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Decode String in Java

I am trying to convert this Python Solution in Java. For some reason, my Java Solution is not working. How can this be done correctly?
https://leetcode.com/problems/decode-string/description/
Given an encoded string, return its decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.
You may assume that the input string is always valid; there are no extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there will not be input like 3a or 2[4].
The test cases are generated so that the length of the output will never exceed 105.
Example 1:
Input: s = "3[a]2[bc]"
Output: "aaabcbc"
Example 2:
Input: s = "3[a2[c]]"
Output: "accaccacc"
Python Solution:
class Solution:
def decodeString(self, s: str) -> str:
stack = []
for char in s:
if char is not "]":
stack.append(char)
else:
sub_str = ""
while stack[-1] is not "[":
sub_str = stack.pop() + sub_str
stack.pop()
multiplier = ""
while stack and stack[-1].isdigit():
multiplier = stack.pop() + multiplier
stack.append(int(multiplier) * sub_str)
return "".join(stack)
Java Attempt:
class Solution {
public String decodeString(String s) {
Deque<String> list = new ArrayDeque<String>();
String subword = "";
String number = "";
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != ']' ) {
list.add(String.valueOf(s.charAt(i)));
}
else {
subword = "";
while (list.size() > 0 && !list.getLast().equals("[") ) {
subword = list.pop() + subword;
}
if (list.size() > 0) list.pop();
number = "";
while (list.size() > 0 && isNumeric(list.getLast())){
number = list.pop() + number;
}
for (int j = 1; (isNumeric(number) && j <= Integer.parseInt(number)); j++) list.add(subword);
}
}
return String.join("", list);
}
public static boolean isNumeric(String str) {
try {
Double.parseDouble(str);
return true;
} catch(NumberFormatException e){
return false;
}
}
}

The reason why your posted code is not working is because the pop() method in python removes the last element by default.
But in Java, the ArrayDeque class's pop() method removes the first element.
In order to emulate the python code with the ArrayDeque, you'll need to use the removeLast() method of the ArrayDeque instance instead.

public class Solution{
public static String decodeString(String s) {
StringBuilder stack = new StringBuilder();
for(char c : s.toCharArray()) {
if(c != ']') {
stack.append(c);
} else {
StringBuilder sub_str = new StringBuilder();
while(stack.charAt(stack.length() - 1) != '[') {
sub_str.insert(0, stack.charAt(stack.length() - 1));
stack.deleteCharAt(stack.length() - 1);
}
stack.deleteCharAt(stack.length() - 1);
StringBuilder multiplier = new StringBuilder();
while(stack.length() > 0 && Character.isDigit(stack.charAt(stack.length() - 1))) {
multiplier.insert(0, stack.charAt(stack.length() - 1));
stack.deleteCharAt(stack.length() - 1);
}
for(int i = 0; i < Integer.parseInt(multiplier.toString()); i++) {
stack.append(sub_str);
}
}
}
return stack.toString();
}
public static void main(String[] args) {
System.out.println( decodeString("3[a2[c]]"));
//Output: "accaccacc"
System.out.println( decodeString("3[a]2[bc]"));
//Output: "aaabcbc"
}
}

Java String sentence - first letter uppercased, second lovercased and so on

Here is a description:
"Write a program that, given an input sentence, alternates the case of every alphabetic character, starting with uppercase. Spaces and non-alphabetical characters should be added to the final output as is, i.e. they should not be taken into account when alternating between upper/lowercase."
Here is what I've tried and does not work (System.out.println in main method should return correct sentence):
public class Main {
public static void main(String[] args) throws IOException {
InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);
BufferedReader in = new BufferedReader(reader);
String line;
while ((line = in.readLine()) != null) {
System.out.println(changeToUppercaseOrLowercase(countLettersWithSpaces(line), line));
}
}
private static int countLettersWithSpaces(String sentence) {
int count = 0;
for (int i = 0; i < sentence.length(); i ++)
{
char c = Character.toUpperCase(sentence.charAt(i));
if (c >= 'A' && c <= 'Z' || c == ' ' )
count ++;
}
return count;
}
private static String changeToUppercaseOrLowercase(int countLetters, String sentence) {
StringBuilder stringBuilder = new StringBuilder();
for(int i=0; i<countLetters; i++) {
if (!sentence.substring(i,i+1).equals(" ")) {
if ((i % 2) == 0) {
stringBuilder.append(sentence.substring(i,i+1).toUpperCase());
}
else {
stringBuilder.append(sentence.substring(i,i+1).toLowerCase());
}
}
if (sentence.substring(i,i+1).equals(" ")) {
stringBuilder.append(" ");
i++;
}
}
return stringBuilder.toString();
}
}
But tests says that:
Input data:
We are the world
Expected result:
We ArE tHe WoRlD
Result:
We Re He OrLd
How to solve that? Thank you in advance!

You can use Character.isAlphabetic and keep a counter that is incremented each time a letter is encountered.
public static String alternateCase(String str){
int count = 0;
StringBuilder sb = new StringBuilder(str.length());
for(int i = 0; i < str.length(); i++){
char c = str.charAt(i);
if(Character.isAlphabetic(c))
sb.append(++count % 2 == 1 ? Character.toUpperCase(c) : Character.toLowerCase(c));
else sb.append(c);
}
return sb.toString();
}

use Character.isLetter() function to check if it's a letter or not. half your problem will be solved.
and your problem description and test case doesnt go with each other. Please try to clarify more.

There are many ways to fix this. This one has minimal impact on your existing code.
Use an evenOdd counter to ensure you are not skipping over characters but still maintaining the alternation.
private static String changeToUppercaseOrLowercase(int countLetters, String sentence) {
StringBuilder stringBuilder = new StringBuilder();
int evenOdd = 0; // init ********HERE*******
for(int i=0; i<countLetters; i++) {
if (!sentence.substring(i,i+1).equals(" ")) {
if ((evenOdd % 2) == 0) { // check ********HERE*******
stringBuilder.append(sentence.substring(i,i+1).toUpperCase());
}
else {
stringBuilder.append(sentence.substring(i,i+1).toLowerCase());
}
}
if (sentence.substring(i,i+1).equals(" ")) {
stringBuilder.append(" ");
evenOdd--; // adjust to preserve proper alternation ********HERE*********
}
evenOdd++; // the normal update ********HERE*******
}
return stringBuilder.toString();
}

Java Programming: Replace all but first and last letters of each word with "_"

The purpose of this method is replace all but the first and last letters of each word with "_". I'm a complete novice when it comes to coding, so I'm certain my code is fairly incorrect. I think where my code starts functioning improperly is with the while loop.
EDIT: How do I make this method without using arrays or extra methods, like the split method?
public static String blankWords(String s1) {
StringBuilder sb = new StringBuilder();
if(s1.length() > 2) {
sb.append(s1.charAt(0));
for(int x = 1; x < s1.length() - 1; x = x + 1) {
char y = ' ';
while(y != s1.charAt(x)) {
sb.append("_");
x = x + 1;
}
}
sb.append(s1.charAt(s1.length() - 1));
return sb.toString();
}
return s1;
}
What my code is outputting:
HW2.blankWords("This is a Test.")
java.lang.StringIndexOutOfBoundsException: String index out of range: 15
at java.lang.String.charAt(Unknown Source)
at HW2.blankWords(HW2.java:73)
What my code should output:
HW2.blankWords("This is a Test.")
"T__s is a T__t."

Here is a pretty simple solution:
class Scratch {
public static void main(String[] args) {
System.out.println(blankWords("My name is sam orozco"));
}
public static String delim = "_";
public static String blankWords(String s1) {
// this split arg on one or more space
String[] words = s1.split("\\s+");
StringBuilder response = new StringBuilder();
for (String val : words) {
val = convertWord(val);
response.append(val).append(" ");
}
return response.toString().trim();
}
public static String convertWord(String val) {
int len = val.length();
StringBuilder bldr = new StringBuilder();
int index = 0;
for (char ch : val.toCharArray()) {
if (index == 0 || index == len - 1) {
bldr.append(ch);
} else {
bldr.append(delim);
}
index++;
}
return bldr.toString();
}
}

You can do this using a StringTokenizer that will extract words based on a list of delimiters. Since you want to keep those delimiters in the output, you'll instruct the tokenizer to return them as tokens:
String blankWords(String s) {
// build a tokenizer for your string, listing all special chars as delimiters. The last argument says that delimiters are going to be returned as tokens themselves (so we can include them in the output string)
StringTokenizer tokenizer = new StringTokenizer(s, " .,;:?!()[]{}", true);
// a helper class to build the output string; think of it as just a more efficient concat utility
StringBuilder sb = new StringBuilder();
while (tokenizer.hasMoreTokens()) {
String blankWord = blank(tokenizer.nextToken());
sb.append(blankWord);
}
return sb.toString();
}
/**
* Replaces all but the first and last characters in a string with '_'
*/
private String blank(String word) {
// strings of up to two chars will be returned as such
// delimiters will always fall into this category, as they are always single characters
if (word.length() <= 2) {
return word;
}
// no need to iterate through all chars, we'll just get the array
final char[] chars = word.toCharArray();
// fill the array of chars with '_', starting with position 1 (the second char) up to the last char (exclusive, i.e. last-but-one)
Arrays.fill(chars, 1, chars.length - 1, '_');
// build the resulting word based on the modified array of chars
return new String(chars);
}
Here is the contents of a test that validates this implementation, using TestNG:
#Test(dataProvider = "texts")
public void testBlankWords(String input, String expectedOutput) {
assertEquals(blankWords(input), expectedOutput);
}
#DataProvider
public Object[][] texts() {
return new Object[][] {
{"This is a test.", "T__s is a t__t."},
{"This one, again, is (yet another) test!", "T__s o_e, a___n, is (y_t a_____r) t__t!"}
};
}
The main drawback of this implementation is that StringTokenizer requires you to list all the delimiters by hand. With a more advanced implementation, you can consider a delimiter any character that returns false for Character.isAlphabetic(c) or however you decide to define your non-word chars.
P.S.
This could be a "more advanced implementation", as I mentioned above:
static String blankWords(String text) {
final char[] textChars = text.toCharArray();
int wordStart = -1; // keep track of the current word start position, -1 means no current word
for (int i = 0; i < textChars.length; i++) {
if (!Character.isAlphabetic(textChars[i])) {
if (wordStart >= 0) {
for (int j = wordStart + 1; j < i - 1; j++) {
textChars[j] = '_';
}
}
wordStart = -1; // reset the current word to none
} else if (wordStart == -1) {
wordStart = i; // alphabetic characters start a new word, when there's none started already
} else if (i == textChars.length - 1) { // if the last character is aplhabetic
for (int j = wordStart + 1; j < i; j++) {
textChars[j] = '_';
}
}
}
return new String(textChars);
}

No while loop necessary!
Look ahead by 1 character to see if it's a space, or if the current character is a space, in that case you append it. Otherwise you make sure to add the next character (skipNext false).
Always add the last character
public static String blankWords(String s1) {
StringBuilder sb = new StringBuilder();
if(s1.length() > 2) {
Boolean skipNext = false;
for(int x = 0; x < s1.length() - 1; x = x + 1) {
if(s1.charAt(x) == ' ' || s1.charAt(x + 1) == ' ') {
sb.append(s1.charAt(x));
skipNext = false;
}
else {
if(skipNext) {
sb.append('_');
}
else {
sb.append(s1.charAt(x));
skipNext = true;
}
}
}
sb.append(s1.charAt(s1.length() - 1));
return sb.toString();
}
return s1;
}

For the more advanced programmer, use regular expression.
public static String blankWords(String s1) {
return s1.replaceAll("\\B\\w\\B", "_");
}
This correctly keeps the final t, i.e. blankWords("This is a Test.") returns "T__s is a T__t.".

How do I reverse the order of only the digits in a string?

Given a string in Java, how can I obtain a new string where all adjacent sequences of digits are reversed?
My code:
import static java.lang.System.*;
public class P2
{
public static void main(String[] args)
{
if(args.length < 1)
{
err.printf("Usage: java -ea P2 String [...]\n");
exit(1);
}
String[] norm = new String[args.length];
for(int i = 0; i<norm.length;i++)
{
norm[i] = args[i];
}
}
public String invertDigits(String[] norm)
{
}
}
And as an example, this is what it should do:
Inputs: 1234 abc9876cba a123 312asd a12b34c56d
1234 -> 4321
abc9876cba -> abc6789cba
a123 -> a321
312asd -> 213asd
a12b34c56d -> a21b43c65d

Although the question is heavily downvoted, the proposed problem seems clear now. I chose to solve it using a regular expression match in a recursive function.
private static String reverseDigits(String s) {
// the pattern will match a sequence of 1 or more digits
Matcher matcher = Pattern.compile("\\d+").matcher(s);
// fetch the position of the next sequence of digits
if (!matcher.find()) {
return s; // no more digits
}
// keep everything before the number
String pre = s.substring(0, matcher.start());
// take the number and reverse it
String number = matcher.group();
number = new StringBuilder(number).reverse().toString();
// continue with the rest of the string, then concat!
return pre + number + reverseDigits(s.substring(matcher.end()));
}
And here's the iterative approach.
private static String reverseDigits(String s) {
//if (s.isEmpty()) return s;
String res = "";
int base = 0;
Matcher matcher = Pattern.compile("\\d+").matcher(s);
while (!matcher.hitEnd()) {
if (!matcher.find()) {
return res + s.substring(base);
}
String pre = s.substring(base, matcher.start());
base = matcher.end();
String number = matcher.group();
number = new StringBuilder(number).reverse().toString();
res += pre + number;
}
return res;
}

String str = "1234";
//indexes
int i = 0, j = str.length()-1;
// find digits (if any)
while (!Character.isDigit(str.charAt(i)) && i < str.length()) {
i++;
}
while (!Character.isDigit(str.charAt(j)) && j >= 0) {
j--;
}
// while we havent searched all the digits
while (i < j) {
// switch digits
str = str.substring(0, i) + str.charAt(j) + str.substring(i + 1, j) + str.charAt(i) + str.substring(j + 1);
i++;
j--;
// find the next digits
while (!Character.isDigit(str.charAt(i)) && i < str.length()) {
i++;
}
while (!Character.isDigit(str.charAt(j)) && j >= 0) {
j--;
}
}
System.out.println(str);

Another dynamic approach without using regex classes:
public static String reverseOnlyNumbers(String s) {
StringBuilder digits = new StringBuilder();
StringBuilder result = new StringBuilder();
boolean start = false;
for (int i = 0; i < s.length(); i++) {
Character c = s.charAt(i);
if (Character.isDigit(c)) {
start = true;
digits.append(c);
}else {
start = false;
if (digits.length() > 0) {
result.append(digits.reverse().toString());
digits = new StringBuilder();
}
result.append(c);
}
}
return start ? result.append(digits.reverse()).toString() : result.toString();
}

mirrorEnds puzzle gives no clue to error

I am attempting to solve a codingbat problem called mirrorEnds. My solution fails but I'm not getting any useful feedback from the site, only a failed test run:
And my code (I changed string to str cause I'm used to the problems with "str"):
public String mirrorEnds(String string) {
String str = string;
StringBuilder sb = new StringBuilder();
int beg = 0;
int end = str.length()-1;
while(beg < end)
{
if(str.charAt(beg)==str.charAt(end))
sb.append(str.substring(beg,beg+1));
else
break;
++beg;
--end;
}
if(beg==end)
return str;
else
return sb.toString();
}

Here's mine, for what it's worth (not much, I know, but I was writing it while you were finding the bug..)
private String mirrorEnds(String string) {
final char[] chars = string.toCharArray();
final int n = chars.length;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++) {
if (chars[i] != chars[n - i - 1])
break;
sb.append(chars[i]);
}
return sb.toString();
}

Bah. I found it. Instance is "abba"
Needed to change "if(beg==end)" to "if(beg>=end)".

public String mirrorEnds(String string) {
String s = "";
String str = "";
for (int i=string.length()-1; i>=0; i--)
{
s = s + string.charAt(i);
}
for (int j=0; j<string.length(); j++)
{
if (s.charAt(j) == string.charAt(j))
{
str = str + string.charAt(j);
}
if (s.charAt(j) != string.charAt(j))
{
break;
}
}
return str;
}

public static String mirrorEnds(String string) {
for (int i = 0; i < string.length(); i++) {
if(string.charAt(i) != string.charAt(string.length()-i-1)){
return string.substring(0,i);
}
else if(i==string.length()-1) return string;
}
return "";
}

Making a helper method is both efficient and makes the job easier, and logic clearer, recommended strategy for beginners, dissect the logic out, then put it together, as seen in codingBat's fizzBuzz questions that build up to the real fizzBuzz. Even though there a shorter solutions, this shows the full extent of logic used.
public String mirrorEnds(String string) {
String reversed = reverseString(string); //the reversed version
String result = "";
for(int a = 0; a < string.length(); a++){
if(string.charAt(a) == reversed.charAt(a)){ //keep going...
result += string.charAt(a);
}
else if(string.charAt(a) != reversed.charAt(a)){
break; //error, stop
}
}
return result;
}
public String reverseString(String s){
String reversed = "";
for(int a = s.length() - 1; a >= 0; a--){
reversed += s.charAt(a);
}
return reversed;
}

Here is mine:
public String mirrorEnds(String str) {
String res = "";
int count = str.length() - 1;
for(int i = 0;i < str.length();i++)
{
if(str.charAt(i) == str.charAt(count))
res += str.substring(i, i + 1);
else
break;
count--;
}
return res;
}

Here's my solution, hope it can help you
public String mirrorEnds(String string) {
int mid = string.length() / 2;
String s = "";
for (int i = 0, j = string.length()-1; i <= mid; i++, j--) {
if (i == mid) {
return string;
}
if (string.charAt(i) == string.charAt(j)) {
s += string.charAt(i) + "";
} else {
break;
}
}
return s;
}

Here's mine. I did mine a little bit different.
public String mirrorEnds(String string) {
//Create a string that we will eventually return.
String ret = "";
//Create a for loop that takes in chars from both ends.
for (int i = 0; i < string.length(); i++)
{
//Create one and two characters in order to simplify it.
char one = string.charAt(i);
char two = string.charAt(string.length() - 1 - i);
//If the front and back character in the iteration
//equal each other, then we add the character to the return string.
if (one == two)
{
ret = ret + one;
}
//Otherwise, we end the loop because we don't want to
//Have a loopback problem.
else
{
break;
}
}
//Return the string that we are working on.
return ret;
}