Im currently trying to create a function where my input is a string such as "AABBCCDDEE" and the function outputs a String array "AA""BB""CC" and so on.
public static char[] stringSplitter(final String input) {
String[] strarray = new String[input.length()];
if (input == null) {
return null;
}
char[] chrarray = input.toCharArray();
char[] outputarray = new char[input.length()];
int j = 0;
for (int i = 0; i < chrarray.length; i++) {
char chr = chrarray[i];
System.out.print(chr);
outputarray[j] = chrarray[i]; //here i need to find a way to add the characters to the index at j if the preceding characters are equal
if (i + 1 < input.length() && chrarray[i + 1] != chr) {
j++;
outputarray[j] = chrarray[i + 1];
System.out.println(" ");
}
}
}
Arrays are fixed-length, so you can't do this with an array unless you allocate one with sufficient extra room up-front (which would require a pass through the string to find out how much extra room you needed).
Instead, consider using a StringBuilder for the output, which you can convert into a char array when you're done.
If I understood correctly, you want to split the characters in a string so that similar-consecutive characters stay together. If that's the case, here is how I would do it:
public static ArrayList<String> splitString(String str) {
ArrayList<String> output = new ArrayList<>();
String combo = "";
//iterates through all the characters in the input
for(char c: str.toCharArray()) {
//check if the current char is equal to the last added char
if(combo.length() > 0 && c != combo.charAt(combo.length() - 1)) {
output.add(combo);
combo = "";
}
combo += c;
}
output.add(combo); //adds the last character
return output;
}
Note that instead of using an array (has a fixed size) to store the output, I used an ArrayList, which has a variable size. Also, note that it's a list of strings (stores strings), not characters. The reason for this is that if it was a list of characters I wouldn't be able to store more than one character in the same index.
In each iteration of the loop, I check for equality between the current character and it's consecutive. The variable combo is used to temporarily store the characters (in a string) before they go to output.
Now, to print the results in a clear way:
public static void main(String[] args)
{
String input = "EEEE BCD DdA";
ArrayList<String> output = splitString(input);
System.out.print("[");
for(int i = 0; i < output.size(); i++) {
System.out.print("\"" + output.get(i) + "\"");
if(i != output.size()-1)
System.out.print(", ");
}
System.out.println("]");
}
The output when running the above code will be:
["EEEE", " ", "B", "C", "D", " ", "D", "d", "A"]
You can use an ArrayList of type String to store the consecutive letter Strings after splitting them. This code should work for you.
import java.util.*;
public class StringSplitter{
static ArrayList<String> splitString(String str)
{
ArrayList<String> result_list= new ArrayList<String>();
int last_index;
if(str == null)
{
return null;
}
else
{
while(str.length() != 0)
{
last_index = str.lastIndexOf(str.charAt(0));
result_list.add(str.substring(0, last_index+1));
str = str.substring(last_index+1);
}
}
return result_list;
}
public static void main(String[] args)
{
ArrayList<String> result = splitString("AABBCCDDEEE");
System.out.println(result);
}
}
I have used an ArrayList because it does not require you to fix a size while declaration.
Related
I am looking for code that produces the following output in standard output from the following string prepared according to a certain format.
Assumptions and rules:
Each letter is used 2 times in the given string and the letters between the same 2 letters are to be considered child letters.
The given string is always given in proper format. The string format
does not need to be checked.
Example:
Input : abccbdeeda
Expected output:
a
--b
----c
--d
----e
Explanation: since the 2 letters "b" occur between the letters "a", the letter b takes 2 hyphens (--b)
Attempt
public static void main(String[] args) {
String input = "abccbdeeda";
System.out.println("input: " + input);
String[] strSplit = input.split("");
String g = "";
String h = "-";
ArrayList<String> list = new ArrayList<String>();
int counter = 1;
boolean secondNumber;
list.add(strSplit[0]);
int dual = 0;
for (int i = 1; i < strSplit.length; i++) {
secondNumber = list.contains(strSplit[i]);
if ((secondNumber)) {
counter--;
dual = counter * 2;
for (int f = 0; f < dual; f++) {
strSplit[i] = h.concat(strSplit[i]);
}
g = "";
dual = 0;
} else {
list.add(strSplit[i]);
counter++;
}
}
Arrays.sort(strSplit);
for (int p = 0; p < strSplit.length; p++) {
System.out.println(strSplit[p]);
}
}
input: abccbdeeda
My output:
----c
----e
--b
--d
a
I wasn't able to sort the output alphabetically. How can I sort alphabetically with those hyphen characters in them?
This task is nicely done with the help of a stack. If the current character is equal to the top of the stack, then the character is closed and can be removed, otherwise we met it for the first time and it must be added to the stack and the resulting string by adding before it stack.size() * 2 dashes.
When we have completely traversed the string we can sort the resulting string.
public static void main(String[] args) {
Stack<Character> stack = new Stack<>();
String string = "abccbdeeda";
StringBuilder result = new StringBuilder();
for(int i = 0; i < string.length(); i++) {
char curChar = string.charAt(i);
if(!stack.isEmpty() && curChar == stack.peek()) {
stack.pop();
} else {
result.append("-".repeat(stack.size() * 2)).append(curChar).append(" ");
stack.add(curChar);
}
}
System.out.println(result);
System.out.println(Arrays.toString(Arrays.stream(result.toString().split(" ")).sorted().toArray()));
}
Output
a --b ----c --d ----e
[----c, ----e, --b, --d, a]
You can go through the strSplit array and extract the charactors in each element to a separate list/array. To check whether the array element contains a letter you can write a regular expression.
Ex: private final Pattern x = Pattern.compile("[a-z]");
Write a separate method to match the patern to each element in the strSplit array. This method will return the charactor in your input string.
private String findCharactor(final StringBuilder element) {
final Matcher matcher = x.matcher(element);
if (matcher.find()) {
final int matchIndex = matcher.start(); //this gives the index of the char in the string
return element.substring(matchIndex);
}
}
Add these returned charactors to a separate array and sort it using sorting function.
Suppose your result list is:
List<String> resultList = Arrays.asList("----c", "----e", "--b", "--d", "a");
You can sort it alphabetically by a single line:
Collections.sort(resultList, (o1, o2) -> new StringBuilder(o1).reverse().toString().compareTo(new StringBuilder(o2).reverse().toString()));
You can use recursion for a depth-first traversal (preorder):
public static String dfs(String string, String prefix) {
if (string.length() == 0) return "";
int i = string.indexOf(string.charAt(0), 1);
return prefix + string.charAt(0) + "\n" // current
+ dfs(string.substring(1, i), prefix + "--") // all nested
+ dfs(string.substring(i + 1), prefix); // all siblings
}
Example call:
public static void main(String[] args) {
System.out.println(dfs("abccbdeeda", ""));
}
There is a way to split a string into repeating characters using a regex function but I want to do it without using it.
for example, given a string like: "EE B" my output will be an array of strings e.g
{"EE", " ", "B"}
my approach is:
given a string I will first find the number of unique characters in a string so I know the size of the array. Then I will change the string to an array of characters. Then I will check if the next character is the same or not. if it is the same then append them together if not begin a new string.
my code so far..
String myinput = "EE B";
char[] cinput = new char[myinput.length()];
cinput = myinput.toCharArray(); //turn string to array of characters
int uniquecha = myinput.length();
for (int i = 0; i < cinput.length; i++) {
if (i != myinput.indexOf(cinput[i])) {
uniquecha--;
} //this should give me the number of unique characters
String[] returninput = new String[uniquecha];
Arrays.fill(returninput, "");
for (int i = 0; i < uniquecha; i++) {
returninput[i] = "" + myinput.charAt(i);
for (int j = 0; j < myinput.length - 1; j++) {
if (myinput.charAt(j) == myinput.charAt(j + 1)) {
returninput[j] += myinput.charAt(j + 1);
} else {
break;
}
}
} return returninput;
but there is something wrong with the second part as I cant figure out why it is not beginning a new string when the character changes.
You question says that you don't want to use regex, but I see no reason for that requirement, other than this is maybe homework. If you are open to using regex here, then there is a one line solution which splits your input string on the following pattern:
(?<=\S)(?=\s)|(?<=\s)(?=\S)
This pattern uses lookarounds to split whenever what precedes is a non whitespace character and what proceeds is a whitespace character, or vice-versa.
String input = "EE B";
String[] parts = input.split("(?<=\\S)(?=\\s)|(?<=\\s)(?=\\S)");
System.out.println(Arrays.toString(parts));
[EE, , B]
^^ a single space character in the middle
Demo
If I understood correctly, you want to split the characters in a string so that similar-consecutive characters stay together. If that's the case, here is how I would do it:
public static ArrayList<String> splitString(String str)
{
ArrayList<String> output = new ArrayList<>();
String combo = "";
//iterates through all the characters in the input
for(char c: str.toCharArray()) {
//check if the current char is equal to the last added char
if(combo.length() > 0 && c != combo.charAt(combo.length() - 1)) {
output.add(combo);
combo = "";
}
combo += c;
}
output.add(combo); //adds the last character
return output;
}
Note that instead of using an array (has a fixed size) to store the output, I used an ArrayList, which has a variable size. Also, instead of checking the next character for equality with the current one, I preferred to use the last character for that. The variable combo is used to temporarily store the characters before they go to output.
Now, here is one way to print the result following your guidelines:
public static void main(String[] args)
{
String input = "EEEE BCD DdA";
ArrayList<String> output = splitString(input);
System.out.print("[");
for(int i = 0; i < output.size(); i++) {
System.out.print("\"" + output.get(i) + "\"");
if(i != output.size()-1)
System.out.print(", ");
}
System.out.println("]");
}
The output when running the above code will be:
["EEEE", " ", "B", "C", "D", " ", "D", "d", "A"]
I am trying to find the common characters in two strings just by using the for loop. The below code is working fine, if I provide two completely different strings ex.one and two but if I provide two strings with same input ex.teen and teen it doesn't work as expected.
import java.util.Scanner;
public class CommonAlphabets {
public static void main(String[] args) {
try(Scanner input = new Scanner(System.in)){
System.out.println("Enter String one ");
String stringOne = input.nextLine();
System.out.println("Enter String two ");
String StringTwo = input.nextLine();
StringBuffer sb = new StringBuffer();
for(int i=0;i<stringOne.length();i++){
for(int j=0;j<StringTwo.length();j++){
if(stringOne.charAt(i)== StringTwo.charAt(j)){
sb.append(stringOne.charAt(i));
}
}
}
System.out.println("Common characters are " +sb.toString());
}
}
}
Should I create another nested for loop to find duplicates in the StringBuffer or is there a better way to handle this scenario.
You do not need an inner for loop but use contains instead
String stringOne = "one";
String stringTwo = "one";
StringBuilder sb = new StringBuilder();
for(int i=0;i<stringOne.length() && i < stringTwo.length ();i++){
if(stringOne.contains(String.valueOf(stringTwo.charAt(i))) &&
!sb.toString().contains(String.valueOf(stringTwo.charAt(i)))){
// check already added
sb.append(stringTwo.charAt(i));
}
}
System.out.println (sb.toString());
edit
check to make sure char to be added does not already exist in StringBuilder -
Could use a Set instead
If using a Set
Set<Character> set = new HashSet<> ();
your logic could be simplified to
if(stringOne.contains(String.valueOf(stringTwo.charAt(i)))){
set.add(stringTwo.charAt(i));
}
You can use Set for it.
Set<Character> set = new HashSet<>();
for(int i = 0; i<stringOne.length(); i++) {
for(int j = 0; j < StringTwo.length(); j++) {
if(stringOne.charAt(i) == StringTwo.charAt(j)){
set.add(stringOne.charAt(i));
}
}
}
StringBuilder sb = new StringBuilder();
for (Character c : set) {
sb.append(c);
}
System.out.println("Common characters are " + sb);
well your approach is fine as the result is showing what you are expecting there fore that code is fine, but you need to stop the duplication , therefore you have to write the code for 'sb' variable so that it will remove duplicates or write code in loop so that it wont provide duplicate.
as your code is becoming complicated to read so i would prefer that you make a method to write code to remove duplicate it will go like
static void removeDuplicate(StringBuilder s){
for(int i=0,i<s.length-1,i++){
for(int j=i+1,j<s.length,j++){
if(s.charAt(i)==s.charAt(j)){
s.deleteCharAt(j);
}
}
}
call this method before printing
Another approach you could try is - combine the two input strings, iterate over the concatenated string and return the characters which exist in both the strings.
Using a Set will ensure you do not add characters which get repeated due to the concatenation of the strings.
Here's what I wrote -
import java.util.HashSet;
public class HelloWorld {
private static Character[] findCommonLetters(String combined, String w1, String w2) {
HashSet<Character> hash = new HashSet<>();
for(char c: combined.toCharArray()) {
if(w1.indexOf(c) != -1 && w2.indexOf(c) != -1) {
hash.add(c);
}
}
return hash.toArray(new Character[hash.size()]);
}
public static void main(String []args){
// System.out.println("Hello World");
String first = "flour";
String second = "four";
String combined = first.concat(second);
Character[] result = findCommonLetters(combined, first, second);
for(char c: result) {
System.out.print(c);
}
System.out.println();
}
}
Demo here.
This is the best way to do this because it's time complexity is n so that why this is the best you could do.
import java.util.Scanner;
public class CommonAlphabets
{
public static void main(String[] args)
{
try (Scanner input = new Scanner(System.in))
{
System.out.println("Enter String one ");
String stringOne = input.nextLine();
System.out.println("Enter String two ");
String StringTwo = input.nextLine();
StringBuffer sb = new StringBuffer();
/**
* Assuming char as index of array where A-Z is from index 0 to 25 and a-z is index 26-51
*/
int[] alphabetArray1 = new int[52];
for(int i = 0, len = stringOne.length(); i < len; i++)
alphabetArray1[stringOne.charAt(i) > 94 ? stringOne.charAt(i) - 71 : stringOne.charAt(i) - 65] = 1;
int[] alphabetArray2 = new int[52];
for(int i = 0, len = StringTwo.length(); i < len; i++)
alphabetArray2[StringTwo.charAt(i) > 94 ? StringTwo.charAt(i) - 71 : StringTwo.charAt(i) - 65] = 1;
// System.out.println(Arrays.toString(alphabetArray1));
// System.out.println(Arrays.toString(alphabetArray2));
for (int i = 0; i < 52; i++)
if (alphabetArray1[i] == 1 && alphabetArray2[i] == 1)
sb.append((char) (i < 26 ? i + 65 : i + 71));
System.out.println("Common characters are " + sb.toString());
}
}
}
I've been really struggling with a programming assignment. Basically, we have to write a program that translates a sentence in English into one in Pig Latin. The first method we need is one to tokenize the string, and we are not allowed to use the Split method usually used in Java. I've been trying to do this for the past 2 days with no luck, here is what I have so far:
public class PigLatin
{
public static void main(String[] args)
{
String s = "Hello there my name is John";
Tokenize(s);
}
public static String[] Tokenize(String english)
{
String[] tokenized = new String[english.length()];
for (int i = 0; i < english.length(); i++)
{
int j= 0;
while (english.charAt(i) != ' ')
{
String m = "";
m = m + english.charAt(i);
if (english.charAt(i) == ' ')
{
j++;
}
else
{
break;
}
}
for (int l = 0; l < tokenized.length; l++) {
System.out.print(tokenized[l] + ", ");
}
}
return tokenized;
}
}
All this does is print an enormously long array of "null"s. If anyone can offer any input at all, I would reallllyyyy appreciate it!
Thank you in advance
Update: We are supposed to assume that there will be no punctuation or extra spaces, so basically whenever there is a space, it's a new word
If I understand your question, and what your Tokenize was intended to do; then I would start by writing a function to split the String
static String[] splitOnWhiteSpace(String str) {
List<String> al = new ArrayList<>();
StringBuilder sb = new StringBuilder();
for (char ch : str.toCharArray()) {
if (Character.isWhitespace(ch)) {
if (sb.length() > 0) {
al.add(sb.toString());
sb.setLength(0);
}
} else {
sb.append(ch);
}
}
if (sb.length() > 0) {
al.add(sb.toString());
}
String[] ret = new String[al.size()];
return al.toArray(ret);
}
and then print using Arrays.toString(Object[]) like
public static void main(String[] args) {
String s = "Hello there my name is John";
String[] words = splitOnWhiteSpace(s);
System.out.println(Arrays.toString(words));
}
If you're allowed to use the StringTokenizer Object (which I think is what the assignment is asking, it would look something like this:
StringTokenizer st = new StringTokenizer("this is a test");
while (st.hasMoreTokens()) {
System.out.println(st.nextToken());
}
which will produce the output:
this
is
a
test
Taken from here.
The string is split into tokens and stored in a stack. The while loop loops through the tokens, which is where you can apply the pig latin logic.
Some hints for you to do the "manual splitting" work.
There is a method String#indexOf(int ch, int fromIndex) to help you to find next occurrence of a character
There is a method String#substring(int beginIndex, int endIndex) to extract certain part of a string.
Here is some pseudo-code that show you how to split it (there are more safety handling that you need, I will leave that to you)
List<String> results = ...;
int startIndex = 0;
int endIndex = 0;
while (startIndex < inputString.length) {
endIndex = get next index of space after startIndex
if no space found {
endIndex = inputString.length
}
String result = get substring of inputString from startIndex to endIndex-1
results.add(result)
startIndex = endIndex + 1 // move startIndex to next position after space
}
// here, results contains all splitted words
String english = "hello my fellow friend"
ArrayList tokenized = new ArrayList<String>();
String m = "";
int j = 0; //index for tokenised array list.
for (int i = 0; i < english.length(); i++)
{
//the condition's position do matter here, if you
//change them, english.charAt(i) will give index
//out of bounds exception
while( i < english.length() && english.charAt(i) != ' ')
{
m = m + english.charAt(i);
i++;
}
//add to array list if there is some string
//if its only ' ', array will be empty so we are OK.
if(m.length() > 0 )
{
tokenized.add(m);
j++;
m = "";
}
}
//print the array list
for (int l = 0; l < tokenized.size(); l++) {
System.out.print(tokenized.get(l) + ", ");
}
This prints, "hello,my,fellow,friend,"
I used an array list since at the first sight the length of the array is not clear.
My code is supposed to separate a given String and covert the chosen letters into # and separate and concatenate the words with the chosen letter. My problem is with one of the methods (allWordsWith) in my code, it won't allow me to return a String array. (p.s, the codes that run this one are irrelevant, I'm not supposed to edit those).
import java.util.Arrays;
import java.util.Scanner;
public class LipogramAnalyzer {
private String line;
public LipogramAnalyzer(String text){
line = text;
}
public String mark (char letter){
String replaceletters = line.replace(letter, '#');
return replaceletters;
}
public String[] allWordsWith (char letter){
String[] arr = line.split(" ");
for ( String ss : arr) {
String ary[] = {ss};
for(int i = 0; i < ary.length; i++){
int numindex = ss.indexOf(letter);
if (numindex != -1){
String result = ary[i];
return result;
}
}
}
}
}
"It won't allow me to return a String array"
Sure it will. You're attempting to return a String:
String result = ary[i];
return result;
Even though the return type is a string array:
public String[] allWordsWith (char letter){
You need to return an array, as allWordsWith implies you want multiple values.
But the bigger problem is that you are initializing the result array to a single element
String ary[] = {ss};
and thes lengths of arrays can't be changed after initialization. This means that ary.length in this loop
for(int i = 0; i < ary.length; i++){
will always equal one. That is not what you want.
In addition, you are searching the strings in the result array (ary), even though you just created it, meaning it has nothing in it--that is, all the values are null.
If you want a list of all the strings in line that have the letter in it, try
public String[] allWordsWith (char letter){
String[] asAllWordsInLine = line.split(" ");
java.util.ArrayList<String> alsAllWordsWithChar = new java.util.ArrayList<String>();
for ( String ss : asAllWordsInLine) {
if(ss.indexOf(letter) != -1) {
alsAllWordsWithChar.add(ss);
continue; //No need to check any more letters.
}
}
return alsAllWordsWithChar.toArray(new String[alsAllWordsWithChar.size()]);
}
I've changed the array to a list, since you can't know how many strings will have letter in it. A list can change size, an array can't. When no strings match, this returns an empty array, which is preferred over null. (more info)
I've also made the variable names more meaningful, and stopped checking a word after it matches a character (with continue, which short-circuits the current for-loop iteration).
Finally, the function is not returning anything until all strings have been analyzed, meaning after the for-loop completes. In your original code the return is inside the loop, meaning only the first string is returned.
A useful testing function:
public static final void main(String[] igno_red) {
testLine('b', "abc def ghi cba def ghi");
}
private static final void testLine(char c_letter, String s_line) {
System.out.println("Line: \"" + s_line + "\"");
String[] asAllWordsWith = (new LipogramAnalyzer(s_line)).allWordsWith(c_letter);
System.out.println("Words with '" + c_letter + "': " + Arrays.toString(asAllWordsWith));
}
Return the ary not ary[i]
for(int i = 0; i < ary.length; i++){
int numindex = ss.indexOf(letter);
if (numindex != -1){
String result = ary[i];
/*here*/ return ary;
}
You can return an array. Try return ary instead of return ary[i].
Also, your function must return something in all cases. In other words, you have to have a return statement after your for loops.
public String[] allWordsWith (char letter){
String[] arr = line.split(" ");
for ( String ss : arr) {
String ary[] = {ss};
for(int i = 0; i < ary.length; i++){
int numindex = ss.indexOf(letter);
if (numindex != -1){
String result = ary[i];
// or here return ary;
}
}
}
return ary;
}
Return the array
You are trying to return a String instead of String[] (String array)
Change the code:
if (numindex != -1) {
String result = ary[i];
return result;
}
To:
if (numindex != -1) {
return new String[]{ary[i]};
}
// afrer first for loop
return arr;
If you want to return a complete String-Array at once, I have modified your function (not tested):
public String[] allWordsWith (char letter){
String[] arr = line.split(" ");
String[] result = new String[ary.length];
for ( String ss : arr) {
String ary[] = {ss};
for(int i = 0; i < ary.length; i++){
int numindex = ss.indexOf(letter);
if (numindex != -1){
result[i] = ary[i];
}
}
}
return result;
}
As already said, you're returning a String instead of a String[]. Also note that your code will only ever return a single "word". I modified your code to add every word that contains your character and add it to a List. At the end of the loop the List is converted to a String[].
public String[] allWordsWith(char letter)
{
String[] arr = line.split(" ");
List<String> result = new ArrayList<>();
for (String ss : arr) {
int numindex = ss.indexOf(letter);
if (numindex != -1) {
result.add(ss);
}
}
return result.toArray(new String[result.size()]);
}