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Remove a Character From String in Java

I'm trying to concatenate a string with itself and remove all capital letters from the resultant string.
Here is my code:
public String removeCapitals(String A) {
StringBuilder B = new StringBuilder(A+A);
int n = B.length();
for(int i=0; i<n; i++){
if(B.charAt(i)>='A' && B.charAt(i)<='Z'){
B.deleteCharAt(i);
}
}
return B.toString();
}
I'm getting Exception saying:
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 6
at java.lang.AbstractStringBuilder.charAt(AbstractStringBuilder.java:237)
at java.lang.StringBuilder.charAt(StringBuilder.java:76)
at Solution.removeCapitals(Solution.java:10)
at Main.main(Main.java:190)
Can someone help me to understand the issue.

If at least one removal succeeds, at some point your code will attempt to access an invalid index that exceeds the length of a StringBuilder.
It happens because the variable n remain unchanged. You should change the condition to be bound to the current size of StringBuilder and decrement the index at each removal, or iterate backwards (as shown in another answer).
Also condition B.charAt(i)>='A' && B.charAt(i)<='Z' can be replaced with:
Character.isUpperCase(b.charAt(i))
Which is more descriptive.
That's how it might look like:
public static String removeCapitals(String a) {
StringBuilder b = new StringBuilder(a + a);
for (int i = 0; i < b.length(); i++) {
if (Character.isUpperCase(b.charAt(i))) {
b.deleteCharAt(i); // which can be combined with the next line `b.deleteCharAt(i--);` - previous value of `i` would be used in the call `deleteCharAt()` and variable `i` will hold a value decremented by 1
i--;
}
}
return b.toString();
}
Method deleteCharAt() runs in a linear time, because it shifts all subsequent characters in the underlying array bytes. Each upper-case letter will trigger these shifts and in the worst case scenario, it would result in the quadratic overall time complexity O(n ^ 2).
You make your method more performant and much more concise without using loops and StringBuilder. This code will run in a linear time O(n).
public static String removeCapitals(String a) {
return a.replaceAll("\\p{Upper}", "").repeat(2);
}

When you delete a character you change the length of the StringBuilder. But n still has the original length. So you will eventually exceed the size of the StringBuilder. So start from the end and move backwards. That way, any deletions will come after (based on relative indices) the next position so the index will be within the modified StringBuilder size. In addition, deleting from the end is more efficient since there is less copying to do in the StringBuilder.
public String removeCapitals(String A) {
StringBuilder B = new StringBuilder(A+A);
int n = B.length();
for(int i=n-1; i>=0; i--){
if(B.charAt(i)>='A' && B.charAt(i)<='Z'){
B.deleteCharAt(i);
}
}
return B.toString();
}

If just remove Capital characters from a string. Alternative solution just create another method replaceAll() + regex
private static String removeCapitals(String A){
if (!A.isEmpty() && !A.equals("")) {
String B = A + A;
String newStr = B.replaceAll("([A-Z])", "");
return newStr;
} else {
return null;
}
}

Shorter solution to your task.
String a = "ABcdEF";
String b = "";
for (int i = 0; i < a.length(); i++) {
if(a.toLowerCase().charAt(i) == a.charAt(i))
b+=a.charAt(i);
}
System.out.println(b);
By changing to .toUpperCase() you'll get rid of the lower case ones.

How do I get rid of the comma at the end? (Java)

I'm working on an assignment and the code is working perfectly, except for one small little problem.
For example at the end it's supposed to say, "t,e,s,t" but instead it prints out "t,e,s,t,".
import javax.swing.JOptionPane;
public class program {
public static int divide(String input) {
int length=0;
String output = "";
for (int i=0; i<input.length(); i++) {
length++;
output += input.charAt(i);
output += ",";
}
JOptionPane.showMessageDialog(null, "Eingabe: " +input+ "\nAnzahl der Zeichen: " +length+ "\nZeichen: \n" +output);
return length;
}
}
I'd just like to say for those who are planning on giving tips, please note that this is an assignment so I'm not really allowed to make any MAJOR changes to it. This program HAS to be:
Solved using a for loop
The output HAS to be in JOptionPane
It has to be in a method (for I'll have to write an inputDialog later in the main-method, but that's unimportant right now).
My only problem with it, for example the output would have to say (I'll translate the output in english, since I'm at an austrian school) if the string was "hello", the program would say "letters: h, e, l, l, o" but instead it says "h, e, l, l, o," with a comma at the end, how do I get rid of it?
(Also sorry if I messed up any variables, I renamed them all from German to English for this post so I hope I didn't mess any of them up)

One of the possibilities is to print comma not after each character, but before every character except the first one.

You only want to add a comma if there is more data to come after. So you can do it in two ways:
add the comma before the text if there's already something in the string:
String output = "";
for (int i=0; i<input.length(); i++) {
length++;
if (output.length() > 0) output += ",";
output += input.charAt(i);
}
add the comma after unless it's the last element:
String output = "";
for (int i=0; i<input.length(); i++) {
length++;
output += input.charAt(i);
if (i < input.length() - 1) output += ",";
}
Personally I like the first way.

Hello I used an if statement to check if we are at the first letter and if we are then we don't write a comma, here is the code:
{
int length = 0;
String output = "";
int i = 0;
for ( i = 0; i < input.length(); i++)
{
if(i == 0)
{
length++;
output += input.charAt(i);
}
else
{
output += ",";
output += input.charAt(i);
length++;
}
}
Console.WriteLine(output);
}

You only need to check whether you are at the last character, and if you are, then break out from the loop.
for (int i=0; i<input.length(); i++) {
length++; //you do not seem to need this
output += input.charAt(i);
if (i==(input.length()-1)) break; //checking whether we are at the last character
output += ",";
}
Two additional notes:
Please follow the Java Naming Conventions and use PascalCase for your class names, this is very important;
It would make your code much more efficient if you'd use StringBuilder instead of String, to concatenate characters and dynamically build your string. String is immutable and per each concatenation, you're actually creating a new instance of it, which is expensive.

I would use a StringJoiner
StringJoiner output = new StringJoiner(",");
for (int i = 0; i < input.length(); i++) {
length++;
output.add(input.substring(i, i + 1));
}
JOptionPane.showMessageDialog(… + output.toString());
so You are not allowed to use StringJoiner
String output = "";
String comma = "";
for (int i = 0; i < input.length(); i++) {
length++;
output += comma;
output += input.charAt(i);
comma = ",";
}

Your loop iterates through every character, appending both the character and a comma. This includes the final character. You need to find a way to avoid adding a comma after the final iteration;
Using String and appending characters one by one is very inefficient. This is what StringBuilder is designed for;
What is the purpose of the length variable? It can be replaced with input.length() - 1;
String.format() makes your code easier to read rather than chaining together string concatenations;
Don't be afraid to use a healthy amount of spacing throughout your code. It is much harder to read otherwise.
Something like this should work well:
public static int divide(String input) {
int length = input.length() - 1;
String output;
if (input.length() == 0) output = input;
else {
StringBuilder outputBuilder = new StringBuilder(input.charAt(0));
for (int i = 1; i < input.length(); i++) {
outputBuilder.append(',').append(input.charAt(i));
}
output = outputBuilder.toString();
}
JOptionPane.showMessageDialog(null, String.format("Eingabe: %s\nAnzahl der Zeichen: %d\nZeichen: \n%s", input, length, output);
return length;
}

Once you learn, use a StringJoiner or similar modern device. My favourite link is at the bottom. There will also be a time when you learn to use a StringBuilder or StringBuffer for assembling a string.
In the meantime, I still like what I learned in my first year of programming: when one iteration through a loop needs to be different, take it out of the loop. In my experience this often gives the clearest code. In your case:
String input = "Wort";
String output = "" + input.charAt(0);
for (int i = 1; i < input.length(); i++) {
output += ",";
output += input.charAt(i);
}
System.out.println("Zeichen: " + output);
Output:
Zeichen: W,o,r,t
In this case I have taken the first character outside the loop and add it to the output (without comma) already in the declaration of output. The loop starts from index 1 (not 0). Inside the loop I have to add the comma before adding the next char. In other cases one may put the last item outside the loop instead, the result will be the same.
Links
Answer by Lii to a similar question demonstrating all of String.join(), Collectors.joining() and StringJoiner.
StringBuilder vs String concatenation in toString() in Java

How to get the common first substrings In two Strings

I need some help, I have two Strings and I want to get the first occurrence of common substrings.
1st String : abacdefghi
2nd String : abaciopiss
I want to get the substring
substring : abac
Thank you everyone.

It maybe isn't the best solution but my attempt would be to find the first matching characters in each string and then continue to check the following characters if they are still the same:
private static String extractFirstEqual(String a, String b) {
//Split your string into an array of characters
String[] arr = a.split("");
String[] brr = b.split("");
StringBuilder result = new StringBuilder();
//Iterate over both arrays
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < brr.length; j++) {
//Find first matching character
if (arr[i].equals( brr[j])) {
//While there are more characters in both arrays and the characters keep matching, append them
// to the result
while (arr[i].equals(brr[j]) && i < arr.length && j < brr.length) {
result.append(arr[i]);
i++;
j++;
}
return result.toString();
}
}
}
return result.toString();
}

How to change a char in a String - replacing letters java

Simular topics were not able to solve my problem.
I need to change the char 'a' to 'x' in an given String str.
Example: "abc" = "xbc". I am only allowed to use substring(), charAt() - no replace() method.
My code so far:
public static String ersetze(String text){
for(int i = 0; i<text.length(); i++){
if(text.substring(i, i+1).charAt(i) == 'a'){
text.substring(i, i+1) = 'x';
}
}
//return statement
}
Now the error is text.substring(i, i+1) = 'x';that the left assignment must be a variable - clear. But how to assigne the letter to a variable now? If I declare a char x; how to put that x in the String to replace the letter?

String is immutable in Java, so you cannot replace a letter of a String. You need to create a new String.
You can convert the String to an array of chars and changing only the needed ones, then create a new String from this array:
public static String ersetze(String text){
char[] letters = text.toCharArray();
for (int i = 0; i < letters.length; i++){
if (letters[i] == 'a') {
letters[i] = 'x';
}
}
return new String(letters);
}

String can not replace with character. First Need to create character array & then replace.
public static String ersetze(String text){
char[] result = text.toCharArray();
for(int i = 0; i < result.length; i++){
if(result [i] == 'a'){
result[i] = 'x';
}
}
return String.valueOf(result);
}

If you REALLLLLY need this and you are limited to the methods you mentioned then you can do this each time you find requested char:
text = text.substring(0, i) + x + text.substring(i + 1);

Convert the string to a character array (included in the java API), iterate through the array and replace all the "a"'s with "x"'s in the char array and then change it back to a string.

How to tackle the Codingbat String-2 oneTwo challenge?

Here is the problem statement:
Given a string, compute a new string by moving the first char to come after the next two chars, so "abc" yields "bca". Repeat this process for each subsequent group of 3 chars, so "abcdef" yields "bcaefd". Ignore any group of fewer than 3 chars at the end.
Here is my code:
// oneTwo("abc") → "bca"
// oneTwo("tca") → "cat"
// oneTwo("tcagdo") → "catdog"
public String oneTwo(String str) {
String x = "";
if (str.length() < 3) {
return "";
// return empty
} else if (str.length() == 3) {
String s = str.substring(1, str.length());
x = s + str.substring(0, 1); // last two + first char
} else if (str.length() > 3) {
int third = 2;
// start with the third element index of 2
for (int i = 0; i < str.length(); i++) {
if (i == third) {
// given three chars substring first char
// substring last two chars and add that to x
x += (str.substring(third - 1, third + 1) +
str.substring(third - 2, third - 2 + 1));
third += 3;
//work with this line but why??????
}
//third +=3;
// doesn't work with this line but why???????
}// end of for loop
}
return x;
// return modified string x
}
With third +=3 inside of if statement work but when I put that outside of if statement I don't get the desired output. I don't really understand why?

Hope this helps:
public String oneTwo(String str) {
String str2 = "";
for(int i=0; i<str.length()-2; i+=3) {
str2 = str2+str.substring(i+1,i+3)+str.charAt(i);
}
return str2;
}

Because putting it outside the loop will cause third to be increased far too often. After the first iteration i is 0, third is 5, next iteration yields i=1, third=8; i=2, third=11; i=3, third=14, etc. -> i will never reach third.
I would improve your code by dropping the entire if-statement, remove third all together and simply increment by 3 in the for-loop:
for( int i = 2; i < str.length(); i+=3){
x += (str.substring(third-1, third+1) +
str.substring(third-2, third-2 + 1));
}
If I am not misinterpreting your code you are missing logic for leaving the last characters alone if they are not part of group of three characters.

If you face such effects take a piece of paper and write down the values of the variables after each line of your code.
The if block creates an alternative execution path if the condition is true which is in every third loop iteration.
Anything behind the if block is executed in every loop iteration.
So when the line in question is inside the if block (before the closing brace) the value in variable third is only changed every third loop iteration.
When you move the line behind the closing brace the assignment is outside the if block and therefore executed every loop iteration.

For the comment = //work with this line but why??????
The value of "third" variable gets changed in the for loop only with i is equal to third character, otherwise the value of third will keep on increasing eg.
when i = 0, third = 2
when i = 1, third = 5
when i = 2, third = 8
so the if statement never gets triggered and hence it doesn't work. Hope this makes sense.
PS - I highly recommend using IDE debugger to understand this properly.
PS - It's better to use charAt method as compared for substring method for performance reason

public String oneTwo(String str) {
String temp = "";
String result = "";
int i = 0;
while (str.substring(i).length() >= 3) {
temp = str.substring(i, i + 3);
result += temp.substring(1) + temp.charAt(0);
i += 3;
}
return result;
}

public String oneTwo(String str) {
String str1 = "";
if(str.length()<3){
return str1;
}else if(str.length()>=3){
for(int i =0; i<str.length()-2; i=i+3){
str1 = str1 + str.substring(i+1,i+3)+ str.substring(i,i+1);
}
}
return str1;
}

public String oneTwo(String str) {
if(str.length()<3)return "";
return str.substring(1,3)+str.substring(0,1)+oneTwo(str.substring(3));
}
this is fairly simple as a recursive problem

public String oneTwo(String str) {
String newThreeChars = "";
if(str.length()<3){
return newThreeChars;
}
for(int i=0; i<str.length()/3; i+=3){
String threeChars = str.substring(i,i+3);
String redesigned = threeChars.substring(1) + threeChars.charAt(0);
newThreeChars +=redesigned;
}
return newThreeChars;
}

Another solution to look at...
public String oneTwo(String str) {
int i = 0;
String result = "";
Character tmpChar = '\0';
while(i <= str.length()-3){
tmpChar = str.charAt(i);
result = result + str.charAt(i+1) + str.charAt(i+2) + tmpChar;
tmpChar = '\0';
i = i + 3;
}
return result;
}

First, We loop through each letter of the given String just stopping shy of the last two letters because the word we are looking for is three letters long. Then, we are returning true if there is two letter "b"'s exactly one character apart.
public boolean bobThere(String str) {
for (int i = 0; i < str.length() - 2; i++) {
if (str.charAt(i) == 'b' && str.charAt(i+2) == 'b')
return true;
}
return false;
}

For string concatenation in a loop use StringBuilder:
public String oneTwo(String str) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length() - 2; i += 3) {
sb.append(str.charAt(i + 1)).append(str.charAt(i + 2)).append(str.charAt(i));
}
return sb.toString();
}