Develop Reference

How to replace the same character with deferent value?

I'm facing a problem in replacing character in a string with its index.
e.g I wanna replace every '?' With its index String:
"a?ghmars?bh?" -> will be "a1ghmars8bh11".
Any help is truly appreciated.
P.s I need to solve this assignment today so I can pass it to my instructor.
Thanks in adv.
So far I get to manage replacing the ? With 0; through this piece of code:
public static void main(String[] args) {
String name = "?tsds?dsds?";
String myarray[] = name.split("");
for (int i = 0; i < myarray.length; i++) {
name = name.replace("?", String.valueOf(i++));
}
System.out.println(name);
output:
0tsds0dsds0
it should be:
0tsds5dsds10

For simple replace operations, String.replaceAll is sufficient. For more complex operations, you have to retrace partly, what this method does.
The documentation of String.replaceAll says that it is equivalent to
Pattern.compile(regex).matcher(str).replaceAll(repl)
whereas the linked documentation of replaceAll contains a reference to the method appendReplacement which is provided by Java’s regex package publicly for exactly the purpose of supporting customized replace operations. It’s documentation also gives a code example of the ordinary replaceAll operation:
Pattern p = Pattern.compile("cat");
Matcher m = p.matcher("one cat two cats in the yard");
StringBuffer sb = new StringBuffer();
while (m.find()) {
m.appendReplacement(sb, "dog");
}
m.appendTail(sb);
System.out.println(sb.toString());
Using this template, we can implement the desired operation as follows:
String name = "?tsds?dsds?";
Matcher m=Pattern.compile("?", Pattern.LITERAL).matcher(name);
StringBuffer sb=new StringBuffer();
while(m.find()) {
m.appendReplacement(sb, String.valueOf(m.start()));
}
m.appendTail(sb);
name=sb.toString();
System.out.println(name);
The differences are that we use a LITERAL pattern to inhibit the special meaning of ? in regular expressions (that’s easier to read than using "\\?" as pattern). Further, we specify a String representation of the found match’s location as the replacement (which is what your question was all about). That’s it.

In previous answer wrong read question, sorry. This code replace every "?" with its index
String string = "a?ghmars?bh?das?";
while ( string.contains( "?" ) )
{
Integer index = string.indexOf( "?" );
string = string.replaceFirst( "\\?", index.toString() );
System.out.println( string );
}
So from "a?ghmars?bh?das?" we got "a1ghmars8bh11das16"

You are (more or less) replacing each target with the cardinal number of the occurrence (1 for 1st, 2 for 2nd, etc) but you want the index.
Use a StringBuilder - you only need a few lines:
StringBuilder sb = new StringBuilder(name);
for (int i = name.length - 1; i <= 0; i--)
if (name.charAt(i) == '?')
sb.replace(i, i + 1, i + "");
Note counting down, not up, allowing for the replacement index to be multiple digits, which if you counted up would change the index of subsequent calls (eg everything would get shuffled to the right by one char when the index of "?" was 10 or more).

I think this may work i have not checked it.
public class Stack{
public static void main(String[] args) {
String name = "?tsds?dsds?";
int newvalue=50;
int countspecialcharacter=0;
for(int i=0;i<name.length();i++)
{
char a=name.charAt(i);
switch(a)
{
case'?':
countspecialcharacter++;
if(countspecialcharacter>1)
{
newvalue=newvalue+50;
System.out.print(newvalue);
}
else
{
System.out.print(i);
}
break;
default:
System.out.print(a);
break;
}
}
}
}

Check below code
String string = "a?ghmars?bh?das?";
for (int i = 0; i < string.length(); i++) {
Character r=string.charAt(i);
if(r.toString().equals("?"))
System.out.print(i);
else
System.out.print(r);
}

How to mimic transliterate in Java?

In Perl, I usually use the transliteration to count the number of characters in a string that match a set of possible characters. Things like:
$c1=($a =~ y[\x{0410}-\x{042F}\x{0430}-\x{044F}]
[\x{0410}-\x{042F}\x{0430}-\x{044F}]);
would count the number of Cyrillic characters in $a. As in the previous example I have two classes (or two ranges, if you prefer), I have some other with some more classes:
$c4=($a =~ y[\x{AC00}-\x{D7AF}\x{1100}-\x{11FF}\x{3130}-\x{318F}\x{A960}-\x{A97F}\x{D7B0}-\x{D7FF}]
[\x{AC00}-\x{D7AF}\x{1100}-\x{11FF}\x{3130}-\x{318F}\x{A960}-\x{A97F}\x{D7B0}-\x{D7FF}]);
Now, I need to do a similar thing in Java. Is there a similar construct in Java? Or I need to iterate over all characters, and check if it is between the limits of each class?
Thank you

Haven't seen anything like tr/// in Java.
You could use something like this to count all the matches tho:
Pattern p = Pattern.compile("[\\x{0410}-\\x{042F}\\x{0430}-\\x{044F}]",
Pattern.CANON_EQ);
Matcher m = p.matcher(string);
int count = 0;
while (m.find())
count++;

For good order: using the Java Unicode support.
int countCyrillic(String s) {
int n = 0;
for (int i = 0; i < s.length(); ) {
int codePoint = s.codePointAt(i);
i += Character.charCount(codePoint);
if (UnicodeScript.of(codePoint) == UnicodeScript.CYRILLIC) {
++n;
}
}
return n;
}
This uses the full Unicode (where two 16 bit chars may represent a Unicode "code point."
And in Java the class Character.UnicodeScript has already everything you need.
Or:
int n = s.replaceAll("\\P{CYRILLIC}", "").length();
Here \\P is the negative of \\p{CYRILLIC} the Cyrillic group.

You can try to play with something like this:
s.replaceAll( "[^\x{0410}-\x{042F}\x{0430}-\x{044F}]*([\x{0410}-\x{042F}\x{0430}-\x{044F}])?", "$1" ).length()
The idea was borrowed from here: Simple way to count character occurrences in a string

how to check string contain any character other than number in java?

I want to check String contain any character or special character other than number.I wrote following code for this
String expression = "[^a-zA-z]";
Pattern pattern = Pattern.compile(expression, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(jTextFieldPurchaseOrder.getText().toString().trim());
It is working fine when i am taking value from jTextField and checking my condition. But giving error when checking String from DTO as below
list.get(0).getChalan_trans_id().toString().trim().matches("[^a-zA-z]");
Where list is arraylist of DTO.
I am not getting where am I going wrong?
Thanks

If you want to check if there is a non-digit character, you can use .*\\D.*:
if (list.get(0).getChalan_trans_id().toString().trim().matches(".*\\D.*")) {
//non-digit found, handle it
}
or, maybe easier, do it the other way around:
if (list.get(0).getChalan_trans_id().toString().trim().matches("\\d*")) {
//only digits found
}

There's probably a more efficient way than regular expressions. Regular expressions are powerful, but can be overkill for a simple task like this.
Something like this ought to work, and I would expect it to be quicker.
static boolean hasNonNumber(String s) {
for (int i = 0; i < s.length(); ++i) {
char c = s.charAt(i);
if (!Character.isDigit(c)) {
return true;
}
}
return false;
}

Efficient way to replace chars in a string (java)?

I'm writing a small JAVA program which:
takes a text as a String
takes 2 arrays of chars
What im trying to do will sound like "find and replace" but it is not the same so i thought its important to clear it.
Anyway I want to take this text, find if any char from the first array match a char in the text and if so, replace it with the matching char (according to index) from the second char array.
I'll explain with an example:
lets say my text (String) is: "java is awesome!";
i have 2 arrays (char[]): "absm" and "!#*$".
The wished result is to change 'a' to '!' , 'b' to '#' and so on..
meaning the resulted text will be:
"java is awesome!" changed to -> "j#v# i* #w*o$e!"
What is the most efficient way of doing this and why?
I thought about looping the text, but then i found it not so efficient.
(StringBuilder/String class can be used)

StringBuilder sb = new StringBuilder(text);
for(int i = 0; i<text.length(); i ++)
{
for (int j = 0; j < firstCharArray.length;j++)
{
if (sb.charAt(i) == firstCharArray[j])
{
sb.setCharAt(i, secondCharArray[j]);
break;
}
}
}
This way is efficient because it uses a StringBuilder to change the characters in place (if you used Strings you would have to create new ones each time because they are immutable.) Also it minimizes the amount of passes you have to do (1 pass through the text string and n passes through the first array where n = text.length())

I guess you are looking for StringUtils.replaceEach, at least as a reference.

How efficient do you need it to be? Are you doing this for hundreds, thousands, millions of words???
I don't know if it's the most efficent, but you could use the string indexOf() method on each of your possible tokens, it will tell you if it's there, and then you can replace that index at the same time with the corresponding char from the other array.
Codewise, something like (this is half pseudo code by the way):
for(each of first array) {
int temp = YourString.indexOf(current array field);
if (temp >=0) {
replace with other array
}
}

Put the 2 arrays you have in a Map
Map<Character, Character> //or Map of Strings
where the key is "a", "b" etc... and the value is the character you want to substitute with - "#" etc....
Then simply replace the keys in your String with the values.

For small stuff like this, an indexOf() search is likely to be faster than a map, while "avoiding" the inner loop of the accepted answer. Of course, the loop is still there, inside String.indexOf(), but it's likely to be optimized to a fare-thee-well by the JIT-compiler, because it's so heavily used.
static String replaceChars(String source, String from, String to)
{
StringBuilder dest = new StringBuilder(source);
for ( int i = 0; i < source.length(); i++ )
{
int foundAt = from.indexOf(source.charAt(i));
if ( foundAt >= 0 )
dest.setCharAt(i,to.charAt(foundAt));
}
return dest.toString();
}
Update: The Oracle/Sun JIT uses SIMD on at least some processors for indexOf(), making it even faster than one would guess.

Since the only way to know if a character should be replaced is to check it, you (or any util method) have to loop through the whole text, character after the other. You can never achieve better complexity than O(n) (n be the number of characters in the text).

This utility class that replaces a char or a group of chars of a String. It is equivalent to bash tr and perl tr///, aka, transliterate.
/**
* Utility class that replaces chars of a String, aka, transliterate.
*
* It's equivalent to bash 'tr' and perl 'tr///'.
*
*/
public class ReplaceChars {
public static String replace(String string, String from, String to) {
return new String(replace(string.toCharArray(), from.toCharArray(), to.toCharArray()));
}
public static char[] replace(char[] chars, char[] from, char[] to) {
char[] output = chars.clone();
for (int i = 0; i < output.length; i++) {
for (int j = 0; j < from.length; j++) {
if (output[i] == from[j]) {
output[i] = to[j];
break;
}
}
}
return output;
}
/**
* For tests!
*/
public static void main(String[] args) {
// Example from: https://en.wikipedia.org/wiki/Caesar_cipher
String string = "THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG";
String from = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String to = "XYZABCDEFGHIJKLMNOPQRSTUVW";
System.out.println();
System.out.println("Cesar cypher: " + string);
System.out.println("Result: " + ReplaceChars.replace(string, from, to));
}
}
This is the output:
Cesar cypher: THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG
Result: QEB NRFZH YOLTK CLU GRJMP LSBO QEB IXWV ALD

How can I check if a single character appears in a string?

In Java is there a way to check the condition:
"Does this single character appear at all in string x"
without using a loop?

You can use string.indexOf('a').
If the char a is present in string :
it returns the the index of the first occurrence of the character in
the character sequence represented by this object, or -1 if the
character does not occur.

String.contains() which checks if the string contains a specified sequence of char values
String.indexOf() which returns the index within the string of the first occurence of the specified character or substring (there are 4 variations of this method)

I'm not sure what the original poster is asking exactly. Since indexOf(...) and contains(...) both probably use loops internally, perhaps he's looking to see if this is possible at all without a loop? I can think of two ways off hand, one would of course be recurrsion:
public boolean containsChar(String s, char search) {
if (s.length() == 0)
return false;
else
return s.charAt(0) == search || containsChar(s.substring(1), search);
}
The other is far less elegant, but completeness...:
/**
* Works for strings of up to 5 characters
*/
public boolean containsChar(String s, char search) {
if (s.length() > 5) throw IllegalArgumentException();
try {
if (s.charAt(0) == search) return true;
if (s.charAt(1) == search) return true;
if (s.charAt(2) == search) return true;
if (s.charAt(3) == search) return true;
if (s.charAt(4) == search) return true;
} catch (IndexOutOfBoundsException e) {
// this should never happen...
return false;
}
return false;
}
The number of lines grow as you need to support longer and longer strings of course. But there are no loops/recurrsions at all. You can even remove the length check if you're concerned that that length() uses a loop.

You can use 2 methods from the String class.
String.contains() which checks if the string contains a specified sequence of char values
String.indexOf() which returns the index within the string of the first occurence of the specified character or substring or returns -1 if the character is not found (there are 4 variations of this method)
Method 1:
String myString = "foobar";
if (myString.contains("x") {
// Do something.
}
Method 2:
String myString = "foobar";
if (myString.indexOf("x") >= 0 {
// Do something.
}
Links by: Zach Scrivena

String temp = "abcdefghi";
if(temp.indexOf("b")!=-1)
{
System.out.println("there is 'b' in temp string");
}
else
{
System.out.println("there is no 'b' in temp string");
}

If you need to check the same string often you can calculate the character occurrences up-front. This is an implementation that uses a bit array contained into a long array:
public class FastCharacterInStringChecker implements Serializable {
private static final long serialVersionUID = 1L;
private final long[] l = new long[1024]; // 65536 / 64 = 1024
public FastCharacterInStringChecker(final String string) {
for (final char c: string.toCharArray()) {
final int index = c >> 6;
final int value = c - (index << 6);
l[index] |= 1L << value;
}
}
public boolean contains(final char c) {
final int index = c >> 6; // c / 64
final int value = c - (index << 6); // c - (index * 64)
return (l[index] & (1L << value)) != 0;
}}

To check if something does not exist in a string, you at least need to look at each character in a string. So even if you don't explicitly use a loop, it'll have the same efficiency. That being said, you can try using str.contains(""+char).

Is the below what you were looking for?
int index = string.indexOf(character);
return index != -1;

Yes, using the indexOf() method on the string class. See the API documentation for this method

String.contains(String) or String.indexOf(String) - suggested
"abc".contains("Z"); // false - correct
"zzzz".contains("Z"); // false - correct
"Z".contains("Z"); // true - correct
"😀and😀".contains("😀"); // true - correct
"😀and😀".contains("😂"); // false - correct
"😀and😀".indexOf("😀"); // 0 - correct
"😀and😀".indexOf("😂"); // -1 - correct
String.indexOf(int) and carefully considered String.indexOf(char) with char to int widening
"😀and😀".indexOf("😀".charAt(0)); // 0 though incorrect usage has correct output due to portion of correct data
"😀and😀".indexOf("😂".charAt(0)); // 0 -- incorrect usage and ambiguous result
"😀and😀".indexOf("😂".codePointAt(0)); // -1 -- correct usage and correct output
The discussions around character is ambiguous in Java world
can the value of char or Character considered as single character?
No. In the context of unicode characters, char or Character can sometimes be part of a single character and should not be treated as a complete single character logically.
if not, what should be considered as single character (logically)?
Any system supporting character encodings for Unicode characters should consider unicode's codepoint as single character.
So Java should do that very clear & loud rather than exposing too much of internal implementation details to users.
String class is bad at abstraction (though it requires confusingly good amount of understanding of its encapsulations to understand the abstraction 😒😒😒 and hence an anti-pattern).
How is it different from general char usage?
char can be only be mapped to a character in Basic Multilingual Plane.
Only codePoint - int can cover the complete range of Unicode characters.
Why is this difference?
char is internally treated as 16-bit unsigned value and could not represent all the unicode characters using UTF-16 internal representation using only 2-bytes. Sometimes, values in a 16-bit range have to be combined with another 16-bit value to correctly define character.
Without getting too verbose, the usage of indexOf, charAt, length and such methods should be more explicit. Sincerely hoping Java will add new UnicodeString and UnicodeCharacter classes with clearly defined abstractions.
Reason to prefer contains and not indexOf(int)
Practically there are many code flows that treat a logical character as char in java.
In Unicode context, char is not sufficient
Though the indexOf takes in an int, char to int conversion masks this from the user and user might do something like str.indexOf(someotherstr.charAt(0))(unless the user is aware of the exact context)
So, treating everything as CharSequence (aka String) is better
public static void main(String[] args) {
System.out.println("😀and😀".indexOf("😀".charAt(0))); // 0 though incorrect usage has correct output due to portion of correct data
System.out.println("😀and😀".indexOf("😂".charAt(0))); // 0 -- incorrect usage and ambiguous result
System.out.println("😀and😀".indexOf("😂".codePointAt(0))); // -1 -- correct usage and correct output
System.out.println("😀and😀".contains("😀")); // true - correct
System.out.println("😀and😀".contains("😂")); // false - correct
}
Semantics
char can handle most of the practical use cases. Still its better to use codepoints within programming environment for future extensibility.
codepoint should handle nearly all of the technical use cases around encodings.
Still, Grapheme Clusters falls out of the scope of codepoint level of abstraction.
Storage layers can choose char interface if ints are too costly(doubled). Unless storage cost is the only metric, its still better to use codepoint. Also, its better to treat storage as byte and delegate semantics to business logic built around storage.
Semantics can be abstracted at multiple levels. codepoint should become lowest level of interface and other semantics can be built around codepoint in runtime environment.

package com;
public class _index {
public static void main(String[] args) {
String s1="be proud to be an indian";
char ch=s1.charAt(s1.indexOf('e'));
int count = 0;
for(int i=0;i<s1.length();i++) {
if(s1.charAt(i)=='e'){
System.out.println("number of E:=="+ch);
count++;
}
}
System.out.println("Total count of E:=="+count);
}
}

static String removeOccurences(String a, String b)
{
StringBuilder s2 = new StringBuilder(a);
for(int i=0;i<b.length();i++){
char ch = b.charAt(i);
System.out.println(ch+" first index"+a.indexOf(ch));
int lastind = a.lastIndexOf(ch);
for(int k=new String(s2).indexOf(ch);k > 0;k=new String(s2).indexOf(ch)){
if(s2.charAt(k) == ch){
s2.deleteCharAt(k);
System.out.println("val of s2 : "+s2.toString());
}
}
}
System.out.println(s1.toString());
return (s1.toString());
}

you can use this code. It will check the char is present or not. If it is present then the return value is >= 0 otherwise it's -1. Here I am printing alphabets that is not present in the input.
import java.util.Scanner;
public class Test {
public static void letters()
{
System.out.println("Enter input char");
Scanner sc = new Scanner(System.in);
String input = sc.next();
System.out.println("Output : ");
for (char alphabet = 'A'; alphabet <= 'Z'; alphabet++) {
if(input.toUpperCase().indexOf(alphabet) < 0)
System.out.print(alphabet + " ");
}
}
public static void main(String[] args) {
letters();
}
}
//Ouput Example
Enter input char
nandu
Output :
B C E F G H I J K L M O P Q R S T V W X Y Z

If you see the source code of indexOf in JAVA:
public int indexOf(int ch, int fromIndex) {
final int max = value.length;
if (fromIndex < 0) {
fromIndex = 0;
} else if (fromIndex >= max) {
// Note: fromIndex might be near -1>>>1.
return -1;
}
if (ch < Character.MIN_SUPPLEMENTARY_CODE_POINT) {
// handle most cases here (ch is a BMP code point or a
// negative value (invalid code point))
final char[] value = this.value;
for (int i = fromIndex; i < max; i++) {
if (value[i] == ch) {
return i;
}
}
return -1;
} else {
return indexOfSupplementary(ch, fromIndex);
}
}
you can see it uses a for loop for finding a character. Note that each indexOf you may use in your code, is equal to one loop.
So, it is unavoidable to use loop for a single character.
However, if you want to find a special string with more different forms, use useful libraries such as util.regex, it deploys stronger algorithm to match a character or a string pattern with Regular Expressions. For example to find an email in a string:
String regex = "^(.+)#(.+)$";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(email);
If you don't like to use regex, just use a loop and charAt and try to cover all cases in one loop.
Be careful recursive methods has more overhead than loop, so it's not recommended.

how about one uses this ;
let text = "Hello world, welcome to the universe.";
let result = text.includes("world");
console.log(result) ....// true
the result will be a true or false
this always works for me

You won't be able to check if char appears at all in some string without atleast going over the string once using loop / recursion ( the built-in methods like indexOf also use a loop )
If the no. of times you look up if a char is in string x is more way more than the length of the string than I would recommend using a Set data structure as that would be more efficient than simply using indexOf
String s = "abc";
// Build a set so we can check if character exists in constant time O(1)
Set<Character> set = new HashSet<>();
int len = s.length();
for(int i = 0; i < len; i++) set.add(s.charAt(i));
// Now we can check without the need of a loop
// contains method of set doesn't use a loop unlike string's contains method
set.contains('a') // true
set.contains('z') // false
Using set you will be able to check if character exists in a string in constant time O(1) but you will also use additional memory ( Space complexity will be O(n) ).