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Alphanumeric Caesar's Cipher in Java

I'm trying to solve Caesar's Cipher in Java but there's a twist to it. The input string has alphanumeric values and I am unable to solve. Here's what I've attempted so far:
String rotationalCipher(String input, int rotationFactor) {
// Write your code here
StringBuilder sb = new StringBuilder();
for(int i = 0; i < input.length(); i++) {
if(Character.isLowerCase(input.charAt(i))) {
char ch = (char)(((int)input.charAt(i) + rotationFactor - 97) % 26 + 97);
} else if (Character.isUpperCase(input.charAt(i))) {
char ch = (char)(((int)input.charAt(i) + rotationFactor - 65) % 26 + 65);
sb.append(ch);
} else {
char ch = (char)(((int)input.charAt(i) + rotationFactor - 48) % 10 + 48);
sb.append(ch);
}
}
return sb.toString();
}
What I'm trying to do is evaluate each case using its ASCII values but I don't seem to get the desired output. Am I using ASCII wrong? Thanks for your help!
Sample input/output:
input = Zebra-493?
rotationFactor = 3
output = Cheud-726?

You have two major problems.
You did not update StringBuilder with an append for lowercase transitions.
You need to handle digits specially using isDigit just like upper and lower case so that you can then ignore characters like - and ?
A couple of suggestions.
just assign ch when you first enter the loop and then use it throughout the loop. No need to keep typing in all the input stuff.
only append ch to the StringBuilder once near the end when you exit the if/else blocks.
Instead of numbers like 97 and 65 use 'a' and 'A'. Less likely to make mistakes that way.
Once you make those changes, your code works just fine.

Below code works for me-
String rotationalCipher(String input, int rotationFactor) {
// Write your code here
StringBuffer sb = new StringBuffer();
for (int i = 0; i < input.length(); i++) {
char x = input.charAt(i);
if (Character.isLowerCase(x)) {
char ch = (char) ((x + rotationFactor - 97) % 26 + 97);
sb.append(ch);
} else if (Character.isUpperCase(x)) {
char ch = (char) ((x + rotationFactor - 65) % 26 + 65);
sb.append(ch);
} else if (Character.isDigit(x)) {
char ch = (char) ((x + rotationFactor - 48) % 10 + 48);
sb.append(ch);
} else {
sb.append(x);
}
}
return sb.toString();
}

String rotationalCipher(String input, int rotationFactor) {
String output = "";
for(char a: input.toCharArray()){
if(Character.isAlphabetic(a)){
char startLetter = Character.isUpperCase(a) ? 'A' : 'a';
output += (char) ((a- startLetter + rotationFactor) % 26 + startLetter);
}
else if(Character.isDigit(a)){
output += (char) ((a + rotationFactor - 48) % 10 + 48);
}
else{
output += a;
}
}
return output;}

How to find out which char in string is a number?

How to find out if the char in string is a letter or a number?
I.e I have a string "abc2e4", I need to find the ints, square them, and put the answer back in the string (no extra operations with the letters), so the new string would be "abc4e16".
Im incredibly lost with this exercise, so any help would be great :D

You can do it using Regular Expression
public static String update(String str) {
final Pattern pattern = Pattern.compile("\\D+|\\d+");
final Matcher matcher = pattern.matcher(str);
StringBuilder buf = new StringBuilder();
int pos = 0;
while (matcher.find(pos)) {
str = matcher.group();
buf.append(Character.isDigit(str.charAt(0)) ? (int)Math.pow(Integer.parseInt(str), 2) : str);
pos = matcher.end();
}
return buf.toString();
}

Java provides a method to check whether a character is a digit. For this you can use Character.isDigit(char).
public static String squareNumbers(String input) {
StringBuilder output = new StringBuilder();
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i); // get char at index
if (Character.isDigit(c)) // check if the char is a digit between 0-9
output.append((int) Math.pow(Character.digit(c, 10), 2)); // square the numerical value
else
output.append(c); // keep if not a digit
}
return output.toString();
}
This will iterate any passed string character by character and square each digit it finds. If for example 2 digits are right next to each other they will be seen as individual numbers and squared each and not as one number with multiple digits.
squareNumbers("10") -> "10"
squareNumbers("12") -> "14"
squareNumbers("abc2e4") -> "abc4e16"

My logic only squares single digit numbers.
For eg - if you provide input he13llo, the output would be he19llo and not he169llo.
Scanner in = new Scanner(System.in) ;
String str = in.next() ;
String ans = str ;
for (int i = 0 ; i < str.length() ; i++)
{
char ch = str.charAt(i) ;
if((ch - '0' >= 0) && (ch - '9' <= 0))
{
int index = i ;
int num = ch - '0' ;
int square = num * num ;
ans = ans.substring(0 ,index) + square + ans.substring(index+1) ;
}
}
System.out.println(ans) ;
}

how to distinguish Unicode characters and ASCII characters

I want to distinguish Unicode characters and ASCII characters from the below string:
abc\u263A\uD83D\uDE0A\uD83D\uDE22123
How can I distinguish characters? Can anyone help me with this issue? I have tried some code, but it crashes in some cases. What is wrong with my code?
The first three characters are abc, and the last three characters are 123. The rest of the string is Unicode characters. I want to make a string array like this:
str[0] = 'a';
str[1] = 'b';
str[2] = 'c';
str[3] = '\u263A\uD83D';
str[4] = '\uDE0A\uD83D';
str[5] = '\uDE22';
str[6] = '1';
str[7] = '2';
str[8] = '3';
Code:
private String[] getCharArray(String unicodeStr) {
ArrayList<String> list = new ArrayList<>();
for (int i = 0; i < unicodeStr.length(); i++) {
if (unicodeStr.charAt(i) == '\\') {
list.add(unicodeStr.substring(i, i + 11));
i = i + 11;
} else {
list.add(String.valueOf(unicodeStr.charAt(i)));
}
}
return list.toArray(new String[list.size()]);
}

ASCII characters exist in Unicode, they are Unicode codepoints U+0000 - U+007F, inclusive.
Java strings are represented in UTF-16, which is a 16-bit byte encoding of Unicode. Each Java char is a UTF-16 code unit. Unicode codepoints U+0000 - U+FFFF use 1 UTF-16 code unit and thus fit in a single char, whereas Unicode codepoints U+10000 and higher require a UTF-16 surrogate pair and thus need two chars.
If the string has UTF-16 code units represented as actual char values, then you can use Java's string methods that work with codepoints, eg:
private String[] getCharArray(String unicodeStr) {
ArrayList<String> list = new ArrayList<>();
int i = 0, j;
while (i < unicodeStr.length()) {
j = unicodeStr.offsetByCodePoints(i, 1);
list.add(unicodeStr.substring(i, j));
i = j;
}
return list.toArray(new String[list.size()]);
}
On the other hand, if the string has UTF-16 code units represented in an encoded "\uXXXX" format (ie, as 6 distinct characters - '\', 'u', ...), then things get a little more complicated as you have to parse the encoded sequences manually.
If you want to preserve the "\uXXXX" strings in your array, you could do something like this:
private boolean isUnicodeEncoded(string s, int index)
{
return (
(s.charAt(index) == '\\') &&
((index+5) < s.length()) &&
(s.charAt(index+1) == 'u')
);
}
private String[] getCharArray(String unicodeStr) {
ArrayList<String> list = new ArrayList<>();
int i = 0, j, start;
char ch;
while (i < unicodeStr.length()) {
start = i;
if (isUnicodeEncoded(unicodeStr, i)) {
ch = (char) Integer.parseInt(unicodeStr.substring(i+2, i+6), 16);
j = 6;
}
else {
ch = unicodeStr.charAt(i);
j = 1;
}
i += j;
if (Character.isHighSurrogate(ch) && (i < unicodeStr.length())) {
if (isUnicodeEncoded(unicodeStr, i)) {
ch = (char) Integer.parseInt(unicodeStr.substring(i+2, i+6), 16);
j = 6;
}
else {
ch = unicodeStr.charAt(i);
j = 1;
}
if (Character.isLowSurrogate(ch)) {
i += j;
}
}
list.add(unicodeStr.substring(start, i));
}
return list.toArray(new String[list.size()]);
}
If you want to decode the "\uXXXX" strings into actual chars in your array, you could do something like this instead:
private boolean isUnicodeEncoded(string s, int index)
{
return (
(s.charAt(index) == '\\') &&
((index+5) < s.length()) &&
(s.charAt(index+1) == 'u')
);
}
private String[] getCharArray(String unicodeStr) {
ArrayList<String> list = new ArrayList<>();
int i = 0, j;
char ch1, ch2;
while (i < unicodeStr.length()) {
if (isUnicodeEncoded(unicodeStr, i)) {
ch1 = (char) Integer.parseInt(unicodeStr.substring(i+2, i+6), 16);
j = 6;
}
else {
ch1 = unicodeStr.charAt(i);
j = 1;
}
i += j;
if (Character.isHighSurrogate(ch1) && (i < unicodeStr.length())) {
if (isUnicodeEncoded(unicodeStr, i)) {
ch2 = (char) Integer.parseInt(unicodeStr.substring(i+2, i+6), 16);
j = 6;
}
else {
ch2 = unicodeStr.charAt(i);
j = 1;
}
if (Character.isLowSurrogate(ch2)) {
list.add(String.valueOf(new char[]{ch1, ch2}));
i += j;
continue;
}
}
list.add(String.valueOf(ch1));
}
return list.toArray(new String[list.size()]);
}
Or, something like this (per https://stackoverflow.com/a/24046962/65863):
private String[] getCharArray(String unicodeStr) {
Properties p = new Properties();
p.load(new StringReader("key="+unicodeStr));
unicodeStr = p.getProperty("key");
ArrayList<String> list = new ArrayList<>();
int i = 0;
while (i < unicodeStr.length()) {
if (Character.isHighSurrogate(unicodeStr.charAt(i)) &&
((i+1) < unicodeStr.length()) &&
Character.isLowSurrogate(unicodeStr.charAt(i+1)))
{
list.add(unicodeStr.substring(i, i+2));
i += 2;
}
else {
list.add(unicodeStr.substring(i, i+1));
++i;
}
}
return list.toArray(new String[list.size()]);
}

It's not entirely clear what you're asking for, but if you want to tell if a specific character is ASCII, you can use Guava's ChatMatcher.ascii().
if ( CharMatcher.ascii().matches('a') ) {
System.out.println("'a' is ascii");
}
if ( CharMatcher.ascii().matches('\u263A\uD83D') ) {
// this shouldn't be printed
System.out.println("'\u263A\uD83D' is ascii");
}

String to Unicode in Java

I have a large string I need to convert all the non alphanumeric chars to unicode
For example
Input string : abc12/dad-das/das_sdj
Output String : abc12:002Fdad:002Ddas:002Fdas:002Fsdj
Currently I am using this function
for (char c : str.toCharArray()) {
System.out.printf(":%04X \n", (int) c);
}
Is there a better way to do it ?

Here are two ways to do it:
// Looping over string characters
private static String convert(String input) {
StringBuilder buf = new StringBuilder(input.length() + 16);
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9'))
buf.append(c);
else
buf.append(String.format(":%04X", (int) c));
}
return buf.toString();
}
// Using regular expression
private static String convert(String input) {
StringBuffer buf = new StringBuffer(input.length() + 16);
Matcher m = Pattern.compile("[^a-zA-Z0-9]").matcher(input);
while (m.find())
m.appendReplacement(buf, String.format(":%04X", (int) m.group().charAt(0)));
return m.appendTail(buf).toString();
}
Test
System.out.println(convert("abc12/dad-das/das_sdj"));
Output
abc12:002Fdad:002Ddas:002Fdas:005Fsdj

How do you convert integer to character in Java? (5 --> '5')

String word = "ABCD";
StringBuffer str = new StringBuffer (word);
int counter = 0;
for (int ch = 0; ch < word.length(); ch ++)
{
int number = word.charAt(ch)- 'A' + 1;
str.setCharAt(counter, (char) number);
if (ch != word.length ()-1)
str.insert(counter +1, '-');
counter += 2;
}
System.out.println (str);
}
I want my output to be 1-2-3-4, so A = 1 and B= 2.... etc. We can assume that all the input are in upper case. But the my code produce random symbols. So how do I fix the code to produce 1-2-3-4 without re writing the whole thing?

You're only missing the conversion from int back to char. The line
str.setCharAt(counter, (char) number);
should be
str.setCharAt(counter, (char) ('0' + number));