacing the if condition and came up with a second version of code.
Code 2:
static int maxMoves(String s, String t) {
int count = 0,i;
while(true)
{
if(s.contains(t))
s.replace(t,""); //Marked Statement
else break;
++count;
}
return count;
}
But for some reason (I don't know why), the "Marked Statement" in the above code gets executed infinitely (this I noticed when I replaced the "Marked Statement" with System.out.println(s.replace(t,""));). I don't the reason for the same.
Since, I am passing only 9/14 test cases, there must be some logical error that is leading to a "Wrong Answer". How do I overcome that if I use Code 1? And if I use Code 2, how do I avoid infinite execution of the "Marked Statement"? Or is there anyone who would like to suggest me a Code 3?
Thank you in advance :)

Try saving the new (returned) string instead of ignoring it.
s = s.replace(t,"");
replace returns a new string; you seemed to think that it alters the given string in-place.

Try adding some simple parameter checks of the strings. The strings shouldn't be equal to null and they should have a length greater than 0 to allow for counts greater than 0.
static int maxMoves(String s, String t) {
int count = 0,i;
if(s == null || s.length() == 0 || t == null || t.length() == 0)
return 0;
while(true)
{
if(s.contains(t) && !s.equals(""))
s = s.replace(t,""); //Marked Statement
else break;
++count;
}
return count;
}

You might be missing on the edge cases in the code 1.
In code 2, you are not storing the new string formed after the replace function.
The replace function replaces each substring of this string that matches the literal target sequence with the specified literal replacement sequence.
Try this out:
public static int findCount(String s, String t){
if( null == s || "" == s || null == t || "" == t)
return 0;
int count =0;
while(true){
if(s.contains(t)){
count++;
int i = s.indexOf(t);
s = s.substring(0, i)+s.substring(i+t.length(), s.length());
// s = s.replace(t,"");
}
else
break;
}
return count;
}

String r1="ramraviraravivimravi";
String r2="ravi";
int count=0,i;
while(r1.contains(r2))
{
count++;
i=r1.indexOf(r2);
StringBuilder s1=new StringBuilder(r1);
s1.delete(i,i+r2.length());
System.out.println(s1.toString());
r1=s1.toString();
}
System.out.println(count);

First of all no logical difference in both the codes.
All the mentioned answers are to rectify the error of code 2 but none told how to pass all (14/14) cases.
Here I am mentioning a test case where your code will fail.
s = "abcabcabab";
t = "abcab"
Your answer 1
Expected answer 2
According to your code:
In 1st step, removig t from index 0 of s,
s will reduce to "cabab", so the count will be 1 only.
But actual answer should be 2
I first step, remove t from index 3 of s,
s will reduced to "abcab", count = 1.
In 2nd step removing t from index 0,
s will reduced to "", count = 2.
So answer would be 2.
If anyone know how to handle such cases, please let me know.

Java - Complex Recursive Backtracking

For Java practice I started working on a method countBinary that accepts an integer n as a parameter that prints all binary numbers that have n digits in ascending order, printing each value on a separate line. Assuming n is non-negative and greater than 0, some example outputs would look like this.
I am getting pretty much nowhere with this. I am able to write a program that finds all possible letter combinations of a String and similar things, but I have been unable to make almost any progress with this specific problem using binary and integers.
Apparently the best way to go about this issue is by defining a helper method that accepts different parameters than the original method and by building up a set of characters as a String for eventual printing.
Important Note: I am NOT supposed to use for loops at all for this exercise.
Edit - Important Note: I need to have trailing 0's so that all outputs are the same length.
So far this is what I have:
public void countBinary(int n)
{
String s = "01";
countBinary(s, "", n);
}
private static void countBinary(String s, String chosen, int length)
{
if (s.length() == 0)
{
System.out.println(chosen);
}
else
{
char c = s.charAt(0);
s = s.substring(1);
chosen += c;
countBinary(s, chosen, length);
if (chosen.length() == length)
{
chosen = chosen.substring(0, chosen.length() - 1);
}
countBinary(s, chosen, length);
s = c + s;
}
}
When I run my code my output looks like this.
Can anyone explain to me why my method is not running the way I expect it to, and if possible show me a solution to my issue so that I might get the correct output? Thank you!

There are more efficient ways to do it, but this will give you a start:
public class BinaryPrinter {
static void printAllBinary(String s, int n) {
if (n == 0) System.out.println(s);
else {
printAllBinary(s + '0', n - 1);
printAllBinary(s + '1', n - 1);
}
}
public static void main(String [] args) {
printAllBinary("", 4);
}
}
I'll let you work out the more efficient way.

Suggestion on equalsIgnoreCase in Java to C implementation

All,
In a bid to improve my C skills, I decided to start implementing various Java libraries/library functions to C code. This would ensure that everyone knows the functionality of my implementation at least. Here is the link to the C source code that simulates the equalsIgnoreCase() of String class in Java : C source code. I have tested the code and it looks fine as per my testing skills are concerned. My aim was to use as much basic operations and datatypes as possible. Though, it would be great if the gurus here can:
1 > Give me any suggestion to improve the code quality
2 > Enlighten me with any missing coding standard/practices
3 > Locate bugs in my logic.

100 lines of code is not too long to post here.
You calculate the string length twice. In C, the procedure to calculate the string length starts at the beginning of the string and runs along all of it (not necessarily in steps of 1 byte) until it finds the terminating null byte. If your strings are 2Mbyte long, you "walk" along 4Mbyte unnecessarily.
in <ctype.h> there are the two functions tolower() and toupper() declared. You can use one of them (tolower) instead of extractFirstCharacterASCIIVal(). The advantage of using the library function is that it is not locked in to ASCII and may even work with foreign characters when you go 'international'.
You use awkward (very long) names for your variables (and functions too). eg: ch1 and ch2 do very well for characters in file 1 and file 2 respectively :-)
return 1; at the end of main usually means something went wrong with the program. return 0; is idiomatic for successful termination.
Edit: for comparison with tcrosley version
#include <ctype.h>
int cmpnocase(const char *s1, const char *s2) {
while (*s1 && *s2) {
if (tolower((unsigned char)*s1) != tolower((unsigned char)*s2)) break;
s1++;
s2++;
}
return (*s1 != *s2);
}

With C++, you could replace performComparison(char* string1, char * string2) with stricmp.
However stricmp is not part of the standard C library. Here is a version adapted to your example. Note you don't need the extractFirstCharacterASCIIVal function, use tolower instead. Also note there is no need to explicitly calculate the string length ahead of time, as strings in C are terminated by the NULL character '\0'.
int performComparison(char* string1, char * string2)
{
char c1, c2;
int v;
do {
c1 = *string1++;
c2 = *string2++;
v = (UINT) tolower(c1) - (UINT) tolower(c2);
} while ((v == 0) && (c1 != '\0') && (c2 != '\0') );
return v != 0;
}
If you do want to use your own extractFirstCharacterASCIIVal function instead of the tolower macro, to make the code more transparent then you should code it like so:
if ((str >= 'a') && (str <= 'z'))
{
returnVal = str - ('a' - 'A');
}
else
{
returnVal = str;
}
to make it more obvious what you are doing. Also you should include a comment that this assumes the characters a..z and A..Z are contiguous. (They are in ASCII, but not always in other encodings.)

Check whether a string is parsable into Long without try-catch?

Long.parseLong("string") throws an error if string is not parsable into long.
Is there a way to validate the string faster than using try-catch?
Thanks

You can create rather complex regular expression but it isn't worth that. Using exceptions here is absolutely normal.
It's natural exceptional situation: you assume that there is an integer in the string but indeed there is something else. Exception should be thrown and handled properly.
If you look inside parseLong code, you'll see that there are many different verifications and operations. If you want to do all that stuff before parsing it'll decrease the performance (if we are talking about parsing millions of numbers because otherwise it doesn't matter). So, the only thing you can do if you really need to improve performance by avoiding exceptions is: copy parseLong implementation to your own function and return NaN instead of throwing exceptions in all correspondent cases.

From commons-lang StringUtils:
public static boolean isNumeric(String str) {
if (str == null) {
return false;
}
int sz = str.length();
for (int i = 0; i < sz; i++) {
if (Character.isDigit(str.charAt(i)) == false) {
return false;
}
}
return true;
}

You could do something like
if(s.matches("\\d*")){
}
Using regular expression - to check if String s is full of digits.
But what do you stand to gain? another if condition?

org.apache.commons.lang3.math.NumberUtils.isParsable(yourString) will determine if the string can be parsed by one of: Integer.parseInt(String), Long.parseLong(String), Float.parseFloat(String) or Double.parseDouble(String)
Since you are interested in Longs you could have a condition that checks for isParsable and doesn't contain a decimal
if (NumberUtils.isParsable(yourString) && !StringUtils.contains(yourString,".")){ ...

This is a valid question because there are times when you need to infer what type of data is being represented in a string. For example, you may need to import a large CSV into a database and represent the data types accurately. In such cases, calling Long.parseLong and catching an exception can be too slow.
The following code only handles ASCII decimal:
public class LongParser {
// Since tryParseLong represents the value as negative during processing, we
// counter-intuitively want to keep the sign if the result is negative and
// negate it if it is positive.
private static final int MULTIPLIER_FOR_NEGATIVE_RESULT = 1;
private static final int MULTIPLIER_FOR_POSITIVE_RESULT = -1;
private static final int FIRST_CHARACTER_POSITION = 0;
private static final int SECOND_CHARACTER_POSITION = 1;
private static final char NEGATIVE_SIGN_CHARACTER = '-';
private static final char POSITIVE_SIGN_CHARACTER = '+';
private static final int DIGIT_MAX_VALUE = 9;
private static final int DIGIT_MIN_VALUE = 0;
private static final char ZERO_CHARACTER = '0';
private static final int RADIX = 10;
/**
* Parses a string representation of a long significantly faster than
* <code>Long.ParseLong</code>, and avoids the noteworthy overhead of
* throwing an exception on failure. Based on the parseInt code from
* http://nadeausoftware.com/articles/2009/08/java_tip_how_parse_integers_quickly
*
* #param stringToParse
* The string to try to parse as a <code>long</code>.
*
* #return the boxed <code>long</code> value if the string was a valid
* representation of a long; otherwise <code>null</code>.
*/
public static Long tryParseLong(final String stringToParse) {
if (stringToParse == null || stringToParse.isEmpty()) {
return null;
}
final int inputStringLength = stringToParse.length();
long value = 0;
/*
* The absolute value of Long.MIN_VALUE is greater than the absolute
* value of Long.MAX_VALUE, so during processing we'll use a negative
* value, then we'll multiply it by signMultiplier before returning it.
* This allows us to avoid a conditional add/subtract inside the loop.
*/
int signMultiplier = MULTIPLIER_FOR_POSITIVE_RESULT;
// Get the first character.
char firstCharacter = stringToParse.charAt(FIRST_CHARACTER_POSITION);
if (firstCharacter == NEGATIVE_SIGN_CHARACTER) {
// The first character is a negative sign.
if (inputStringLength == 1) {
// There are no digits.
// The string is not a valid representation of a long value.
return null;
}
signMultiplier = MULTIPLIER_FOR_NEGATIVE_RESULT;
} else if (firstCharacter == POSITIVE_SIGN_CHARACTER) {
// The first character is a positive sign.
if (inputStringLength == 1) {
// There are no digits.
// The string is not a valid representation of a long value.
return null;
}
} else {
// Store the (negative) digit (although we aren't sure yet if it's
// actually a digit).
value = -(firstCharacter - ZERO_CHARACTER);
if (value > DIGIT_MIN_VALUE || value < -DIGIT_MAX_VALUE) {
// The first character is not a digit (or a negative sign).
// The string is not a valid representation of a long value.
return null;
}
}
// Establish the "maximum" value (actually minimum since we're working
// with negatives).
final long rangeLimit = (signMultiplier == MULTIPLIER_FOR_POSITIVE_RESULT)
? -Long.MAX_VALUE
: Long.MIN_VALUE;
// Capture the maximum value that we can multiply by the radix without
// overflowing.
final long maxLongNegatedPriorToMultiplyingByRadix = rangeLimit / RADIX;
for (int currentCharacterPosition = SECOND_CHARACTER_POSITION;
currentCharacterPosition < inputStringLength;
currentCharacterPosition++) {
// Get the current digit (although we aren't sure yet if it's
// actually a digit).
long digit = stringToParse.charAt(currentCharacterPosition)
- ZERO_CHARACTER;
if (digit < DIGIT_MIN_VALUE || digit > DIGIT_MAX_VALUE) {
// The current character is not a digit.
// The string is not a valid representation of a long value.
return null;
}
if (value < maxLongNegatedPriorToMultiplyingByRadix) {
// The value will be out of range if we multiply by the radix.
// The string is not a valid representation of a long value.
return null;
}
// Multiply by the radix to slide all the previously parsed digits.
value *= RADIX;
if (value < (rangeLimit + digit)) {
// The value would be out of range if we "added" the current
// digit.
return null;
}
// "Add" the digit to the value.
value -= digit;
}
// Return the value (adjusting the sign if needed).
return value * signMultiplier;
}
}

You can use java.util.Scanner
Scanner sc = new Scanner(s);
if (sc.hasNextLong()) {
long num = sc.nextLong();
}
This does range checking etc, too. Of course it will say that "99 bottles of beer" hasNextLong(), so if you want to make sure that it only has a long you'd have to do extra checks.

This case is common for forms and programs where you have the input field and are not sure if the string is a valid number. So using try/catch with your java function is the best thing to do if you understand how try/catch works compared to trying to write the function yourself. In order to setup the try catch block in .NET virtual machine, there is zero instructions of overhead, and it is probably the same in Java. If there are instructions used at the try keyword then these will be minimal, and the bulk of the instructions will be used at the catch part and that only happens in the rare case when the number is not valid.
So while it "seems" like you can write a faster function yourself, you would have to optimize it better than the Java compiler in order to beat the try/catch mechanism you already use, and the benefit of a more optimized function is going to be very minimal since number parsing is quite generic.
If you run timing tests with your compiler and the java catch mechanism you already described, you will probably not notice any above marginal slowdown, and by marginal I mean it should be almost nothing.
Get the java language specification to understand the exceptions more and you will see that using such a technique in this case is perfectly acceptable since it wraps a fairly large and complex function. Adding on those few extra instructions in the CPU for the try part is not going to be such a big deal.

I think that's the only way of checking if a String is a valid long value. but you can implement yourself a method to do that, having in mind the biggest long value.

There are much faster ways to parse a long than Long.parseLong. If you want to see an example of a method that is not optimized then you should look at parseLong :)
Do you really need to take into account "digits" that are non-ASCII?
Do you really need to make several methods calls passing around a radix even tough you're probably parsing base 10?
:)
Using a regexp is not the way to go: it's harder to determine if you're number is too big for a long: how do you use a regexp to determine that 9223372036854775807 can be parsed to a long but that 9223372036854775907 cannot?
That said, the answer to a really fast long parsing method is a state machine and that no matter if you want to test if it's parseable or to parse it. Simply, it's not a generic state machine accepting complex regexp but a hardcoded one.
I can both write you a method that parses a long and another one that determines if a long can be parsed that totally outperforms Long.parseLong().
Now what do you want? A state testing method? In that case a state testing method may not be desirable if you want to avoid computing twice the long.
Simply wrap your call in a try/catch.
And if you really want something faster than the default Long.parseLong, write one that is tailored to your problem: base 10 if you're base 10, not checking digits outside ASCII (because you're probably not interested in Japanese's itchi-ni-yon-go etc.).

Hope this helps with the positive values. I used this method once for validating database primary keys.
private static final int MAX_LONG_STR_LEN = Long.toString(Long.MAX_VALUE).length();
public static boolean validId(final CharSequence id)
{
//avoid null
if (id == null)
{
return false;
}
int len = id.length();
//avoid empty or oversize
if (len < 1 || len > MAX_LONG_STR_LEN)
{
return false;
}
long result = 0;
// ASCII '0' at position 48
int digit = id.charAt(0) - 48;
//first char cannot be '0' in my "id" case
if (digit < 1 || digit > 9)
{
return false;
}
else
{
result += digit;
}
//start from 1, we already did the 0.
for (int i = 1; i < len; i++)
{
// ASCII '0' at position 48
digit = id.charAt(i) - 48;
//only numbers
if (digit < 0 || digit > 9)
{
return false;
}
result *= 10;
result += digit;
//if we hit 0x7fffffffffffffff
// we are at 0x8000000000000000 + digit - 1
// so negative
if (result < 0)
{
//overflow
return false;
}
}
return true;
}

Try to use this regular expression:
^(-9223372036854775808|0)$|^((-?)((?!0)\d{1,18}|[1-8]\d{18}|9[0-1]\d{17}|92[0-1]\d{16}|922[0-2]\d{15}|9223[0-2]\d{14}|92233[0-6]\d{13}|922337[0-1]\d{12}|92233720[0-2]\d{10}|922337203[0-5]\d{9}|9223372036[0-7]\d{8}|92233720368[0-4]\d{7}|922337203685[0-3]\d{6}|9223372036854[0-6]\d{5}|92233720368547[0-6]\d{4}|922337203685477[0-4]\d{3}|9223372036854775[0-7]\d{2}|922337203685477580[0-7]))$
It checks all possible numbers for Long.
But as you know in Java Long can contain additional symbols like +, L, _ and etc. And this regexp doesn't validate these values. But if this regexp is not enough for you, you can add additional restrictions for it.

Guava Longs.tryParse("string") returns null instead of throwing an exception if parsing fails. But this method is marked as Beta right now.

You could try using a regular expression to check the form of the string before trying to parse it?

A simple implementation to validate an integer that fits in a long would be:
public static boolean isValidLong(String str) {
if( str==null ) return false;
int len = str.length();
if (str.charAt(0) == '+') {
return str.matches("\\+\\d+") && (len < 20 || len == 20 && str.compareTo("+9223372036854775807") <= 0);
} else if (str.charAt(0) == '-') {
return str.matches("-\\d+") && (len < 20 || len == 20 && str.compareTo("-9223372036854775808") <= 0);
} else {
return str.matches("\\d+") && (len < 19 || len == 19 && str.compareTo("9223372036854775807") <= 0);
}
}
It doesn't handle octal, 0x prefix or so but that is seldom a requirement.
For speed, the ".match" expressions are easy to code in a loop.