Language : Java Key Notes: *Needs to loop through a String using either a For loop or While loop *It removes the duplicate letter(s) of the String and returns the word without the dupilcates.
Eg: The string is HELLO - The method then loops through and removes any duplicates, in this case " L " and returns in the end HELO
i have this so far
private String removeAnyDuplicates(String userWord)
{
//Code goes here?
return "" ; // Need to return the new string
}
You can do that with regular expressions. e.g.:
private static final Pattern REGEX_PATTERN =
Pattern.compile("(.)\\1*");
public static void main(String[] args) {
String input = "HELLO, AABBCC";
System.out.println(
REGEX_PATTERN.matcher(input).replaceAll("$1")
); // prints "HELO, ABC"
}
I'm assuming that removing duplicates means that the result contains at most one occurrence of any character. (Some of the other answers assume that adjacent duplicates only need to be reduced to single occurrences.) The basic algorithm would be:
A naive (and very inefficient) implementation would be:
private String removeAnyDuplicates(String userWord)
{
String result = "";
for (int i = 0; i < userWord.length(); ++i) {
char c = result.charAt(i);
if (result.indexOf(c) < 0) {
// negative index indicates not present
result += String.valueOf(c);
}
}
return result;
}
This has two major sources of inefficiency: it creates many intermediate String objects and it has to scan the entire result so far for each character of the input. These problems can be solved by using some other built-in Java classes—a StringBuilder to more efficiently accumulate the result and a Set implementation to efficiently record and test which characters have already been seen:
private String removeAnyDuplicates(String userWord)
{
int len = userWord.length();
StringBuilder result = new StringBuilder(len);
Set<Character> unique = new HashSet<Character>();
for (int i = 0; i < len; ++i) {
char c = result.charAt(i);
// try to add c to set of unique characters
if (unique.add(c)) {
// if it succeeds, this is the first time seeing c
result.append(c);
}
}
return result.toString();
}
private String removeAnyDuplicates(String userWord)
{
CharSequence inputStr = userWord;
int length = inputStr.length();
Set<Character> uniqueChars = new HashSet<Character>();
for(int i=0; i < length; ++i) {
uniqueChars.add(inputStr.charAt(i));
}
return uniqueChars.size() >= 3;
}
check out this answer
String not boolean.Convert the string to an array of char, and store it in a LinkedHashSet. That will preserve your ordering, and remove duplicates.
Like this:
private static String removeAnyDuplicates(String userWord)
{
char[] chars = userWord.toCharArray();
Set<Character> charSet = new LinkedHashSet<Character>();
for (char c : chars) {
charSet.add(c);
}
StringBuilder sb = new StringBuilder();
for (Character character : charSet) {
sb.append(character);
}
return sb.toString();
}
Remember:
import java.util.LinkedHashSet;
import java.util.Set;
You can try this
public static void main(String args[]){
System.out.println(removeAnyDuplicates("HELLO"));
}
private static String removeAnyDuplicates(String userWord)
{
char[] arr=userWord.toCharArray();
List<String> list=new ArrayList<>();
for(int i=0;i<arr.length;i++){
if(!list.contains(String.valueOf(arr[i]))){
list.add(String.valueOf(arr[i]));
}
}
return list.toString().replaceAll("\\[|\\]|\\,","") ;
}
Try this one liner:
private String removeAnyDuplicates(String userWord) {
return userWord.replaceAll("(.)\\1+", "$1");
}
This uses a regular expression to find repeated (2 or more) letters and replaces them with a single instance of the letter.
It is unclear if "repeated" means appearing immediately after or anywhere after. For anywhere, use this:
private String removeAnyDuplicates(String userWord) {
return userWord.replaceAll("(.)(?=.*\\1)", "");
}
