I am trying to solve this . I passed 18 tests from 20 . 2 of them are hidden and I cant understand what are those 2 scenarios. Please tell me if you see something wrong here . I saw many solutions on the WEB, but I want to understand what is wrong with my own code.
For inputString = "aabb", the output should be
palindromeRearranging(inputString) = true.
We can rearrange "aabb" to make "abba", which is a palindrome.
Here is my code:
boolean palindromeRearranging(String str) {
Stack<Character> stack=new Stack<>();
for (int i = 0; i < str.length(); i++) {
if(stack.contains(str.charAt(i))){
stack.pop();
continue;
}
stack.push(str.charAt(i));
}
if(stack.size()>1) return false;
return true;
}
When you're by using condition stack.contains(str.charAt(i)),
find out that current checking character str.charAt(i) is already present in Stack,
you need to remove that particular repeating character instead of the last character present in Stack.
so only stack.pop() is a mistake in code.
One of those scenarios in my case :
"abcabc" . Should return true, but instead it returns false. I sorted the String first then my algorithm worked :)
Related
while (scan_file.hasNext()) {
String b = scan_file.nextLine();
// checks if string b contains the tag <h>
if (b.contains("<h>")) {
char arrayString[] = b.toCharArray();
for (int i = 0; i < arrayString.length; i++) {
if (arrayString[i] == '<') {
arrayString[i] = arrayString[i + 2];
}
System.out.print(arrayString[i]);
}
}
}
What I was expecting the program to do was(for now) iterate through the while loop and store each line as string 'b'.
I want to check if that string b contains a certain string like <h> for this example. And I want to convert string b into an array if it contains said string like <h> and iterate through that array to check for '<' and move the array up 2 spaces.
For example, string b had <h>hello, I wanted to eventually print hello because the program would have moved up 2 elements.
I feel like I got the loops and general idea on how I want to tackle the problem.. but when I ran the program, nothing printed so I don't know if I did the loops and if statements correctly.
I really don't know how to word my problem well, so bear with me and I'm sorry in advance.
All feedbacks are greatly appreciated (:
System.out.print(arrayString[i]); just print the ith character of arrayString, it's definitely not what you want.
In fact you don't have to convert a String to char[], String has many utils method can help you with your goal.
I won't give you full code , but I can give you some tips.
You can use String.indexof('<') to find the index of '<'.
You can use String.subString(startIndex) to get the subString start with the specified index.
Suppose your code scan_file.hasNext() and scan_file.nextLine() is work well. You can try code below to remove all from current line:
if (b != null && b.contains("<h>")) {
System.out.println(b.replaceAll("<h>", ""));
}
i need to get a string and rearrange it with recursion by getting char and by that char i have to move that char everywhere on the string to the end
like "Hello world!" ,'l' => "Heo word!lll"
i have problems understading the recursion way of thinking
so i started with this:
public static String ChToLast (String str, char ch){
if(str.indexOf(ch)== -1){
return str;
}else{
if(str.indexOf(0) == ch){
return str;
}
}
thank you for your help :)
Recursion is the practise of reusing your method inside itself. In this case, I will provide a solution to explain what happens:
public static String chrToLast(String str, char ch) {
//This if statement details the end condition
if(str.length() < 1) {
return "";
}
String newString = str.substring(1); //Create new string without first character
if(str.indexOf(ch) == 0) { //This happens when your character is found
return chrToLast(newString, ch) + ch;
} else { //This happens with all other characters
return str.charAt(0) + chrToLast(newString, ch);
}
}
If you execute:
chrToLast("Hello, World!", 'l')
This will result in the desired result: Heo, Word!lll
Process
In general, this method works by checking which character is currently the first in the given string, and then deciding what to do. If the first character is the same as the one your looking for (l), it will then remove that character from the string and use chrToLast on that new string. But, it also adds the character it found to the end of the result by using + ch. It continues to do this until there are no more characters left, which is what the end condition is for.
The end condition
The end condition returns an empty string "" because that is what is called the base case of the algorithm. You can think of a recursive algorithm as something solving a problem by calling itself a number of times. By calling themselves, recursive algorithms move towards a base. In this particular case, it does that by subtracting one character off the string each time the method is executed. Once there are no characters left, it reaches the base case which is "", where the string is finally empty and no characters can be subtracted anymore. (Hence it returns nothing as it's final state)
I hope this answers your question. It's important to understand this concept, as it is very powerful. Try to study the code and comment if something's not clear.
Something that can also help is by executing this code in an IDE and using the debugger to walk through its execution. You can then see for yourself what the flow of the program is, and see the value of the variables in play.
If you use recursion, it will be pretty expensive call for the result you are expecting. Lot of movement of String or charArray elements, eitherway you do. I don't see its a wiser choice. I would do it this way, it will be of space complexity O(2n) & performance complexity O(n).
public class Solve {
public static void main(String[] args) {
System.out.println(ChToLast("Hello world!", 'l'));
}
public static String ChToLast(String str, char ch) {
char[] chars = str.toCharArray();
char[] modChars = new char[chars.length];
int i = 0;
for(char element : chars){
if(ch != element){
modChars[i++] = element;
}
}
Arrays.fill(modChars, i, chars.length , ch);
return new String(modChars);
}
}
If you use while loop and write a method to check if that string means perfect statement then that may work for you
Here you would need some help of NLP concept to check everytime if arranged chars are making any statement or are grammatically correct.
This will help
Return true if the string "cat" and "dog" appear the same number of times in the given string.
catDog("catdog") → true
catDog("catcat") → false
catDog("1cat1cadodog") → true
public boolean catDog(String str) {
int countCat=0;
int countDog=0;
for(int i=0;i<str.length()-3;i++){
if(str.substring(i).startsWith("cat")){
countCat++;
}
if(str.substring(i).startsWith("dog")){
countDog++;
}
}
if(countCat==countDog){
return true;
}
else{
return false;
}
}
I am having trouble writing this method. Does anybody know why my code doesn't work correctly?
Edit: The code compiles, however it gives the wrong output. For example if i put in "catdog" it returns false.
with the examples you posted, its because of your for loop which should be for(int i=0;i<str.length();i++){. you can also use str.length()-1 and str.length()-2 to get the right result. -3 will give the wrong result. little example: string is catdog1dog. result should be false. lets have a look at the substring which will be created with -3:
catdog1dog
atdog1dog
tdog1dog
dog1dog
og1dog
g1dog
1dog
as you can see with -3 the last substring is wrong and therefore the result too. that is because if you look at substring, you will see that the start is at char 0 and not at 1 therefore str.length()-1 is the last character in your string. sorry if my explanations isn't that good
As suggestsed in this question, I would recomment using StringUtils.countMatches from Apache Commons Lang?
Credit to #A_M
Although your problem can be solved through other strategies, I think you might only subtract 2 to str.length() instead of 3.
I hope be useful!
I would do something like this:
public boolean catDog(String str){
return countWords(str, "cat") == countWords(str, "dog");
}
private int countWords(String original, String word){
int counter = 0;
boolean searching = true;
while(searching){
if(original.indexOf(word) >= 0){
counter++;
original = original.substring(original.indexOf(word) + word.length());
}
else{
searching = false;
}
}
return counter;
}
I need to create a loop that adds "o" after each consonant
I am going to walk you through what I corrected and changed in your code to make it work in order to make it quick and easy for you to comprehend why your code doesn't work and why my answer fixes it.
The mistakes you made are basic ones and frankly you shouldn't have to much of a hard time correcting them yourself if you would use a debugger that walks you step by step in how your code works. You should look on how to use a debugger (for example the debugger used in Eclipse, hopefully you are using an IDE to make your life easier).
Firstly, when you are looking for a consonant in your code, you are only walking through the half of it because of your condition for(int x = 0; x<20; x++) since your string holding the consonants if of a length of 40 characters. This means you are missing consonants like the letter s.
Then you are correctly the consonants you find according to your Swedish language game. But you are never handling characters that are not of these found consonants. You should make a case where you handle these "non consonant" letters, may they be vowels or any kind of character (like punctuation marks and so on). I am fixing this with the use of a simple boolean here.
Keep in mind that my goal here is to change your code as little as I can, thus I went for adding a boolean to handle your cases (checking the presence of a consonant). There are, obviously, many other ways to implement what you are trying to do.
Here come the changes you should add to your code:
/*This comes after your print "På rövarspråk:"*/
boolean isConsonant = false; //Boolean to check wether there is a consonant or not
for(int i = 0; i<length; i++) {
//You didn't go through the whole consonants list you made with your prevision condition
for(int x = 0; x<consonants.length; x++){
if(array[i] == consonants[x])
{
isConsonant = true; //Set the boolean accordingly
String add = array[i]+"o"+array[i];
slang = slang + add;
break;
}
}
if(!isConsonant){ //If we don't have a consonant, add the char to the result string
slang += array[i];
}
isConsonant = false; //Reset the boolean for the next character
}
/*Here you can print the result (slang) if you want, as you did*/
so the idea is to dublicate consonants and put "o" between them, like t becomes tot, s becomes sos. Vocals are just copied. So you need a method that tells you if a given character is a vocal or consonant to base your decision on that.
public static boolean isConsonant(char inputChar){
final String consonantsx = "bBcCdDfFgGhHjJkKlLmMnNpPqQrRsStTvVwWxXzZ";
char consonants[] = consonantsx.toCharArray(); // String to charr
for(int i=0; i < consonants.length;i++){
if(inputChar == consonants[i]){ //note that in Strings u use the equals method instead of "=="
return true;
}
}
return false;
}
Given this method you can use it in the "translator method".
public String rovarSpraket(String normalString) {
char[] array = normalString.toCharArray(); // Input to a char array
System.out.println("På rövarspråk:");
String slang = "";
for (int i = 0; i < normalString.length(); i++) {
String add = "" + array[i];
if(Goran.isConsonant(array[i])){
add += "o" + array[i];
}
slang += add;
}
return slang;
}
This translates stubborn to sostotubobboborornon like in the wikipedia article https://en.wikipedia.org/wiki/R%C3%B6varspr%C3%A5ket.
The above question might seems vague but it's actually a very simple idea which i can't seem to figure out.
It basically is a 4 digit letter code containing letters from A to F for example: ABDF, BAAF, DBAF etc.
Now I'm trying to do some post input-handling where it must become impossible to enter a letter that is already in the code cause it has to be a unique 4 digit code with no repeating letter. I've been trying to make it work but none of my code seems to work so i'm back to scratch asking you guys for help :)
I hope this is somewhat clear otherwise i'll be happy to clear it up.
Thanks in advance.
Kind of a pseudocode but it would work.
String uniquePass="";
while(uniquePass.length<4){
String userInput=getUserInputChar()
if(uniquePass.contains(userInput))
rejectInputAndNotifyUser
else
uniquePass=uniquePass+userInput
}
public static boolean hasDuplicateChars(String string) {
Set<Character> chars = new HashSet<Character>();
for (char c : string.toCharArray()) {
if (!chars.add(c)) return false;
}
return true;
}
Set is a collection that contains no duplicate elements. We will use add method which returns true if this set did not already contain the specified element.
hasDuplicateChars functions iterates over characters in the input string using toCharArray function and for loop; each character is added to the chars set which is initially empty. If add method returns false it means that we have already encountered same character before. So we return false from our function.
Otherwise input is valid and method returns true.
using this function you'll be able to see if the string contains unique characters
public static boolean checkForUnique(String str){
boolean containsUnique = false;
for(char c : str.toCharArray()){
if(str.indexOf(c) == str.lastIndexOf(c)){
containsUnique = true;
} else {
containsUnique = false;
}
}
return containsUnique;
}
Update:
This will be ran everytime a user enters a character and if it fails, this would mean there is a duplicate. You have the choice of discarding that input or showing an error.
If you're validating the complete input, you can lean on the set semantics, and a few tricks
String s = "ABAF";
int count = new HashSet<>(Arrays.asList(s.split(""))).size();
if (count - 1 == 4) {
System.out.println("All ok");
} else {
System.out.println("Repeated letters");
}
the split("") will split the string to a an array like {"","A", "B", "A", "F"}.
The new HashSet<>(Arrays.asList(s.split(""))) will create a Set with String elements, and as the Set will bounce back the elements already contained, the size of the set for e.g. "AAAF" will be 3 (it'll contain the "", "A" and "F"). This way you can use the size of the set to figure out if all letters of a String are unique
If its while typing you'll than the solution depends on the input method, but you can have something like (pseudo stuff)
if (pass.contains(letter)) {
breakAndNotifyUser();
} else {
pass+=letter;
}