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Surrounded Character Method Java

I'm working on a java method that checks whether a character in an array of characters is surrounded by a character. Ex: abcdc, d is surrounded by c. Ex: abccc, has no letters that are surrounded. Here is what I have so far.
public static boolean surroundedCharacter(char[] letters){
boolean result = false;
for(char letter : letters)
if(letters[letter-1] == letters[letter+1]){
result = true;
}
return result;
} }
So I basically have a for each loop going through letter in letters and checks whether the letter before the position is equal to letter after the position. If it is, it means that the letter is surrounded and it should change result to true. The junit test says that the if statement is wrong, but I don't know how to fix it. Any help is appreciated.

You have to use an Integer for the index:
public static boolean surroundedCharacter(char[] letters){
boolean result = false;
for(int i = 1; i < letters.length - 1; i++) {
// You said that if the string is "abccc", should return false.
// So, we check if the previous or the next letter is different to
//the actual value of i
if((letters[i-1] == letters[i+1]) && (letters[i-1] != letters[i])) {
result = true;
}
}
return result;
}

Try this one:
public static boolean surroundedCharacter(char[] letters){
boolean result = false;
for( int i=1;i<letters.length-1;i++)
if(letters[i-1] == letters[i+1]){
result = true;
}
}
return result;}

You must use for loop instead of short for. Try this for loop:
for(int i = 1; i< letters.length-1; i++){
if(letters[i-1] == letters[i+1]){
result = true;
}
}

A Java foreach is designed to iterate element after element.
In case of need to get two distinct element at each iteration, you should use a classic for using a int value.
Besides, you don't need to have intermediary variable. When the condition is matched you can return true. Otherwise you return false after the loop.
At last, according to your needs :
Ex: abccc, has no letters that are surrounded.
you should accept the match only if the surrounded char differs from the chars that surround it.
public static boolean surroundedCharacter(char[] letters){
for(int i=1; i<letters.length-1; i++){
var beforeLetter = letters[i-1];
var afterLetter = letters[i+1];
if(beforeLetter == afterLetter && beforeLetter != letters[i]){
return true;
}
}
return false;
}

Java - Boolean method always returning as false

public static boolean isValidNumber(String a1)
{
String x = ("0123456789");
boolean valid = false;
for (int i = 0; i < 4; i++) {
char c = a1.charAt(i);
for (int j = 0; j < 10; j++) {
if ( c == x.charAt(j)) {
valid = true;
}
else {
valid = false;
}
}
}
return valid;
}
The above method checks to see whether an input of a four character string is composed of the characters 0123456789. However, regardless of what the input is, the method always returns as false.
If I were to change the valid value in the else statement to true, the method would always return as true.
What is the error that I have made in this method?

As soon as you find a non matching character, break the loop otherwise the next matching character will set valid to true.
e.g. "123a456" is considered valid.
for (int j = 0; j < 10; j++) {
if ( c == x.charAt(j)) {
valid = true;
}
else {
valid = false;
break;
}
}
If for some reason you don't want to break the loop, you could keep an "invalid counter" and make sure that is 0 at the end.
Of course for what you are doing here, Integer.parseInt() might be your best bet ;-)

a String.equals method will check these two strings in a single statement if you are permitted to use that.
public static boolean isValidNumber(String a1)
{
String x = ("0123456789");
return x.equals(a1);
}

I would rewrite your function as given below,
String x = ("0123456789");
boolean valid = false;
for (int i = 0; i < 4; i++) {
char c = a1.charAt(i);
boolean isCharOK = false;
for (int j = 0; j < 10; j++) {
if ( c == x.charAt(j)) {
isCharOK = true;
break;
}
}
if (!isCharOK) {
valid = false;
break;
}
}
return valid;

John3136 is quite correct, but I would like to propose even better solution to your whole task:
final static String x = "0123456789";
public static boolean isValidNumber(String a1) {
for (int i = 0; i < a1.length(); ++i) {
if (x.indexOf(a1.charAt(i)) == -1) return false;
}
return true;
}
In short: the above code "looks up" every character in your parameter string a1 in the string composed of digits. If it can find it, continues. If it can't, it means a1 consist not only digits and returns false. If it passes through all a1 characters then it returns true :)
And as asked and described in the comments - handling of duplicate characters in argument string:
final static String x = "0123456789";
public static boolean isValidNumber(String a1) {
for (int i = 0; i < a1.length(); ++i) {
final char currentChar = a1.charAt(i);
if (x.indexOf(currentChar) == -1 || a1.indexOf(currentChar, i+1) != -1)
return false;
}
return true;
}
The function call a1.indexOf(currentChar, i+1) essentially checks if there is any duplicate character in the rest of the string (from position i+1 and farther). Which means if it will be able to find duplicate char, the method return false :) Hope this helps, here is more info on String.indexOf(int, int) function if you want:
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf(int, int)

You can use this one liner function to check for validity of a String as Number using Regular Expression
public static boolean isValidNumber(String a1)
{
return a1.matches("[\\d]+");
}
Hope this helps.

Looking Up Quotation marks with a array of Strings

So I have the following String: dog "and" cat
I split it up into an array named: array1 with,
{'d','o','g',' ','"','c','a',t','"'}
boolean works = false;
for (int i=0; i < array1.length;i++){
if (array1[i].equals("d"){
if (array1[i+1].equals("o"){
if(array1[i+2].equals("g"){
if (array1[i+3].equals(" "){
if (array1[i+4].equals("""){ //does work here
if (array1[i+5].equals("c"){
if (array1[i+6].equals("a"){
if (array1[i+7].equals("t"){
works = true;
}
}
}
}
}
}
}
}
}
System.out.println(works);
It doesnt work at the equals with quotations. Does anyone have any ideas?

I would try to simplify your code, for example you could write
String s = "dog \"and\" cat";
boolean works = s.contains("dog \"cat");
This makes it much more obvious that works will always be false.

You have ambiguous data types.
I assume that your array1 is of type char[]. If it is, then you should have == comparators in your code (note the single quotes 'x' where x is the char you are testing):
if (array1[i] == 'd'){
....
}
if the arrays are of type String[], then you need to escape the " character using a backslash in the comparison:
if (array1[i+4].equals("\""){ //does work here
....
}

{'d','o','g',' ','"','c','a',t','"'} this is a char array
you need below code
You can no equal Char with String.
boolean works = false;
for (int i=0; i < array1.length;i++){
if (array1[i]=='d'){
if (array1[i+1]=='o'){
if(array1[i+2]=='g'){
if (array1[i+3]==' '){
if (array1[i+4]=='"'){ //does work here
if (array1[i+5]=='c'){
if (array1[i+6]=='a'){
if (array1[i+7]=='t'){
works = true;`
}
}
}
}
}
}
}
}

boolean works = false;
String[] pattern = { "d","o","g"," ","\"","c","a","t","\"" };
for (int i = 0; i < array1.length; i++) {
// Loop over the items in the pattern array, and check if they match array1
boolean inner = true;
for (int j = 0; j < pattern.length; j++) {
if (i + j >= array1.length) {
// Don't go beyond the end of array1
break;
}
if (!pattern[j].equals(array1[i+j])) {
// We found an item that doesn't match.
inner = false;
break;
}
}
if (inner) {
// All items matched
works = true;
break;
}
}
System.out.println(works);

You need to escape the " character inside the "". Like this:
(array1[i+4].equals("\"")

check password for digits and letters

I have problem with two of my methods for password validation.
The method hasDigitsAndLetters is supposed to check whether all the characters of the String are digits and letters and the second method hasTwoDigits is supposed to check whether there are at least two digits in the pass, but the problem is that for expected result true they are ruturning false. If someone can help. here is the code.
//check if the whole string consists of digits and letters
public static boolean hasDigitsAndLetters(String pass)
{
for(int i=0; i<pass.length(); i++)
{
if(!Character.isLetterOrDigit((i)))
{
return false;
}
}
return true;
}
// check whether the password has at least 2 digits
public static boolean hasTwoDigits(String pass)
{
int counter = 0;
for(int i=0; i<pass.length(); i++)
{
if(Character.isDigit(i))
{
counter ++;
}
}
System.out.println("Number of digits: " + counter);
if(counter >= 2)
{
return true;
}
return false;
}

You need to pass character at position i for that String.
Character.isLetterOrDigit((pass.charAt(i)))
same for digit also
Character.isDigit((pass.charAt(i)))

You want to check the character in the string at index i, not the index variable itself:
Character.isLetterOrDigit(pass.charAt(i))

You aren't checking against characters in your pass, you need to change your checks to:
if(!Character.isLetterOrDigit((pass.charAt(i)))
and
if(Character.isDigit(pass.charAt(i)))

Right now you are checking if i is a digit or letter and i is an int. You need to check the character at position i.
if(Character.isDigit(pass.charAt(i)))

The error is that you're comparing the position into the string rather than the character at that position in the string. I'd probably not use charAt, however... there's no point in keeping explicit management of the position here. I suggest you use String.toCharArray instead.
public static boolean isAlphanumeric(final String str) {
for (char c : str.toCharArray()) {
if (!Character.isLetterOrDigit(c)) {
return false;
}
}
return true;
}
public static boolean isBidigital(final String str) {
int n = 0;
for (char c : str.toCharArray()) {
if (Character.isDigit(c)) {
++n;
}
}
return n >= 2;
}

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.