We'll say that a String is xy-balanced if for all the 'x' chars in the string, there exists a 'y' char somewhere later in the string. So "xxy" is balanced, but "xyx" is not. One 'y' can balance multiple 'x's. Given a string str, Print “xy balanced”if the given string is xy-balanced otherwise print “not xy balanced”.
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
String str = scan.next();
int x = str.length();
boolean xyBalanced = false;
for(int i = x - 1; i >= 0; i--)
{
char y = str.charAt(i);
if(y == 'y')
xyBalanced = true;
else if(y == 'x')
xyBalanced = false;
else
xyBalanced = true;
}
if(xyBalanced = true)
System.out.print("xy balanced");
else if(xyBalanced = false)
System.out.print("not xy balanced");
}
It's giving me "xy balanced" when I input aabbx when in fact it should give "not xy balanced". What is wrong with this code?
Why do you check from the end of the string to the start? It is very difficult to understand.
Apart from the if/else issue you have at the end of your program, which you can change to below, your program still doesn't work after I corrected it.
System.out.print(xyBalanced ? "xy balanced" : "not xy balanced");
To check if there is always a 'y' on the right of an 'x', you can use the following loop - basically, an 'x' always set the xyBalanced flag to false and a 'y' always set it to be true.
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String str = scan.next();
int length = str.length();
boolean xyBalanced = true;
for (int i = 0; i < length; ++i) {
if (str.charAt(i) == 'x') {
xyBalanced = false;
} else if (str.charAt(i) == 'y') {
xyBalanced = true;
}
}
System.out.print(xyBalanced ? "xy balanced" : "not xy balanced");
}
The statement:
if(xyBalanced = true)
Is an assignment operator, it should be using the == operator:
if(xyBalanced == true){
System.out.print("xy balanced");
} else {
System.out.print("not xy balanced");
}
Since it's a boolean you can also say:
System.out.print( (!xyBalanced? "not" : "") + "xy balanced");
However, from your logic, it seems like you can also say
boolean xyBalanced = str.lastIndexOf('x') < (str.length()-1);
Because you are considering non-x/non-y characters.
So really you can simplify all the code into two lines.
Also there is an issue with your loop. You're actually going from the right to left. Is that intentional?
for(int i = x - 1; i >= 0; i--)
would seem like it should be:
for (int i=0;i<x;i++)
you can do following for identifying balanced string
boolean xyBalanced = inputString.lastIndexOf('x') < inputString.indexOf('y');
indexOf function always returns first occurrence of string. We can check if last index of x is less than first index of y, and variable 'xyBalanced' will be accordingly get true or false.
public boolean xyBalance(String str) {
if(str.endsWith("x")){
System.out.println("false");
return false;
}
if(str.endsWith("y")){
System.out.println("true");
return true;
}
if(str.lastIndexOf("x")>str.lastIndexOf("y")) {
System.out.println("false");
return false;
}
System.out.println("true");
return true;
}
Related
What I have so far:
public boolean allSameLetter(String str)
{
for (int i = 1; i < str.length(); i++)
{
int charb4 = i--;
if ( str.charAt(i) != str.charAt(charb4))
{
return false;
}
if ( i == str.length())
{
return true;
}
}
}
Please excuse any inefficiencies if any; still relatively new to coding in general. Am I lacking some knowledge in terms of using operators and .charAt() together? Is it illogical? Or is my error elsewhere?
Using regex:
return str.matches("^(.)\\1*$");
Using streams:
str.chars().allMatch(c -> c == str.charAt(0));
Other:
return str.replace(String.valueOf(str.charAt(0), "").length() == 0;
You can follow the below steps:
(1) Get the first character (i.e., 0th index)
(2) Check the first character is the same with subsequent characters, if not return false (and comes out from method)
(3) If all chars match i.e., processing goes till the end of the method and returns true
public boolean allSameLetter(String str) {
char c1 = str.charAt(0);
for(int i=1;i<str.length;i++) {
char temp = str.charAt(i);
if(c1 != temp) {
//if chars does NOT match,
//just return false from here itself,
//there is no need to verify other chars
return false;
}
}
//As it did NOT return from above if (inside for)
//it means, all chars matched, so return true
return true;
}
As Andrew said, you are decreasing i within your for loop. You can fix this by changing it to int charb4 = i - 1;. As for making your code more efficient you could condense it down to this.
public boolean allSameLetter(String str) {
for(char c : str.toCharArray())
if(c != str.charAt(0)) return false;
return true;
}
Comment if you don't understand a part of it :)
public boolean allSameLetter(String str)
{
for (int i = 1; i < str.length() -1; i++)
{
if ( str.charAt(i) != str.charAt(i+1))
{
return false;
}
}
return true
}
-1 is there since I am checking the current value in the array, then the next value in the array, thus I need to stop a place earlier.
If the loop if statement is never entered, it will make it far enough into the code to return true
You have to create a for loop that searches through the length of the String - 1. this way the program will not crash because of a 3 letter word with the program trying to get the 4th letter. This is what works for me:
public boolean allSameLetter(String str)
{
for(int i = 0; i< str.length()-1; i++){
if (str.charAt(i) != str.charAt(i+1)){
return false;
}
}
return true;
}
if((new HashSet<Character>(Arrays.asList(s.toCharArray()))).size()==1)
return true;
return false;
This should be enough
The bug is caused by
int charb4 = i--;
this line is equal to
int charb4 = i-1;
i=i-1;
Because of this, your loop will never stop.
The easiest way to fix this
public boolean allSameLetter(String str)
{
for (int i = 1; i < str.length(); i++)
{
if ( str.charAt(i) != str.charAt(i-1))
{
return false;
}
}
}
I am having trouble rewriting the following code as a recursive method rather than using the for loop. The for loop tests to see if the String 'noSpaces' is a palindrome (the same forwards as it is backwards). The noSpaces String has no punctuation, spaces, or differences in capitalization.
Thanks for the help
public boolean isRegularPalindrome(String noSpaces) {
noSpaces = noSpaces.toUpperCase();
String[] letters = new String[noSpaces.length()];
for (int i = 0; i < letters.length; i++) {
letters[i] = Character.toString(noSpaces.charAt(i));
}
for (int i = 0; i < letters.length / 2; i++) {
if (!letters[i].equals(letters[letters.length - i - 1])) {
return false;
}
}
return true;
}
There you go:
public static boolean isPalindrome(String input) {
if (input.charAt(0) != input.charAt(input.length() - 1)) {
// Missmatch. Not a palindrome!
return false;
} else if (input.length() > 1){
// If there is more to test, continue.
return isPalindrome(input.substring(1, input.length() - 1));
} else {
// All chars were tested, or 1 char input. Palindrome!
return true;
}
}
Writing a recursive algorithm for anything requires base cases. For a palindrome, this would be a string of length 0 or length 1 -- if the string is length 0 or 1, it is a palindrome.
If the base cases aren't met, you check the first character against the last character.
If the characters aren't the same, return false.
If the characters are the same, return the recursive call to the string except for the first and last characters.
The code should look something like this.
public boolean isPalindrome(string str){
if (str.length == 0)
return true;
else if (str.length == 1)
return true;
else if(str.charAt(0) != str.charAt(str.length - 1)
return false;
else
return isPalindrome(str.substring(1, length - 1));
}
I have a text file with a bunch of words. One part of my program is supposed to, based on the user input, scan this list and insert into the output file the words that contain the requested number of vowels. I have a isVowel method that returns a boolean, but it doesn't seem to work--the output is just an empty list.
Here's the method:
public boolean isVowel(char c)
{
if(c=='a' || c=='A' || c=='e' || c=='E' || c=='i' || c=='I' || c=='o' || c=='O' || c=='u' || c=='U')
{
return true;
}
else
{
return false;
}
}
}
Here's the part of the program where it is used:
public ArrayList<String> vowelHeavy(int n, int m)
{
int vowels = 0;
while(input.hasNextLine())
{
word = input.nextLine();
if (word.length() == n)
{
for(int i = 0; i < word.length(); i++)
{
if(isVowel(word.charAt(i)) == true)
{
vowels++;
}
}
if (vowels == m)
{
output.add(word);
}
}
word = input.nextLine();
}
return output;
}
Your biggest problem is you are not resetting you vowels counting variable when you test a new word. The value just keeps accumulating the total vowels in the file.
Change:
int vowels = 0;
while(input.hasNextLine()) {
...
to:
while(input.hasNextLine()) {
int vowels = 0;
...
As an aside, most of your code, including the isVowel() method, could be eliminated with a single line:
int vowels = word.replaceAll("[^aeiouAEIOU]", "").length();
This works by eliminating from the word all characters that aren't vowels; what's left are the vowels, so just take the length to get the count.
Theres a much easier way to write this code:
public boolean isVowel(char c){
return(your boolean tests);
}
2 things:
If the word.length() == n test is never true, then output is never changed. Please make sure that test is correct.
This statement exists at the top AND bottom of the while loop, causing you to ignore every other input line:
word = input.nextLine();
Here is the problem: Return true if the string "cat" and "dog" appear the same number of times in the given string. Examples: 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();i++)
{
if (str.charAt(i)== 'c'&& str.length()>=3)
{
if (str.substring(i,i+3).equals("cat"))
countCat++;
}
}
for (int i=0; i<str.length();i++)
{
if (str.charAt(i)== 'd' && str.length()>=3)
{
if (str.substring(i,i+3).equals("dog"))
countDog++;
}
}
if (countCat == countDog)
return true;
else
return false;
}
In your for loops conditions you are checking if entire String has length greater or equal 3 instead of checking only part from i till end. Try maybe with
str.length() - i >= 3
instead of
str.length() >= 3
str.substring(i,i+3).equals("cat")
i might be the last and i+3 will give an error
Why don't you simply use StringUtils#countMatches?
StringUtils.countMatches(myStr, "cat") == StringUtils.countMatches(myStr, "dog");
Don't get lost with the indexes. However, if you don't want to use this method, debugging your code is the best thing you can do.
Okay, this is what I might do:
The problem was with your check str.length() >= 3. It should have been i + str.length().
I also suggest some changes to your code to get rid of duplication. Here I extracted the part that counts the number of appearances of a substring and moved it to its own method. The part that checks if count of cat equals count of dog now calls said method twice.
public static void main(String[] args) {
System.out.println(catDog("catdog"));
System.out.println(catDog("catcat"));
System.out.println(catDog("1cat1cadodog"));
System.out.println(catDog("catdogcatc"));//Would previously throw error.
}
public static boolean catDog(String str) {
int countCat = countAppearances(str, "cat");
int countDog = countAppearances(str, "dog");
return countCat == countDog;
}
private static int countAppearances(String str, String key) {
int count = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == key.charAt(0) && i + key.length() <= str.length()) {
if (str.substring(i, i + key.length()).equals(key)) {
count++;
}
}
}
return count;
}
You need to update your first condition before spiting you string like:
if (str.charAt(i)== 'c' && (str.length() - i) >= 3)
{
if (str.substring(i,i+3).equals("cat"))
countCat++;
}
public boolean catDog(String str) {
int catCount = 0, dogCount = 0;
//run a for loop to check cat count
//run loop till 2nd last character
for (int i = 0; i < str.length() - 2; i++) {
//now check if the charaters at positions matches "cat"
//if matches then increment cat count
if (str.charAt(i) == 'c' && str.charAt(i + 1) == 'a' && str.charAt(i + 2) == 't') {
catCount++;
} else if (str.charAt(i) == 'd' && str.charAt(i + 1) == 'o' && str.charAt(i + 2) == 'g') {
//else check if the charaters at positions matches "dog"
//if matches then increment dog count
dogCount++;
}
}
//check cat count and dog count
if (catCount == dogCount) {
return true;
} else {
return false;
}
}
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
I am a beginner of programming, and am writing a Java method to remove vowel in Strings, but I do not know how to fix this error: ";" expected :
public String disemvowel(String s) {
boolean isVowel(char c);
if (c == 'a') {
return true;
} else if if (c == 'e') {
return true;
} else if if (c == 'i') {
return true;
} else if if (c == 'o') {
return true;
} else if if (c == 'u') {
return true;
}
String notVowel = "";
int l = s.length();
for (int z = 0; z <= l; z++) {
if (isVowel == "false") {
char x = s.charAt(z);
notVowel = notVowel + x;
}
}
return notVowel;
}
String str= "Your String";
str= str.replaceAll("[AEIOUaeiou]", "");
System.out.println(str);
A much simpler approach would be to do the following:
String string = "A really COOL string";
string = string.replaceAll("[AaEeIiOoUu]", "");
System.out.println(string);
This will apply the regular expression, [AaEeIiOoUu] to string. This expression will match all vowels in the character group [AaEeIiOoUu] and replace them with "" empty string.
You've got a lot of syntax errors.
boolean isVowel(char c); - not sure what you're doing with this. if you want it as a separate method, separate it out (and don't place a semicolon after it, which would be invalid syntax.
else if if is invalid syntax. If you're doing an else if, then you only need the one if.
Even if the code would compile, for (int z = 0; z <= l; z++) will cause you to step off of the String. Remove the <= in favor of <.
isVowel == "false" is never going to work. You're comparing a String to a boolean. You want !isVowel instead.
Putting the syntax errors aside, think of it like this.
You have a string that contains vowels. You wish to have a string that doesn't contain vowels.
The most straightforward approach is to iterate over the String, placing all non-vowel characters into a separate String, which you then return.
Interestingly enough, the half-method you have there can accomplish the logic of determining whether something is or isn't a vowel. Extract that to its own method. Then, call it in your other method. Do take into account capital letters though.
I leave the rest as an exercise to the reader.
Here is your code, without changing any logic, but unscrambling the isVowel method:
public String disemvowel(String s) {
// Removed the "isVowel" method from here and moved it below
String notVowel = "";
int l = s.length();
for (int z = 0; z <= l; z++) {
// Note that the "isVowel" method has not been called.
// And note that, when called, isVowel returns a boolean, not a String.
// (And note that, as a general rule, you should not compare strings with "==".)
// So this area needs a lot of work, but we'll start with this
boolean itIsAVowel = isVowel(s.charAt(z));
// (I made the variable name "itIsAVowel" to emphasize that it's name has nothing to do with the method name.
// You can make it "isVowel" -- the same as the method -- but that does not in any way change the function.)
// Now take it from there...
if (isVowel == "false") {
char x = s.charAt(z);
notVowel = notVowel + x;
}
}
return notVowel;
}
// You had this line ending with ";"
boolean isVowel(char c) {
if (c == 'a') {
return true;
// Note that you coded "if if" on the lines below -- there should be only one "if" per line, not two
} else if if (c == 'e') {
return true;
} else if if (c == 'i') {
return true;
} else if if (c == 'o') {
return true;
} else if if (c == 'u') {
return true;
}
// You were missing this final return
return false;
}
(Yes, I know this should be a comment, but you can't put formatted code in a comment.)
You could try something like this:
public static String removeVowels(final String string){
final String vowels = "AaEeIiOoUu";
final StringBuilder builder = new StringBuilder();
for(final char c : string.toCharArray())
if(vowels.indexOf(c) < 0)
builder.append(c);
return builder.toString();
}