Develop Reference

Pig it method that I am trying to make trouble checking punctuation at the end java

I am trying to answer this question.
Move the first letter of each word to the end of it, then add "ay" to the end of the word. Leave punctuation marks untouched.
This is what I did so far:
public static String pigIt(String str) {
//Populating the String argument into the String Array after splitting them by spaces
String[] strArray = str.split(" ");
System.out.println("\nPrinting strArray: " + Arrays.toString(strArray));
String toReturn = "";
for (int i = 0; i < strArray.length; i++) {
String word = strArray[i];
for (int j = 1; j < word.length(); j++) {
toReturn += Character.toString(word.charAt(j));
}
//Outside of inner for loop
if (!(word.contains("',.!?:;")) && (i != strArray.length - 1)) {
toReturn += Character.toString(word.charAt(0)) + "ay" + " ";
} else if (word.contains("',.!?:;")) {
toReturn += Character.toString(word.charAt(0)) + "ay" + " " + strArray[strArray.length - 1];
}
}
return toReturn;
}
It is supposed to return the punctuation mark without adding "ay" + "". I think I am overthinking but please help. Please see the below debugger.

One of the problems here is that your else if statement is never being invoked. The .contains method will not work with multiple characters like that unless you are trying to match them all. In your conditions you are essentially asking if the word matches that entire string "',.!?:;". If you just keep the exclamation point in there it will work invoke it. I don't know how else you can use contains besides making a condition for each one like word.contains("!")|| word.contains(",")|| word.contains("'"), etc.. You can also use regex for this problem.
Alternatively, you can use something like,
Character ch = new Character(yourString.charAt(i));
if(!Character.isAlphabetic(yourString.charAt(i))) {
to determine if a character is not an alphabetical one, and is a symbol or punctuation.

I think the best way is not relay on str.split("\\s++"), because you could have punctuation in any plase. The best one is to look through the string and find all not letter or digit symbols. After that you can define a word borders and translate it.
public static String pigIt(String str) {
StringBuilder buf = new StringBuilder();
for (int i = 0, j = 0; j <= str.length(); j++) {
char ch = j < str.length() ? str.charAt(j) : '\0';
if (Character.isLetterOrDigit(ch))
continue;
if (i < j) {
buf.append(str.substring(i + 1, j));
buf.append(str.charAt(i));
buf.append("ay");
}
if (ch != '\0')
buf.append(ch);
i = j + 1;
}
return buf.toString();
}
Output:
System.out.println(pigIt(",Hello, !World")); // ,elloHay, !orldWay

Regex may be difficult to start with but is very powerful:
public static String pigIt(String str) {
return str.replaceAll("([a-zA-Z])([a-zA-Z]*)", "$2$1ay");
}
The () specify groups. So I have one group with the first alphabet character and a second group with the remaining alphabet characters.
In the replace parameter you can refer to these groups ($1, $2).
String.replaceAll will search all matching string parts and apply the replacement. Non matching characters like the punctuations are left untouched.
public static void main(String[] args) {
System.out.println("Hello, World, ! -->"+ pigIt("Hello, World, !"));
System.out.println("Hello?, Wo$, F, ! -->"+ pigIt("Hello?, Wo$, F, !"));
}
The output of this method is:
Hello, World, ! -->elloHay, orldWay, !
Hello?, Wo$, F, ! -->elloHay?, oWay$, Fay, !

How to check if a String can be formed from the characters of another String in Java?

A string is good if it can be formed by characters from chars. I want to return the sum of lengths of all good strings in words.
Input: words = ["cat","bt","hat","tree"], chars = "atach"
Output: 6
Explanation:
The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.
Below is the code that I have written.
class Solution
{
public int countCharacters(String[] words, String chars)
{
int k = 0, count = 0;
Set<Character> set = new HashSet<>();
for(int i = 0; i < chars.length(); i++)
{
set.add(chars.charAt(i));
}
StringBuilder chrs = new StringBuilder();
for(Character ch : set)
{
chrs.append(ch);
}
for(int i = 0; i < words.length; i++)
{
char[] ch = words[i].toCharArray();
for(int j = 0; j < ch.length; j++)
{
if(chrs.contains("" + ch[j]))
{
k++;
}
}
if(k == words[i].length())
{
count+= k;
}
}
return count;
}
}
Output:
Line 24: error: cannot find symbol
if(chrs.contains("" + ch[j]))
Can someone help me? What am I doing wrong in accessing the character?

The issues which I noticed is you are using contains() to compare a String and a character. But the contains() method is a Java method to check if String contains another substring or not.
So you can solve this by converting the character to a string.
Ex 1:
if(chars.contains(Character.toString(ch[j]))){
k++;
} else {
}
Ex 2:
f(chars.contains(""+ch[j]))
{
k++;
} else {
}
Otherwise, You can compare if the string contains a char by using indexOf(). If the string isn't containing the char it return -1. Please refer bellow example.
Ex:
if(chars.indexOf(ch[j])!=-1){
k++;
} else {
}

contains tells you if a string is contained in another string. But in your case ch[j] is not a string but a char, so you can't use contains.
Instead, use indexOf, it returns -1 if the char is not present in the string.

the most simple way is
chars.contains("" + ch[i]);

Here, ch[j] is not a string but a char, so you can't use contains as you've done. Instead, make the following change.
chars.contains(String.valueOf(ch[j]));

How to reverse a String after a comma and then print the 1st half of the String Java

For example String grdwe,erwd becomes dwregrdwe
I have most of the code I just have trouble accessing all of ch1 and ch2 in my code after my for loop in my method I think I have to add all the elements to ch1 and ch2 into two separate arrays of characters but I wouldn't know what to initially initialize the array to it only reads 1 element I want to access all elements and then concat them. I'm stumped.
And I'd prefer to avoid Stringbuilder if possible
public class reverseStringAfterAComma{
public void reverseMethod(String word){
char ch1 = ' ';
char ch2 = ' ';
for(int a=0; a<word.length(); a++)
{
if(word.charAt(a)==',')
{
for(int i=word.length()-1; i>a; i--)
{
ch1 = word.charAt(i);
System.out.print(ch1);
}
for (int j=0; j<a; j++)
{
ch2 = word.charAt(j);
System.out.print(ch2);
}
}
}
//System.out.print("\n"+ch1);
//System.out.print("\n"+ch2);
}
public static void main(String []args){
reverseStringAfterAComma rsac = new reverseStringAfterAComma();
String str="grdwe,erwd";
rsac.reverseMethod(str);
}
}

You can use string builder as described here:
First split the string using:
String[] splitString = yourString.split(",");
Then reverse the second part of the string using this:
splitString[1] = new StringBuilder(splitString[1]).reverse().toString();
then append the two sections like so:
String final = splitString[1] + splitString[0];
And if you want to print it just do:
System.out.print(final);
The final code would be:
String[] splitString = yourString.split(",");
splitString[1] = new StringBuilder(splitString[1]).reverse().toString();
String final = splitString[1] + splitString[0];
System.out.print(final);
Then, since you are using stringbuilder all you need to do extra, is import it by putting this at the top of your code:
import java.lang.StringBuilder;

It appears you currently have working code, but are looking to print/save the value outside of the for loops. Just set a variable before you enter the loops, and concatenate the chars in each loop:
String result = "";
for (int a = 0; a < word.length(); a++) {
if (word.charAt(a) == ',') {
for (int i = word.length() - 1; i > a; i--) {
ch1 = word.charAt(i);
result += ch1;
}
for (int j = 0; j < a; j++) {
ch2 = word.charAt(j);
result += ch2;
}
}
}
System.out.println(result);
Demo

Let propose a solution that doesn't use a StringBuilder
You should knoz there is no correct reason not to use that class since this is well tested
The first step would be to split your String on the first comma found (I assumed, in case there is more than one, that the rest are part of the text to reverse). To do that, we can you String.split(String regex, int limit).
The limit is define like this
If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n and the array's last entry will contain all input beyond the last matched delimiter.
If n is non-positive then the pattern will be applied as many times as possible and the array can have any length.
If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.
Example :
"foobar".split(",", 2) // {"foobar"}
"foo,bar".split(",", 2) // {"foo", "bar"}
"foo,bar,far".split(",", 2) // {"foo", "bar,far"}
So this could be used at our advantage here :
String text = "Jake, ma I ,dlrow olleh";
String[] splittedText = text.split( ",", 2 ); //will give a maximum of a 2 length array
Know, we just need to reverse the second array if it exists, using the simplest algorithm.
String result;
if ( splittedText.length == 2 ) { //A comma was found
char[] toReverse = splittedText[1].toCharArray(); //get the char array to revese
int start = 0;
int end = toReverse.length - 1;
while ( start < end ) { //iterate until needed
char tmp = toReverse[start];
toReverse[start] = toReverse[end];
toReverse[end] = tmp;
start++; //step forward
end--; //step back
}
result = new String( toReverse ) + splittedText[0];
}
This was the part that should be done with a StringBuilder using
if ( splittedText.length == 2 ){
result = new StringBuilder(splittedText[1]).reverse().toString() + splittedText[0];
}
And if there is only one cell, the result is the same as the original text
else { //No comma found, just take the original text
result = text;
}
Then we just need to print the result
System.out.println( result );
hello world, I am Jake

Java - make new string based on old one and lag

I need to get a new string based on an old one and a lag. Basically, I have a string with the alphabet (s = "abc...xyz") and based on a lag (i.e. 3), the new string should replace the characters in a string I type with the character placed some positions forward (lag). If, let's say, I type "cde" as my string, the output should be "fgh". If any other character is added in the string (apart from space - " "), it should be removed. Here is what I tried, but it doesn't work :
String code = "abcdefghijklmnopqrstuvwxyzabcd"; //my lag is 4 and I added the first 4 characters to
char old; //avoid OutOfRange issues
char nou;
for (int i = 0; i < code.length() - lag; ++i)
{
old = code.charAt(i);
//System.out.print(old + " ");
nou = code.charAt(i + lag);
//System.out.println(nou + " ");
// if (s.indexOf(old) != 0)
// {
s = s.replace(old, nou);
// }
}
I commented the outputs for old and nou (new, but is reserved word) because I have used them only to test if the code from position i to i + lag is working (and it is), but if I uncomment the if statement, it doesn't do anything and I leave it like this, it keeps executing the instructions inside the for statmement for code.length() times, but my string doesn't need to be so long. I have also tried to make the for statement like below, but I got lost.
for (int i = 0; i < s.length(); ++i)
{
....
}
Could you help me with this? Or maybe some advices about how I should think the algorithm?
Thanks!

It doesn't work because, as the javadoc of replace() says:
Returns a new string resulting from replacing all occurrences of oldChar in this string with newChar.
(emphasis mine)
So, the first time you meet an 'a' in the string, you replace all the 'a's by 'd'. But then you go to the next char, and if it's a 'd' that was an 'a' before, you replace it once again, etc. etc.
You shouldn't use replace() at all. Instead, you should simply build a new string, using a StringBuilder, by appending each shifted character of the original string:
String dictionary = "abcdefghijklmnopqrstuvwxyz";
StringBuilder sb = new StringBuilder(input.length());
for (int i = 0; i < input.length(); i++) {
char oldChar = input.charAt(i);
int oldCharPositionInDictionary = dictionary.indexOf(oldChar);
if (oldCharPositionInDictionary >= 0) {
int newCharPositionInDictionary =
(oldCharPositionInDictionary + lag) % dictionary.length();
sb.append(dictionary.charAt(newCharPositionInDictionary));
}
else if (oldChar == ' ') {
sb.append(' ');
}
}
String result = sb.toString();

Try this:
Convert the string to char array.
iterate over each char array and change the char by adding lag
create new String just once (instead of loop) with new String passing char array.
String code = "abcdefghijklmnopqrstuvwxyzabcd";
String s = "abcdef";
char[] ch = s.toCharArray();
char[] codes = code.toCharArray();
for (int i = 0; i < ch.length; ++i)
{
ch[i] = codes[ch[i] - 'a' + 3];
}
String str = new String(ch);
System.out.println(str);
}

My answer is something like this.
It returns one more index to every character.
It reverses every String.
Have a good day!
package org.owls.sof;
import java.util.Scanner;
public class Main {
private static final String CODE = "abcdefghijklmnopqrstuvwxyz"; //my lag is 4 and I added the first 4 characters to
#SuppressWarnings("resource")
public static void main(String[] args) {
System.out.print("insert alphabet >> ");
Scanner scanner = new Scanner(System.in);
String s = scanner.next();
char[] char_arr = s.toCharArray();
for(int i = 0; i < char_arr.length; i++){
int order = CODE.indexOf(char_arr[i]) + 1;
if(order%CODE.length() == 0){
char_arr[i] = CODE.charAt(0);
}else{
char_arr[i] = CODE.charAt(order);
}
}
System.out.println(new String(char_arr));
//reverse
System.out.println(reverse(new String(char_arr)));
}
private static String reverse (String str) {
char[] char_arr = str.toCharArray();
for(int i = 0; i < char_arr.length/2; i++){
char tmp = char_arr[i];
char_arr[i] = char_arr[char_arr.length - i - 1];
char_arr[char_arr.length - i - 1] = tmp;
}
return new String(char_arr);
}
}

String alpha = "abcdefghijklmnopqrstuvwxyzabcd"; // alphabet
int N = alpha.length();
int lag = 3; // shift value
String s = "cde"; // input
StringBuilder sb = new StringBuilder();
for (int i = 0, index; i < s.length(); i++) {
index = s.charAt(i) - 'a';
sb.append(alpha.charAt((index + lag) % N));
}
String op = sb.toString(); // output

Delete next two characters in string with indexOf and map

i have a problem with a algorithm.
I have a Map (Each key int its a hex unicode character) and a String with unicode characters.
I want to delete the next character in the string when i found a character that exists as key in my map.
for example my map contains those keys: 0x111,0x333,0x444,0x555,0x666 and my string its:
0x111+0xffff+0x444+0xEEEEE+0x666
I want to convert it to:
0x111+0x444+0x666
I have this but this doesnt work:
private String cleanFlags(String text) {
int textLong = text.length();
for (int i = 0; i < textLong; i++) {
if (flagCountryEmojis.containsKey(text.codePointAt(text.charAt(i)))) {
text = text.replace(text.substring(i + 1, i + 2), "");
textLong-=2;
}
}
return text;
}
How can i do it this?

Since you didn't mention anything about space complexity, I went ahead and took the liberties of using an array to solve the question:
public String cleanFlags(String text){
String [] arr = text.split("+");
String newText = "";
for(int i = 0; i < arr.length; i++){
if(flagCountryEmojis.containsKey(arr[i])){
newText += arr[i];
i++; // skips the next character
}
if(i < arr.length - 1)
newText += "+";
}
return newText;
}
Not sure if this solution solves your problem, since strings are immutable anyways, and calling "replace" simply creates a new string in the background, I went ahead and created a new string for you and returned the result when it is populated correctly.
Lemme know if there is something I am missing or other restrictions that were unmentioned.