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How do I replace more than one type of Character in Java String

newbie here. Any help with this problem would be appreciated:
You are given a String variable called data that contain letters and spaces only. Write the Java class to print a modified version of the String where all lowercase letters are replaced by ? and all whitespaces are replaced by +. An example is shown below: I Like Java becomes I+L???+J???.
What I have so far:
public static void main (String[] args) {
Scanner input = new Scanner(System.in);
String data;
//prompt
System.out.println("Enter a sentence: ");
//input
data = input.nextLine();
for (int i = 0; i < data.length(); i++) {
if (Character.isWhitespace(data.charAt(i))) {
data.replace("", "+");
if (Character.isLowerCase(data.charAt(i))) {
data.replace(i, i++, ); //not sure what to include here
}
} else {
System.out.print(data);
}
}
}
any suggestions would be appreciated.

You can do it in two steps by chaining String#replaceAll. In the first step, replace the regex, [a-z], with ?. The regex, [a-z] means a character from a to z.
public class Main {
public static void main(String[] args) {
String str = "I Like Java";
str = str.replaceAll("[a-z]", "?").replaceAll("\\s+", "+");
System.out.println(str);
}
}
Output:
I+L???+J???
Alternatively, you can use a StringBuilder to build the desired string. Instead of using a StringBuilder variable, you can use String variable but I recommend you use StringBuilder for such cases. The logic of building the desired string is simple:
Loop through all characters of the string and check if the character is a lowercase letter. If yes, append ? to the StringBuilder instance else if the character is whitespace, append + to the StringBuilder instance else append the character to the StringBuilder instance as it is.
Demo:
public class Main {
public static void main(String[] args) {
String str = "I Like Java";
StringBuilder sb = new StringBuilder();
int len = str.length();
for (int i = 0; i < len; i++) {
char ch = str.charAt(i);
if (Character.isLowerCase(ch)) {
sb.append('?');
} else if (Character.isWhitespace(ch)) {
sb.append('+');
} else {
sb.append(ch);
}
}
// Assign the result to str
str = sb.toString();
// Display str
System.out.println(str);
}
}
Output:
I+L???+J???
If the requirement states:
The first character of each word is a letter (uppercase or lowercase) which needs to be left as it is.
Second character onwards can be any word character which needs to be replaced with ?.
All whitespace characters of the string need to be replaced with +.
you can do it as follows:
Like the earlier solution, chain String#replaceAll for two steps. In the first step, replace the regex, (?<=\p{L})\w, with ?. The regex, (?<=\p{L})\w means:
\w specifies a word character.
(?<=\p{L}) specifies a positive lookbeghind for a letter i.e. \p{L}.
In the second step, simply replace one or more whitespace characters i.e. \s+ with +.
Demo:
public class Main {
public static void main(String[] args) {
String str = "I like Java";
str = str.replaceAll("(?<=\\p{L})\\w", "?").replaceAll("\\s+", "+");
System.out.println(str);
}
}
Output:
I+l???+J???
Alternatively, again like the earlier solution you can use a StringBuilder to build the desired string. Loop through all characters of the string and check if the character is a letter. If yes, append it to the StringBuilder instance and then loop through the remaining characters until all characters are exhausted or a space character is encountered. If a whitespace character is encountered, append + to the StringBuilder instance else append ? to it.
Demo:
public class Main {
public static void main(String[] args) {
String str = "I like Java";
StringBuilder sb = new StringBuilder();
int len = str.length();
for (int i = 0; i < len; i++) {
char ch = str.charAt(i++);
if (Character.isLetter(ch)) {
sb.append(ch);
while (i < len && !Character.isWhitespace(ch = str.charAt(i))) {
sb.append('?');
i++;
}
if (Character.isWhitespace(ch)) {
sb.append('+');
}
}
}
// Assign the result to str
str = sb.toString();
// Display str
System.out.println(str);
}
}
Output:
I+l???+J???

package com.company;
import java.util.*;
public class dat {
public static void main(String[] args) {
System.out.println("enter the string:");
Scanner ss = new Scanner(System.in);
String data = ss.nextLine();
for (int i = 0; i < data.length(); i++) {
char ch = data.charAt(i);
if (Character.isWhitespace(ch))
System.out.print("+");
else if (Character.isLowerCase(ch))
System.out.print("?");
else
System.out.print(ch);
}
}
}
enter the string:
i Love YouU
?+L???+Y??U

Firstly, you are trying to make changes to String object which is immutable. Simple way to achieve what you want is convert string to character array and loop over array items:
Scanner input = new Scanner(System.in);
String data;
//prompt
System.out.println("Enter a sentence: ");
//input
data = input.nextLine();
char[] dataArray = data.toCharArray();
for (int i = 0; i < dataArray.length; i++) {
if (Character.isWhitespace(dataArray[i])) {
dataArray[i] = '+';
} else if (Character.isLowerCase(dataArray[i])) {
dataArray[i] = '?';
}
}
System.out.print(dataArray);

See the below code and figure out what's wrong in your code. To include multiple regex put the char within square brackets:
import java.util.Scanner;
public class mainClass {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter a sentence: ");
String data = input.nextLine();
String one = data.replaceAll(" ", "+");
String two = one.replaceAll("[a-z]", "?");
System.out.println(two);
}
}

You can use String.codePoints method to get a stream over int values of characters of this string, and process them:
private static String replaceCharacters(String str) {
return str.codePoints()
.map(ch -> {
if (Character.isLowerCase(ch))
return '?';
if (Character.isWhitespace(ch))
return '+';
return ch;
})
.mapToObj(Character::toString)
.collect(Collectors.joining());
}
public static void main(String[] args) {
System.out.println(replaceCharacters("Lorem ipsum")); // L????+?????
System.out.println(replaceCharacters("I Like Java")); // I+L???+J???
}
See also: Replace non ASCII character from string

Hacker Rank Java String Tokens

I'm working on the Java String Tokens on HackerRank. My code is as following:
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
// Write your code here.
scan.close();
// if length is 0
if (s.length() == 0 || s == null) {
System.out.println(0);
return;
}
// It seems we need to remove some spaces
s = s.trim();
String[] words = s.split("[ |!|,|\\?|\\.|_|'|#|]+");
System.out.println(words.length);
for (String word: words){
System.out.println(word);
}
}
}
It has passed most tests but failed when the input is null. I've attached a screenshot in this question:
So, can anyone plz explain what happened here? And how can I fix it?
Thank you so much!

Regex "[ |!|,|\\?|\\.|_|'|#|]+" is extraneous.
Do not separate characters in a [ ] character class by the | OR pattern, since that pattern only applies outside a character class.
There is no need to escape ? and . in a character class, since they are not special characters there.
Correct regex would be [ !,?._'#]+ or [^A-Za-z]+.
The main problem with the code in the question is that split() may return an array where the first element is an empty string.
Example 1: Input ",X," will return ["", "X"]. The empty string before the leading , is included, and the empty string after the trailing , is excluded, because the javadoc says so: "Trailing empty strings are therefore not included in the resulting array".
Example 2: Input "" will return [""], because the javadoc explicitly says so: "If the expression does not match any part of the input then the resulting array has just one element, namely this string". Note how the "trailing empty string" rule is not applied to this specific use case.
Example 3: Input ",," will return [], because trailing empty strings are excluded.
In examples 1 and 2, that leading empty string should be ignored.
I'll leave the actual fixing of the code to you, since this is your challenge to solve.

if (scan.hasNext()) {
s = scan.nextLine();
} else {
System.out.println(0);
return;
}
I edited your code a bit. For no String you have to use Scanner.hasNext(). If there is a String you will read it, otherwise just return and print 0.

The solution success with all test case
package com.example;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
if (s.length() > 1 && s.length() < 400000) {
String textFiltered = s.trim();
if (textFiltered.length() > 0) {
String[] words = textFiltered.split("[!,?.*_'#\\ ]+");
int count = words.length;
System.out.println(count);
for (String word : words) {
System.out.println(word);
}
} else {
System.out.println(0);
}
} else {
System.out.println(0);
}
scan.close();
}
}

The issue is with .trim(), move it before s.length()
my working code
import java.util.*;
public class StringTokens {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
// Write your code here.
s = s.trim();
// if length is 0
if (s.length()>400000){
return ;
}else if (s.length()==0 || s == null){
System.out.println(0);
return ;
}else{
// It seems we need to remove some spaces
String[] words = s.split("[!,?.*_'#\\ ]+");
int count = words.length;
System.out.println(count);
for (String word : words){
System.out.println(word);
}
scan.close();
}
}

Need remove leading empty element in tokens array.
private static String[] getTokens(String s) {
if (s == null || s.isEmpty()) {
return new String[0];
}
String[] tokens = s.trim().split("[^a-zA-Z]+"); // or [^A-Za-z]+
if (tokens != null && tokens.length > 0 ) {
if (tokens[0].isEmpty()) {
return Arrays.copyOfRange(tokens, 1, tokens.length);
} else {
return tokens;
}
}
return new String[0];
}

how to capitalize first letter after period in each sentence using java?

I'm currently learning on how to manipulate strings and i think it'll take awhile for me to get used to it. I wanted to know how to capitalize a letter after a period in each sentence.
The output is like this:
Enter sentences: i am happy. this is genius.
Capitalized: I am happy. This is genius.
I have tried creating my own code but its not working, feel free to correct and change it. Here is my code:
package Test;
import java.util.Scanner;
public class TestMain {
public static void main(String[]args) {
String sentence = getSentence();
int position = sentence.indexOf(".");
while (position != -1) {
position = sentence.indexOf(".", position + 1);
sentence = Character.toUpperCase(sentence.charAt(position)) + sentence.substring(position + 1);
System.out.println("Capitalized: " + sentence);
}
}
public static String getSentence() {
Scanner hold = new Scanner(System.in);
String sent;
System.out.print("Enter sentences:");
sent = hold.nextLine();
return sent;
}
}
The tricky part is how am i gonna capitalize a letter after the period(".")? I don't have a lot of string manipulation knowledge so I'm really stuck in this area.

Try this:
package Test;
import java.util.Scanner;
public class TestMain {
public static void main(String[]args){
String sentence = getSentence();
StringBuilder result = new StringBuilder(sentence.length());
//First one is capital!
boolean capitalize = true;
//Go through all the characters in the sentence.
for(int i = 0; i < sentence.length(); i++) {
//Get current char
char c = sentence.charAt(i);
//If it's period then set next one to capital
if(c == '.') {
capitalize = true;
}
//If it's alphabetic character...
else if(capitalize && Character.isAlphabetic(c)) {
//...we turn it to uppercase
c = Character.toUpperCase(c);
//Don't capitalize next characters
capitalize = false;
}
//Accumulate in result
result.append(c);
}
System.out.println(result);
}
public static String getSentence(){
Scanner hold = new Scanner(System.in);
String sent;
System.out.print("Enter sentences:");
sent = hold.nextLine();
return sent;
}
}
What this is doing it advancing sequentially through all of the characters in the string and keeping state of when the next character needs to be capitalized.
Follow the comments for a deeper exaplanations.

You could implement a state machine:
It starts in the capitalize state, as each character is read it emits it and then decides what state to go to next.
As there are just two states, the state can be stored in a boolean.
public static String capitalizeSentence(String sentence) {
StringBuilder result = new StringBuilder();
boolean capitalize = true; //state
for(char c : sentence.toCharArray()) {
if (capitalize) {
//this is the capitalize state
result.append(Character.toUpperCase(c));
if (!Character.isWhitespace(c) && c != '.') {
capitalize = false; //change state
}
} else {
//this is the don't capitalize state
result.append(c);
if (c == '.') {
capitalize = true; //change state
}
}
}
return result.toString();
}

Here is solution with regular expressions:
public static void main(String[]args) {
String sentence = getSentence();
Pattern pattern = Pattern.compile("^\\W*([a-zA-Z])|\\.\\W*([a-zA-Z])");
Matcher matcher = pattern.matcher(sentence);
StringBuffer stringBuffer = new StringBuffer("Capitalized: ");
while (matcher.find()) {
matcher.appendReplacement(stringBuffer, matcher.group(0).toUpperCase());
}
matcher.appendTail(stringBuffer);
System.out.println(stringBuffer.toString());
}

Seems like your prof is repeating his assignments. This has already been asked:
Capitalize first word of a sentence in a string with multiple sentences
Use a pre-existing lib:
http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/text/WordUtils.html#capitalize(java.lang.String,%20char...)
and guava
public static void main(String[] args) {
String sentences = "i am happy. this is genius.";
Iterable<String> strings = Splitter.on('.').split(sentences);
List<String> capStrings = FluentIterable.from(strings)
.transform(new Function<String, String>()
{
#Override
public String apply(String input){
return WordUtils.capitalize(input);
}
}).toList();
System.out.println(Joiner.on('.').join(capStrings));
}

Just use
org.apache.commons.lang3.text.WordUtils.capitalizeFully(sentence);

You can use below code to capitalize first letter after period in
each sentence.
String input = "i am happy. this is genius.";
String arr[] = input.split("\\.");
for (int i = 0; i < arr.length; i++) {
System.out.print(Character.toUpperCase(arr[i].trim().
charAt(0)) + arr[i].trim().substring(1) + ". ");
}

I'd go for regex as it is fast to use:
Split your string by ".":
String[] split = input.split("\\.");
Then capitalize the first letter of the resulting substrings and reunite to result string. (Be careful for spaces between periods and letters, maybe split by "\. "):
String result = "";
for (int i=0; i < split.length; i++) {
result += Character.toUpperCase(split[i].trim());
}
System.out.println(result);
Should do it.

The correct method to do it with core java using regex will be
String sentence = "i am happy. this is genius.";
Pattern pattern = Pattern.compile("[^\\.]*\\.\\s*");
Matcher matcher = pattern.matcher(sentence);
String capitalized = "", match;
while(matcher.find()){
match = matcher.group();
capitalized += Character.toUpperCase(match.charAt(0)) + match.substring(1);
}
System.out.println(capitalized);

Try this:
1. Capitalize the first letter.
2. If the character is '.' set the flag true so that you can capitalize the next character.
public static String capitalizeSentence(String str)
{
if(str.length()>0)
{
char arr[] = str.toCharArray();
boolean flag = true;
for (int i = 0; i < str.length(); i++)
{
if (flag)
{
if (arr[i] >= 97 && arr[i] <= 122)
{
arr[i] = (char) (arr[i] - 32);
flag = false;
}
} else
{
if (arr[i] == '.')
flag = true;
}
}
return new String(arr);
}
return str;
}

Removing contiguous spaces in a String without trim() and replaceAll()

I have to remove leading and trailing spaces from the given string as well as combine the contiguous spaces. For example,
String str = " this is a string containing numerous whitespaces ";
and I need to return it as:
"this is a string containing numerous whitespaces";
But the problem is I can't use String#trim(). (This is a homework and I'm not allowed to use such methods.) I'm currently trying it by accessing each character one-by-one but quite unsuccessful.
I need an optimized code for this. Could anybody help? I need it to be done by today :(

EDIT: Answer posted before we were told we couldn't use replaceAll. I'm leaving it here on the grounds that it may well be useful to other readers, even if it's not useful to the OP.
I need an optimized code for this.
Do you really need it to be opimtized? Have you identified this as a bottleneck?
This should do it:
str = str.replaceAll("\\s+", " ");
That's a regular expression to say "replace any contintiguous whitespace with a single space". It may not be the fastest possible, but I'd benchmark it before trying anything else.
Note that this will replace all whitespace with spaces - so if you have tabs or other whitespace characters, they will be replaced with spaces too.

I'm not permitted to use these methods. I've to do this with loops
and all.
So i wrote for you some little snipet of code if you can't use faster and more efficient way:
String str = " this is a string containing numerous whitespaces ";
StringBuffer buff = new StringBuffer();
String correctedString = "";
boolean space = false;
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if (c == ' ') {
if (!space && i > 0) {
buff.append(c);
}
space = true;
}
else {
buff.append(c);
space = false;
}
}
String temp = buff.toString();
if (temp.charAt(temp.length() - 1) == ' ') {
correctedString = temp.substring(0, buff.toString().length() - 1);
System.out.println(correctedString);
}
System.out.println(buff.toString())
Note:
But this is "harcoded" and only for "learning".
More efficient way is for sure use approaches pointed out by #JonSkeet and #BrunoReis

What about str = str.replaceAll(" +", " ").trim();?
If you don't want to use trim() (and I really don't see a reason not to), replace it with:
str = str.replaceAll(" +", " ").replaceAll("^ ", "").replaceAll(" $", "");`

Remove White Spaces without Using any inbuilt library Function
this is just a simple example with fixed array size.
public class RemWhite{
public static void main(String args[]){
String s1=" world qwer ";
int count=0;
char q[]=new char[9];
char ch[]=s1.toCharArray();
System.out.println(ch);
for(int i=0;i<=ch.length-1;i++)
{
int j=ch[i];
if(j==32)
{
continue;
}
else
q[count]=ch[i];
count++;
}
System.out.println(q);
}}

To remove single or re-occurrence of space.
public class RemoveSpace {
public static void main(String[] args) {
char space = ' ';
int ascii = (int) space;
String str = " this is a string containing numerous whitespaces ";
char c[] = str.toCharArray();
for (int i = 0; i < c.length - 1; i++) {
if (c[i] == ascii) {
continue;
} else {
System.out.print(c[i]);
}
}
}
}

If you don't want to use any inbuilt methods here's what you refer
private static String trim(String s)
{
String s1="";boolean nonspace=false;
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)!=' ' || nonspace)
{
s1 = s1+s.charAt(i);
nonspace = true;
}
}
nonspace = false;
s="";
for(int i=s1.length()-1;i>=0;i--)
{
if(s1.charAt(i)!=' ' || nonspace)
{
s = s1.charAt(i)+s;
nonspace = true;
}
}
return s;
}

package removespace;
import java.util.Scanner;
public class RemoveSpace {
public static void main(String[] args) {
Scanner scan= new Scanner(System.in);
System.out.println("Enter the string");
String str= scan.nextLine();
String str2=" ";
char []arr=str.toCharArray();
int i=0;
while(i<=arr.length-1)
{
if(arr[i]==' ')
{
i++;
}
else
{
str2= str2+arr[i];
i++;
}
}
System.out.println(str2);
}
}

This code is used for removing the white spaces and re-occurrence of alphabets in the given string,without using trim(). We accept a string from user. We separate it in characters by using charAt() then we compare each character with null(' '). If null is found we skip it and display that character in the else part. For skipping the null we increment the index i by 1.
try this code to get the solution of your problem.
String name = " abc ";
System.out.println(name);
for (int i = 0; i < name.length(); i++) {
char ch = name.charAt(i);
if (ch == ' ') {
i = 2 + i - 2;
} else {
System.out.print(name.charAt(i));
}
}

Check if String contains only letters

The idea is to have a String read and to verify that it does not contain any numeric characters. So something like "smith23" would not be acceptable.

What do you want? Speed or simplicity? For speed, go for a loop based approach. For simplicity, go for a one liner RegEx based approach.
Speed
public boolean isAlpha(String name) {
char[] chars = name.toCharArray();
for (char c : chars) {
if(!Character.isLetter(c)) {
return false;
}
}
return true;
}
Simplicity
public boolean isAlpha(String name) {
return name.matches("[a-zA-Z]+");
}

Java 8 lambda expressions. Both fast and simple.
boolean allLetters = someString.chars().allMatch(Character::isLetter);

Or if you are using Apache Commons, [StringUtils.isAlpha()].

First import Pattern :
import java.util.regex.Pattern;
Then use this simple code:
String s = "smith23";
if (Pattern.matches("[a-zA-Z]+",s)) {
// Do something
System.out.println("Yes, string contains letters only");
}else{
System.out.println("Nope, Other characters detected");
}
This will output:
Nope, Other characters detected

I used this regex expression (".*[a-zA-Z]+.*"). With if not statement it will avoid all expressions that have a letter before, at the end or between any type of other character.
String strWithLetters = "123AZ456";
if(! Pattern.matches(".*[a-zA-Z]+.*", str1))
return true;
else return false

A quick way to do it is by:
public boolean isStringAlpha(String aString) {
int charCount = 0;
String alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
if (aString.length() == 0) {
return false; //zero length string ain't alpha
}
for (int i = 0; i < aString.length(); i++) {
for (int j = 0; j < alphabet.length(); j++) {
if (aString.substring(i, i + 1).equals(alphabet.substring(j, j + 1))
|| aString.substring(i, i + 1).equals(alphabet.substring(j, j + 1).toLowerCase())) {
charCount++;
}
}
if (charCount != (i + 1)) {
System.out.println("\n**Invalid input! Enter alpha values**\n");
return false;
}
}
return true;
}
Because you don't have to run the whole aString to check if it isn't an alpha String.

private boolean isOnlyLetters(String s){
char c=' ';
boolean isGood=false, safe=isGood;
int failCount=0;
for(int i=0;i<s.length();i++){
c = s.charAt(i);
if(Character.isLetter(c))
isGood=true;
else{
isGood=false;
failCount+=1;
}
}
if(failCount==0 && s.length()>0)
safe=true;
else
safe=false;
return safe;
}
I know it's a bit crowded. I was using it with my program and felt the desire to share it with people. It can tell if any character in a string is not a letter or not. Use it if you want something easy to clarify and look back on.

Faster way is below. Considering letters are only a-z,A-Z.
public static void main( String[] args ){
System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
}
public static boolean bettertWay(String name) {
char[] chars = name.toCharArray();
long startTimeOne = System.nanoTime();
for(char c : chars){
if(!(c>=65 && c<=90)&&!(c>=97 && c<=122) ){
System.out.println(System.nanoTime() - startTimeOne);
return false;
}
}
System.out.println(System.nanoTime() - startTimeOne);
return true;
}
public static boolean isAlpha(String name) {
char[] chars = name.toCharArray();
long startTimeOne = System.nanoTime();
for (char c : chars) {
if(!Character.isLetter(c)) {
System.out.println(System.nanoTime() - startTimeOne);
return false;
}
}
System.out.println(System.nanoTime() - startTimeOne);
return true;
}
Runtime is calculated in nano seconds. It may vary system to system.
5748//bettertWay without numbers
true
89493 //isAlpha without numbers
true
3284 //bettertWay with numbers
false
22989 //isAlpha with numbers
false

Check this,i guess this is help you because it's work in my project so once you check this code
if(! Pattern.matches(".*[a-zA-Z]+.*[a-zA-Z]", str1))
{
String not contain only character;
}
else
{
String contain only character;
}

String expression = "^[a-zA-Z]*$";
CharSequence inputStr = str;
Pattern pattern = Pattern.compile(expression);
Matcher matcher = pattern.matcher(inputStr);
if(matcher.matches())
{
//if pattern matches
}
else
{
//if pattern does not matches
}

Try using regular expressions: String.matches

public boolean isAlpha(String name)
{
String s=name.toLowerCase();
for(int i=0; i<s.length();i++)
{
if((s.charAt(i)>='a' && s.charAt(i)<='z'))
{
continue;
}
else
{
return false;
}
}
return true;
}

Feels as if our need is to find whether the character are only alphabets.
Here's how you can solve it-
Character.isAlphabetic(c)
helps to check if the characters of the string are alphabets or not.
where c is
char c = s.charAt(elementIndex);

While there are many ways to skin this cat, I prefer to wrap such code into reusable extension methods that make it trivial to do going forward. When using extension methods, you can also avoid RegEx as it is slower than a direct character check. I like using the extensions in the Extensions.cs NuGet package. It makes this check as simple as:
Add the https://www.nuget.org/packages/Extensions.cs package to your project.
Add "using Extensions;" to the top of your code.
"smith23".IsAlphabetic() will return False whereas "john smith".IsAlphabetic() will return True. By default the .IsAlphabetic() method ignores spaces, but it can also be overridden such that "john smith".IsAlphabetic(false) will return False since the space is not considered part of the alphabet.
Every other check in the rest of the code is simply MyString.IsAlphabetic().

To allow only ASCII letters, the character class \p{Alpha} can be used. (This is equivalent to [\p{Lower}\p{Upper}] or [a-zA-Z].)
boolean allLettersASCII = str.matches("\\p{Alpha}*");
For allowing all Unicode letters, use the character class \p{L} (or equivalently, \p{IsL}).
boolean allLettersUnicode = str.matches("\\p{L}*");
See the Pattern documentation.

I found an easy of way of checking a string whether all its digit is letter or not.
public static boolean isStringLetter(String input) {
boolean b = false;
for (int id = 0; id < input.length(); id++) {
if ('a' <= input.charAt(id) && input.charAt(id) <= 'z') {
b = true;
} else if ('A' <= input.charAt(id) && input.charAt(id) <= 'Z') {
b = true;
} else {
b = false;
}
}
return b;
}
I hope it could help anyone who is looking for such method.

Use StringUtils.isAlpha() method and it will make your life simple.