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How to delete specific digit from integer?

In the n integer where n = 1237534 (for example) I have to delete digit 3 so I can get the biggest value possible. n can be negative number also.
I can get 127534 or 123754.
The bigger is of course 127534, but how can I return it?
I've tried something like this:
int n = 1237534;
String newNum = String.valueOf(n);
int[] newGuess = new int[newNum.length()];
for (int i = 0; i < newNum.length(); i++) {
newGuess[i] = newNum.charAt(i) - '0';
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < newGuess.length; i++) {
if (!(newGuess[i] % 3 == 0)) {
sb.append(newGuess[i]);
}
}
System.out.println(sb);
I get 12754 which is not correct answer. Anyone maybe have an idea how to solve it?

This does not require arrays nor loops:
int n = 1237534;
String newNum = String.valueOf(n);
char c = '3';
StringBuilder sb1 = new StringBuilder( newNum );
StringBuilder sb2 = new StringBuilder( newNum );
int newGuess1 = Integer.parseInt(sb1.deleteCharAt(newNum.indexOf(c)).toString());
int newGuess2 = Integer.parseInt(sb2.deleteCharAt(newNum.indexOf(c, newNum.indexOf(c)+1)).toString());
System.out.println( newGuess1 > newGuess2 ? newGuess1: newGuess2 );

Here is an alternative approach that also checks to ensure that the number actually contains the digit. Otherwise, an exception could be thrown if the index returns -1.
The method simply deletes the first or last occurrence of the digit depending on whether the number is negative or not.
int n = 1234321;
int bad_digit = 3;
StringBuilder sb = new StringBuilder(Integer.toString(n));
String bad = Integer.toString(bad_digit);
int idxOfFirst = sb.indexOf(bad);
// first, make certain the number contains the digit.
if (idxOfFirst >= 0) {
if (sb.charAt(0) == '-') {
// if negative, delete last character to give the larger value
sb.deleteCharAt(sb.lastIndexOf(bad));
} else {
// else delete the first character to give the larger value
sb.deleteCharAt(idxOfFirst);
}
}
System.out.println(sb);
This prints the string rather than convert to an integer since printing an integer results in conversion to a string anyway. If the digit does not appear in the original number, then the original is printed. You can alter that handling to meet your requirements.

How about this:
int n = 123454321;
int bad_digit = 3;
StringBuilder s = new StringBuilder(n + "");
s.reverse();
System.out.println("Reversed String: " + s);
for (int i = 0; i < s.length(); i++) {
// Check if the character at index 'i' and the
// digit are equal, by converting both to strings
if (("" + s.charAt(i)).equals("" + bad_digit)) {
s.replace(i, i + 1, "");
break;
}
}
System.out.println("Reversed String after removing the digit (if found): " + s);
s.reverse();
System.out.println("Greatest number without the bad digit: " + s);
Basically, we need to remove the last occurrence of the digit to be removed (the bad_digit). So we convert the number to a string and reverse it. Now we need to remove the first occurrence of the digit. So we iterate over the characters, and when we find that digit, we remove it from the string, and exit the loop. Now we reverse the string again, and the output is what you want.

StringIndexOutOfBoundsException:

I know this error has been asked a lot. I have looked at them but still don't really understand how it applies to my problem. I am trying to write a program that creates a diamond pattern from a string.
public static String nameDiamond(String s) {
int len = s.length();
int i = 0;
String out = "";
while(i < len) {
out = out + s.substring(0, i + 1) + "\n";
i++;
}
while(i > 0) {
int numSpaces = len - i + 1;
for(int j = 0; j < numSpaces; j++)
out = out + " ";
out = out + s.substring(len - i + 1, i - 1) + "\n";
i--;
}
return out;
}
Now my understanding of substring is, that it is structured like s.substring(startIndex, endIndex) and a cause of this error can be when the end index is less then the start but I don't think this is the problem here. I also saw some causes of the error is when something does not exist but again I don't think this is the problem. What am I missing?
When it is tested with "Marty" it says -1 and with "Bill Nye" it says -2. Would the space be causing this?
I looked at lots of other versions of the questions but the ones that seemed relevent were:
StringIndexOutOfBoundsException: string index out of range
StringIndexOutOfBoundsException: String index out of range: -1
But I don't think they apply.
For reference, this is the problem:
https://www.codestepbystep.com/problem/view/java/strings/nameDiamond

Here is one you can use as an example. It doesn't do exactly what you want but it will illustrate one important thing. For fixed width fonts, each row must be an odd number of characters. Otherwise they won't line up properly.
If n is an even number, this particular version will print the same diamond for n and n+1. It is also limited by the string of spaces.
public static void printDiamond(int len) {
String str = "*";
String space = " ";
for (int i = 0; i < len/2 + 1; i++) {
System.out.print(space.substring(0,len/2-i));
System.out.println(str);
str = str+"**";
}
str = str.substring(2);
for(int i = 0; i < len/2; i++) {
str = str.substring(2);
System.out.print(space.substring(0,i+1));
System.out.println(str);
}
}
As you debug your program, use ubiquitous print statements to print the different variables such as indices and string lengths.

Java Code output If input is "aabbba" then output should be "a2b3a1" [closed]

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 years ago.
Improve this question
I am doing one simple java code where if
input is : "aabbba"
then
output should be: "a2b3a1"
I have done the below coding but missing somewhere. So let me know my mistake.
public class Test {
public static void main(String[] args) {
String str = "aabbba";
int count = 1;
for (int i = 0; i < str.length(); i = i + count) {
count = 1;
for (int j = i + 1; j < str.length(); j++) {
if (str.charAt(i) == str.charAt(j)) {
count = count + 1;
} else {
System.out.println(str.charAt(i) + "" + count);
break;
}
}//end of inner for
}//end of outer for
}//end of main
}//end of class

Actually you have too much code, You only need one loop, and you should be comparing the letter to the previous one, not attempting to compare each letter to every letter after it.
If you are confused about what your program is doing, the best place to start is to use your debugger to step through the code.
for(int i = 0, count = 1; i < str.length(); i++, count++) {
char ch = str.charAt(i);
char next = i + 1 < str.length() ? str.charAt(i + 1) : (char) -1;
if (ch != next) {
System.out.print("" + ch + count);
count = 0;
}
}

Using your effort and code, you simply did put the print to the wrong place
String str = "aabbba";
int count = 1;
for(int i = 0; i <str.length();i=i+count){
count =1;
for(int j = i+1; j<str.length();j++){
if(str.charAt(i) == str.charAt(j)){
count = count+1;
}
else{
break;
}
}
// Print here otherwise you will miss the last group of letters
// Also if you just want one line use .print instead of println
System.out.print(str.charAt(i)+""+count);
}

Using Java-8 and my StreamEx library it's a one-liner:
String input = "aabbba";
String result = IntStreamEx.ofChars(input).mapToObj(ch -> (char)ch)
.runLengths().join("").joining();
Step-by step:
IntStreamEx.ofChars(input): create IntStreamEx (enhanced IntStream) where each element is the corresponding character of input line.
.mapToObj(ch -> (char)ch): transform to StreamEx<Character> (enhanced Stream<Character>) where each element is the Character object.
.runLengths(): convert to EntryStream<Character, Long> (enhanced Stream<Map.Entry<Character, Long>>) where keys are Character objects and values are counts of equal adjacent characters.
.join(""): convert to StreamEx<String>, joining keys (characters) and values (counts) via given empty separator.
.joining(): final reduction to the resulting string without additional separators.

You're just missing the print of the last group of letters. you only print inside the loop once you found a different letter, you should take into account the last group of letters that has no "different letter" after it

I would suggest using a StringBuilder:
public String myOutput(String str) {
if (str == null || str.length() == 0)
return str;
StringBuilder sb = new StringBuilder();
int count = 1;
char currentChar;
for (int i = 0; i < str.length() - 1; i++) {
currentChar = str.charAt(i);
if (currentChar == str.charAt(i+1)) {
count++;
} else {
sb.append(currentChar);
sb.append(String.valueOf(count));
count = 1;
}
}
sb.append(str.charAt(str.length()-1));
sb.append(String.valueOf(count));
return sb.toString();
}

You only need 1 loop
System.out.println() will cause your output to have line break. You better use System.out.print(). Now your currrent code is resulting :
a2
b3

Unsure how to implement for loop

Hello I am having trouble implementing this function
Function:
Decompress the String s. Character in the string is preceded by a number. The number tells you how many times to repeat the letter. return a new string.
"3d1v0m" becomes "dddv"
I realize my code is incorrect thus far. I am unsure on how to fix it.
My code thus far is :
int start = 0;
for(int j = 0; j < s.length(); j++){
if (s.isDigit(charAt(s.indexOf(j)) == true){
Integer.parseInt(s.substring(0, s.index(j))

Assuming the input is in correct format, the following can be a simple code using for loop. Of course this is not a stylish code and you may write more concise and functional style code using Commons Lang or Guava.
StringBuilder builder = new StringBuilder();
for (int i = 0; i < s.length(); i += 2) {
final int n = Character.getNumericValue(s.charAt(i));
for (int j = 0; j < n; j++) {
builder.append(s.charAt(i + 1));
}
}
System.out.println(builder.toString());

Here is a solution you may like to use that uses Regex:
String query = "3d1v0m";
StringBuilder result = new StringBuilder();
String[] digitsA = query.split("\\D+");
String[] letterA = query.split("[0-9]+");
for (int arrIndex = 0; arrIndex < digitsA.length; arrIndex++)
{
for (int count = 0; count < Integer.parseInt(digitsA[arrIndex]); count++)
{
result.append(letterA[arrIndex + 1]);
}
}
System.out.println(result);
Output
dddv
This solution is scalable to support more than 1 digit numbers and more than 1 letter patterns.
i.e.
Input
3vs1a10m
Output
vsvsvsammmmmmmmmm

Though Nami's answer is terse and good. I'm still adding my solution for variety, built as a static method, which does not use a nested For loop, instead, it uses a While loop. And, it requires that the input string has even number of characters and every odd positioned character in the compressed string is a number.
public static String decompress_string(String compressed_string)
{
String decompressed_string = "";
for(int i=0; i<compressed_string.length(); i = i+2) //Skip by 2 characters in the compressed string
{
if(compressed_string.substring(i, i+1).matches("\\d")) //Check for a number at odd positions
{
int reps = Integer.parseInt(compressed_string.substring(i, i+1)); //Take the first number
String character = compressed_string.substring(i+1, i+2); //Take the next character in sequence
int count = 1;
while(count<=reps)//check if at least one repetition is required
{
decompressed_string = decompressed_string + character; //append the character to end of string
count++;
};
}
else
{
//In case the first character of the code pair is not a number
//Or when the string has uneven number of characters
return("Incorrect compressed string!!");
}
}
return decompressed_string;
}

Doubling each letter in a String

I'm doing a project for Java 1, and I'm completely stuck on this question.
Basically I need to double each letter in a string.
"abc" -> "aabbcc"
"uk" -> "uukk"
"t" -> "tt"
I need to do it in a while loop in what is considered "Java 1" worthy. So i'm guessing that this means more of a problematic approach.
I know that the easiest way for me to do this, from my knowledge, would be using the charAt method in a while loop, but for some reason my mind can't figure out how to return the characters to another method as a string.
Thanks
[EDIT] My Code (wrong, but maybe this will help)
int index = 0;
int length = str.length();
while (index < length) {
return str.charAt(index) + str.charAt(index);
index++;
}

String s="mystring".replaceAll(".", "$0$0");
The method String.replaceAll uses the regular expression syntax which is described in the documentation of the Pattern class, where we can learn that . matches “any character”. Within the replacement, $number refers to numbered “capturing group” whereas $0 is predefined as the entire match. So $0$0 refers to the matching character two times. As the name of the method suggests, it is performed for all matches, i.e. all characters.

Yeah, a for loop would really make more sense here, but if you need to use a while loop then it would look like this:
String s = "abc";
String result = "";
int i = 0;
while (i < s.length()){
char c = s.charAt(i);
result = result + c + c;
i++;
}

You can do:
public void doubleString(String input) {
String output = "";
for (char c : input.toCharArray()) {
output += c + c;
}
System.out.println(output);
}

Your intuition is very good. charAt(i) will return the character in the string at location i, yes?
You also said you wanted to use a loop. A for loop, traversing the length of the list, string.length(), will allow you to do this. At every single node in the string, what do you need to do? Double the character.
Let's take a look at your code:
int index = 0;
int length = str.length();
while (index < length) {
return str.charAt(index) + str.charAt(index); //return ends the method
index++;
}
Problematically for your code, you are returning two characters immediately upon entering the loop. So for a string abc, you are returning aa. Let's store the aa in memory instead, and then return the completed string like so:
int index = 0;
int length = str.length();
String newString = "";
while (index < length) {
newString += str.charAt(index) + str.charAt(index);
index++;
}
return newString;
This will add the character to newString, allowing you to return the entire completed string, as opposed to a single set of doubled characters.
By the way, this may be easier to do as a for loop, condensing and clarifying your code. My personal solution (for a Java 1 class) would look something like this:
String newString = "";
for (int i = 0; i < str.length(); i++){
newString += str.charAt(i) + str.charAt(i);
}
return newString;
Hope this helps.

try this
String a = "abcd";
char[] aa = new char[a.length() * 2];
for(int i = 0, j = 0; j< a.length(); i+=2, j++){
aa[i] = a.charAt(j);
aa[i+1]= a.charAt(j);
}
System.out.println(aa);

public static char[] doubleChars(final char[] input) {
final char[] output = new char[input.length * 2];
for (int i = 0; i < input.length; i++) {
output[i] = input[i];
output[i + 1] = input[i];
}
return output;
}

Assuming this is inside a method, you should understand that you can only return once from a method. After encountering a return statement, the control goes back to the calling method. Thus your approach of returning char every time in a loop is faulty.
int index = 0;
int length = str.length();
while (index < length) {
return str.charAt(index) + str.charAt(index); // only the first return is reachable,other are not executed
index++;
}
Change your method to build a String and return it
public String modify(String str)
{
int index = 0;
int length = str.length();
String result="";
while (index < length) {
result += str.charAt[index]+str.charAt[index];
index++;
}
return result;
}