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StringBuilder deleteCharAt() function " String index out of range : 6 "

Code and Error
Error 2
hey Guys see the Image and help me out to sort out this issue
It is the first code and it runs perfectly but when i use the same approach in the blow code it have some error
" For error detail open image link "
String str = "Samaarth";
StringBuilder sb = new StringBuilder(str);
sb.deleteCharAt(3);
System.out.println(sb.toString());
This is where the error start and the error is because of DeleteCharAt() function but in the above code this function works perfectly but here it is not
IDK why so please help me out to sort our this issue
String str= "aaabccddd";
StringBuilder sb = new StringBuilder(str);
for (int i = 0; i < str.length() -1; i++) {
if (sb.charAt(i) == sb.charAt(i + 1)) {
sb.deleteCharAt(i);
//sb.deleteCharAt(i+1);
}
}
System.out.println(sb.toString());

Samarath, you both modify the string and advance the counter.
This is wrong. Consider the string "aaaa"
This is what your code does:
i = 0: you find the duplicate, remove it. The string becomes "aaa".
Then you advance the position: i becomes 1
i = 1: the string is "a|aa" (the vertical bar shows the position).
You find the duplicate at position 1. You kill it, the string
becomes "aa", but you advance the position one again: i becomes 2
At this step the for loop ends and your string is "aa".
Instead the algorithm should use while loop: "while there are duplicates, kill them!"
String str= "aaabccddd";
StringBuilder sb = new StringBuilder(str);
int i = 0;
while (i < sb.length()-1) {
if (sb.charAt(i) == sb.charAt(i + 1)) {
sb.deleteCharAt(i);
// Do not increment -- kill all duplicates
} else {
// Either not a duplicate, or all duplicated killed
// Advance one char
i++;
}
}
System.out.println(sb.toString());
The output is abcd.
If you are inclined to use for loop, then iterate in the reverse order:
String str= "aaabccddd";
StringBuilder sb = new StringBuilder(str);
for (int i = sb.length()-1; i > 0; i--) {
if (sb.charAt(i) == sb.charAt(i - 1)) {
// Note charAt(i - 1) - we compare with the preceding character
sb.deleteCharAt(i);
// The string squeezes by one char, but the decremented position
// will follow
}
}
System.out.println(sb.toString());
The output is abcd

The problem is you are using for loop and you are actually changing/mutating StringBuilder instance at the same time, so the .length() will not be fixed and eventually you will try to reach non-existing index in your for loop and exception will be thrown.
EDIT:
Add these two lines inside your for loop if statement, just before you invoke deleteCharAt() method:
System.out.println("Value of i is: " + i);
System.out.println("StringBuilder length is: " + sb.length());
"i" represents index you are trying to delete, and sb.length() will display actual length of the StringBuilder.

Performance (JAVA) ~ String concatenation in a loop with prepending and appending

I'm having performance issues. Does anyone have a faster/better solution for doing the following:
String main = "";
for (String proposition : propositions) {
if (main.length() == 0) {
main = proposition;
} else {
main = "|(" + proposition + "," + main + ")";
}
}
I know concat and stringbuilder are faster, but i don't see how i can use these methods. Because of the following line of code:
main = "|(" + proposition + "," + main + ")";
Thanks in advance!

So from what I can tell there are 3 problems here:
Values are primarily prepended to the string.
For each value a character is appended.
If only one value is present, nothing should be appended or prepended.
With 2 or more items, the 0th item is handled differently:
0:""
1:"A"
2:"|(B,A)"
3:"|(C,|(B,A))"
It can be made quicker by making a few changes:
Reverse the algorithm, this means the majority of the work involves appending, allowing you to use StringBuilders.
Count the number of closing )'s and append those after the loop is finished.
Special case for 0 or 1 items in the list.
With those changes the algorithm should be able to use a StringBuilder and be a lot quicker.
Attempt at an algorithm:
int length = propositions.size();
if (length == 0) {
main = "";
} else {
StringBuilder sb = new StringBuilder();
int nestingDepth = 0;
// Reverse loop, ignoring 0th element due to special case
for (int i = length - 1; i > 0; i--) {
sb.append("|(").append(propositions.get(i)).append(',');
nestingDepth++;
}
// Append last element due to special casing
sb.append(propositions.get(0));
for (int i = 0; i < nestingDepth; i++) {
sb.append(')');
}
main = sb.toString();
}
I believe this should produce the correct results, but it should give the right idea.

The problem is that you're prepending and appending to the string as you go. String and StringBuilder dont handle this well (and give quadratic performance). But you can use a dequeue which supports insertion at start and end to store all the pieces. Then finally you can join the bits in the dequeue.
ArrayDeque bits = new ArrayDeque();
for (String proposition : propositions) {
if (bits.size() == 0) {
bits.push(proposition);
} else {
// Add prefix
main.offerFirst("|(" + proposition + "," );
// Add suffix
main.push(")");
}
}
StringBuilder sb = new StringBuilder();
for( String s : bits) {
sb.append(s);
}
main = sb.toString();

Assuming this is an array of propositions, you could first sum the length of the String(s) in the array. Add 4 for your additional characters, and subtract 4 because you don't use those separators on the first element. That should be the perfect size for your output (this is optional, because StringBuilder is dynamically sized). Next, construct a StringBuilder. Add the first element. All subsequent elements follow the same pattern, so the loop is simplified with a traditional for. Something like,
int len = Stream.of(propositions).mapToInt(s -> s.length() + 4).sum() - 4;
StringBuilder sb = new StringBuilder(len); // <-- len is optional
sb.append(propositions[0]);
for (int i = 1; i < propositions.length; i++) {
sb.insert(0, ",").insert(0, propositions[i]).insert(0, "|(").append(")");
}
System.out.println(sb);

How to insert a space in a charArray in an exact position [Java]

My problem is that I'm getting a String and I need to check if there is a space in the 4th position but starting from the end. If in this position there is not a space, I should insert it.
For example:
I get this String: TW12EF, need to get it like this: TW1 2EF
First of all I get the 4 last characters in a char array because I also need to check if they are numbers or letters.
With this method I check if there is a space:
public static boolean isSpace(){
return String.valueOf(charArray[0]).matches("[ \\t\\n\\x0B\\f\\r]");
}
charArray contains the last 4 characters of the input String
If charArray[0] wouldn't be a space, I want to insert a space in the 2nd place (charArray[1])
If there is something that I can correct in the question to make it easier to understand, just let me know and I will try to make it better for next questions.

A simple and direct solution (most likely faster than using a regular expression) is to get the 4th to the last character (if it exists), and if it isn't a white-space, insert a space at that position.
public static void main(String[] args) {
String str = "TW12EF";
int insertPos = str.length() - 4;
if (insertPos >= 0) {
char ch = str.charAt(insertPos);
if (!Character.isWhitespace(ch)) {
str = new StringBuilder(str).insert(insertPos + 1, ' ').toString();
}
}
System.out.println(str);
}
A whitespace is determined by invoking isWhitespace, which returns true for space but also tabs or line feeds, like you did in your question. The character is inserted by leveraging the StringBuilder#insert method, which is more direct that taking 2 substrings and concatenating them.

A quick, dirty regex will help :
String p = "TW12EF";
System.out.println(p.replaceAll("(.)\\s*(\\S.{2})$", "$1 $2")); // Select a character followed by 0 or more spaces and followed by 3 non-space characters. And replace multiple spaces if they exist with a single space
O/P :
TW1 2EF
Also works if there are one or more spaces after the 3rd char (from the left)

As char is a primitive data type, the comparison can be done simply with
if (charArray[0] == ' ') {
char[] temp = new char[5];
temp[0] = ' ';
for (int i = 1; i <= 4; i++) {
temp[i] = charArray[i - 1];
}
charArray = temp;
}

You could use something like:
public static void main(String[] args) {
String str = "TW12EF";
processStr(str);
}
public static final int SPACE_POS = 4, OFFSET = 1;
public static String processStr(String str)
{
if(!Character.isWhitespace(str.charAt(str.length() - SPACE_POS)))
{
str = String.format("%s %s", str.substring(0, str.length() - SPACE_POS + OFFSET), str.substring(SPACE_POS - OFFSET));
}
return str;
}

Like this？
` String s="TW12EF";
String result="";
int length=s.length();
for(int i=length-1;i>-1;i--){
if(i==length-4&&s.charAt(i)!=' '){
result+=" ";
}
result+=s.charAt(length-i-1);
}
System.out.println(result);`

How to specify the number of alphabets (and not blank spaces) to include in a substring?

I am trying to print a substring using index value. I need to exclude the blank space while counting but it should print the output along with blank space. I want to display, say, n alphabets from the main string. The blank spaces will be as they are but the number of alphabets from the lower bound to upper bound index should be n. My code is
public class Test {
public static void main(String args[])
{
String Str=new String("Welcome to the class");
System.out.println("\nReturn value is:");
System.out.println(Str.substring(2,9));
}
}
Output:
lcome t
In the above mentioned code, it counts the space between the "Welcome" and "to". i need not want to count the space between them. My expected output is lcome to

You could use simple mathematics. Just substring it, remove all whitespaces and compare the original length to the String without whitespaces. Afterwards add the difference in size to your end index for the substring.
public static void main(String args[]) {
String Str = "Welcome to the class";
System.out.println("\nReturn value is:");
String sub = Str.substring(2, 9);
String wsRemoved = sub.replaceAll(" ", "");
String wsBegginingRemoved = sub.replaceAll("^ *", "");
String outputSub = Str.substring(2+(sub.length()-wsBegginingRemoved.length()), 9+(sub.length()-wsRemoved.length()+(sub.length() - wsBegginingRemoved.length())));
System.out.println(outputSub);
}
Edit: not ignoring leading whitespaces anymore
O/P
lcome to
O/P "My name is Earl"
name is E

One way would be to extract it to using a regex ^.{2}([^ ] *){7}.
Another option is to use a simple for loop to traverse the string and calculate the end point to use for substring.
int non_whitespace = 0; int i;
for(i = 2; non_whitespace < 7; ++non_whitespace, ++i) {
while (str.charAt(i) == ' ') ++i;
}
return str.substring(2, i);
It is up to you which method do you consider more readable, and assess which one leads to better performance if speed is a concern.

What you want to do is display n number of characters from the string including the spaces but n doesn't include the no. of blank spaces. For that, you could simply be using a loop instead of a library function.
The Logic: Keep displaying characters of the String str from index = 2 to index = 9-1 in a while loop. If the current character is a blank space, then increase the value of n, which is the upper bound of the string index for the sub string, by 1, i.e., the program will now display an extra character beyond the upper bound for each blank space encountered.
Consider the code below.
String str = "Welcome to the class";
int index = 2, n = 9;
while(index < n){
char c = str.charAt(index);
System.out.print(c);
if(c==' ')
n++;
index++;
}
Output: lcome to
Hope you can understand this code.
EDIT
As #Finbarr O'B said, a check to prevent StringIndexOutOfBoundsException would be necessary for the program for which, the loop will have to be defined as:
while(index < n && index < str.length()){
...
}

If you don't want to use regex, you can implement your own version of substring. The straightforward solution:
private static String substring(int begin, int end, String str) {
StringBuilder res = new StringBuilder();
while (begin < end) {
if (str.charAt(begin) == ' ') {
end++;
}
res.append(str.charAt(begin));
begin++;
}
return res.toString();
}
The trick here is to ignore the "count" of a space, by incrementing end when it's encountered, forcing the loop to make one extra iteration.
The code complexity is O(n).
System.out.println(substring(2, 9, "Welcome to the class"));
>> lcome to

You could use replaceFirst for this:
String Str = "Welcome to the class"; // remove new String()
Str = Str.replaceFirst("^ *", "");
System.out.println("\nReturn value is:");
System.out.println(Str.substring(2, 10)); // increment end index by 1
Output:
lcome to

Algorithm for duplicated but overlapping strings

I need to write a method where I'm given a string s and I need to return the shortest string which contains s as a contiguous substring twice.
However two occurrences of s may overlap. For example,
aba returns ababa
xxxxx returns xxxxxx
abracadabra returns abracadabracadabra
My code so far is this:
import java.util.Scanner;
public class TwiceString {
public static String getShortest(String s) {
int index = -1, i, j = s.length() - 1;
char[] arr = s.toCharArray();
String res = s;
for (i = 0; i < j; i++, j--) {
if (arr[i] == arr[j]) {
index = i;
} else {
break;
}
}
if (index != -1) {
for (i = index + 1; i <= j; i++) {
String tmp = new String(arr, i, i);
res = res + tmp;
}
} else {
res = res + res;
}
return res;
}
public static void main(String args[]) {
Scanner inp = new Scanner(System.in);
System.out.println("Enter the string: ");
String word = inp.next();
System.out.println("The requires shortest string is " + getShortest(word));
}
}
I know I'm probably wrong at the algorithmic level rather than at the coding level. What should be my algorithm?

Use a suffix tree. In particular, after you've constructed the tree for s, go to the leaf representing the whole string and walk up until you see another end-of-string marker. This will be the leaf of the longest suffix that is also a prefix of s.

As #phs already said, part of the problem can be translated to "find the longest prefix of s that is also a suffix of s" and a solution without a tree may be this:
public static String getShortest(String s) {
int i = s.length();
while(i > 0 && !s.endsWith(s.substring(0, --i)))
;
return s + s.substring(i);
}

Once you've found your index, and even if it's -1, you just need to append to the original string the substring going from index + 1 (since index is the last matching character index) to the end of the string. There's a method in String to get this substring.

i think you should have a look at the Knuth-Morris-Pratt algorithm, the partial match table it uses is pretty much what you need (and by the way it's a very nice algorithm ;)

If your input string s is, say, "abcde" you can easily build a regex like the following (notice that the last character "e" is missing!):
a(b(c(d)?)?)?$
and run it on the string s. This will return the starting position of the trailing repeated substring. You would then just append the missing part (i.e. the last N-M characters of s, where N is the length of s and M is the length of the match), e.g.
aba
^ match "a"; append the missing "ba"
xxxxxx
^ match "xxxxx"; append the missing "x"
abracadabra
^ match "abra"; append the missing "cadabra"
nooverlap
--> no match; append "nooverlap"

From my understanding you want to do this:
input: dog
output: dogdog
--------------
input: racecar
output: racecaracecar
So this is how i would do that:
public String change(String input)
{
StringBuilder outputBuilder = new StringBuilder(input);
int patternLocation = input.length();
for(int x = 1;x < input.length();x++)
{
StringBuilder check = new StringBuilder(input);
for(int y = 0; y < x;y++)
check.deleteCharAt(check.length() - 1);
if(input.endsWith(check.toString()))
{
patternLocation = x;
break;
}
}
outputBuilder.delete(0, input.length() - patternLocation);
return outputBuilder.toString();
}
Hope this helped!