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How I return specific string from recursive function in java?

Input for method :
word = AbcDef
input = 3
w = AbcDef
The Following code showing the ouput :
fAbcDe
efAbcD
DefAbc
I want to return only DefAbc.
How can I do the coding for return keyword.
public static String match2(String word,int input,String w)
{
StringBuilder st = new StringBuilder();
StringBuilder st1 = new StringBuilder();
String str = "";
int count = 0;
st.append(word.charAt(i));
for(int j = 0;j < i;j++)
{
st.append(word.charAt(j));
}
if(input!=0)
{
str = st.toString();
System.out.println(str);
int input2 = input-1;
match2(str,input2,w);
}
return null;
}

Please try this one. Hope it might solve your problem using recursion-> tillValue should be->3 if you want to rotate for 3 characters
static String recurse(String word, int count, int tillValue) {
if (count == tillValue) {
return word;
}
String recurse = recurse(word, count + 1,tillValue);
word = swap(recurse);
return word;
}
private static String swap(String word) {
if (word.length() > 0) {
String c = word.substring(word.length() - 1);
String dc = c + word.substring(0, word.length() - 1);
System.out.println(dc);
return dc;
}
return "";
}

parsing/converting task with characters and numbers within

It is necessary to repeat the character, as many times as the number behind it.
They are positive integer numbers.
case #1
input: "abc3leson11"
output: "abccclesonnnnnnnnnnn"
I already finish it in the following way:
String a = "abbc2kd3ijkl40ggg2H5uu";
String s = a + "*";
String numS = "";
int cnt = 0;
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
if (Character.isDigit(ch)) {
numS = numS + ch;
cnt++;
} else {
cnt++;
try {
for (int j = 0; j < Integer.parseInt(numS); j++) {
System.out.print(s.charAt(i - cnt));
}
if (i != s.length() - 1 && !Character.isDigit(s.charAt(i + 1))) {
System.out.print(s.charAt(i));
}
} catch (Exception e) {
if (i != s.length() - 1 && !Character.isDigit(s.charAt(i + 1))) {
System.out.print(s.charAt(i));
}
}
cnt = 0;
numS = "";
}
}
But I wonder is there some better solution with less and cleaner code?

Could you take a look below? I'm using a library from StringUtils from Apache Common Utils to repeat character:
public class MicsTest {
public static void main(String[] args) {
String input = "abc3leson11";
String output = input;
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(input);
while (m.find()) {
int number = Integer.valueOf(m.group());
char repeatedChar = input.charAt(m.start()-1);
output = output.replaceFirst(m.group(), StringUtils.repeat(repeatedChar, number));
}
System.out.println(output);
}
}
In case you don't want to use StringUtils. You can use the below custom method to achieve the same effect:
public static String repeat(char c, int times) {
char[] chars = new char[times];
Arrays.fill(chars, c);
return new String(chars);
}

Using java basic string regx should make it more terse as follows:
public class He1 {
private static final Pattern pattern = Pattern.compile("[a-zA-Z]+(\\d+).*");
// match the number between or the last using regx;
public static void main(String... args) {
String s = "abc3leson11";
System.out.println(parse(s));
s = "abbc2kd3ijkl40ggg2H5uu";
System.out.println(parse(s));
}
private static String parse(String s) {
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
int num = Integer.valueOf(matcher.group(1));
char prev = s.charAt(s.indexOf(String.valueOf(num)) - 1);
// locate the char before the number;
String repeated = new String(new char[num-1]).replace('\0', prev);
// since the prev is not deleted, we have to decrement the repeating number by 1;
s = s.replaceFirst(String.valueOf(num), repeated);
matcher = pattern.matcher(s);
}
return s;
}
}
And the output should be:
abccclesonnnnnnnnnnn
abbcckdddijkllllllllllllllllllllllllllllllllllllllllggggHHHHHuu

String g(String a){
String result = "";
String[] array = a.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)");
//System.out.println(java.util.Arrays.toString(array));
for(int i=0; i<array.length; i++){
String part = array[i];
result += part;
if(++i == array.length){
break;
}
char charToRepeat = part.charAt(part.length() - 1);
result += repeat(charToRepeat+"", new Integer(array[i]) - 1);
}
return result;
}
// In Java 11 this could be removed and replaced with the builtin `str.repeat(amount)`
String repeat(String str, int amount){
return new String(new char[amount]).replace("\0", str);
}
Try it online.
Explanation:
The split will split the letters and numbers:
abbc2kd3ijkl40ggg2H5uu would become ["abbc", "2", "kd", "3", "ijkl", "40", "ggg", "2", "H", "5", "uu"]
We then loop over the parts and add any strings as is to the result.
We then increase i by 1 first and if we're done (after the "uu") in the array above, it will break the loop.
If not the increase of i will put us at a number. So it will repeat the last character of the part x amount of times, where x is the number we found minus 1.

Here is another solution:
String str = "abbc2kd3ijkl40ggg2H5uu";
String[] part = str.split("(?<=\\d)(?=\\D)|(?=\\d)(?<=\\D)");
String res = "";
for(int i=0; i < part.length; i++){
if(i%2 == 0){
res = res + part[i];
}else {
res = res + StringUtils.repeat(part[i-1].charAt(part[i-1].length()-1),Integer.parseInt(part[i])-1);
}
}
System.out.println(res);

Yet another solution :
public static String getCustomizedString(String input) {
ArrayList<String > letters = new ArrayList<>(Arrays.asList(input.split("(\\d)")));
letters.removeAll(Arrays.asList(""));
ArrayList<String > digits = new ArrayList<>(Arrays.asList(input.split("(\\D)")));
digits.removeAll(Arrays.asList(""));
for(int i=0; i< digits.size(); i++) {
int iteration = Integer.valueOf(digits.get(i));
String letter = letters.get(i);
char c = letter.charAt(letter.length()-1);
for (int j = 0; j<iteration -1 ; j++) {
letters.set(i,letters.get(i).concat(String.valueOf(c)));
}
}
String finalResult = "";
for (String str : letters) {
finalResult += str;
}
return finalResult;
}
The usage:
public static void main(String[] args) {
String testString1 = "abbc2kd3ijkl40ggg2H5uu";
String testString2 = "abc3leson11";
System.out.println(getCustomizedString(testString1));
System.out.println(getCustomizedString(testString2));
}
And the result:
abbcckdddijkllllllllllllllllllllllllllllllllllllllllggggHHHHHuu
abccclesonnnnnnnnnnn

Using StringBuilder getting null as output

I am doing one coding question in which I try to decrypt the input string. The procedure for the decryption is:
from 0 to 9 it represent alphabets from a to i.
then 10# represent j, 11# represent k and so.
import java.util.HashMap;
public class Julia {
public static void main(String[] args) {
String s="10#21#12#91";
Julia obj=new Julia();
String result=obj.decrypt(s);
System.out.println(result);
}
public String decrypt(String msg)
{
HashMap<String,Character> hs=new HashMap<>();
hs.put("1",'a');
hs.put("2",'b');
hs.put("3",'c');
hs.put("4",'d');
hs.put("5",'e');
hs.put("6",'f');
hs.put("7",'g');
hs.put("8",'h');
hs.put("9",'i');
hs.put("10",'j');
hs.put("11",'k');
hs.put("12",'l');
hs.put("13",'m');
hs.put("14",'n');
hs.put("15",'o');
hs.put("16",'p');
hs.put("17",'q');
hs.put("18",'r');
hs.put("19",'s');
hs.put("20",'t');
hs.put("21",'u');
hs.put("22",'v');
hs.put("23",'w');
hs.put("24",'x');
hs.put("25",'y');
hs.put("26",'x');
StringBuilder n=new StringBuilder();
for(int i=msg.length()-1;i>=0;i--)
{
if(msg.charAt(i)=='#' && i>=2)
{
StringBuilder s=new StringBuilder().append(msg.charAt(i-2)).append(msg.charAt(i-1));
System.out.println(s);
n.append(hs.get(s));
System.out.println(n);
i=i-2;
}
else
{
n.append(hs.get(msg.charAt(i)));
}
}
return n.toString();
}
}
That is code I wrote. But the output I am getting is nullnullnullnullnull.
I think the issue is with StringBuilder. Can anyone help me with that and explain the concept? If someone has better solution please guide.

You should not use data (a map) when you could have used a simple formula.
My suggestion:
import java.util.ArrayList;
import java.util.List;
public final class Julia {
public static void main(final String[] args) {
final String s = "10#21#12#91";
final String result = decrypt(s);
System.out.println(result);
}
private static String decrypt(final String s) {
final List<Integer> crypt = new ArrayList<>();
final String[] groups = s.split("#");
for (int i = 0; i < groups.length; i++) {
final String group = groups[i];
int j = 0;
// Special case for last group
if ((i == (groups.length - 1)) && !s.endsWith("#")) {
j = group.length();
}
if (group.length() > 2) {
j = group.length() - 2;
}
for (int k = 0; k < j; k++) {
crypt.add(Integer.valueOf(group.substring(k, k + 1)));
}
if (j < group.length()) {
crypt.add(Integer.valueOf(group.substring(j, group.length())));
}
}
final StringBuilder n = new StringBuilder(crypt.size());
for (final Integer c : crypt) {
final char d = (char) (('a' + c) - 1);
n.append(d);
}
return n.toString();
}
}
Please note that there are two mistakes in the question: The letter a is 1, not zero, and the value for 26 is z, not x. The latter error is typical when you use data where a formula would do.
Since you are learning, I would note that the decrypt methods - both my suggestion and yours - should be static since they do not use any fields, so the instantiation is not necessary.

This is Pattern Matching problem which can be solved by Regex.
Your code has some bugs and those are already pointed out by others. I don't see any solution which looks better than a simple regex solution.
Below regex code will output 'julia' for input '10#21#12#91'.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Julia {
public static void main(String[] args) {
String s="10#21#12#91";
Julia obj=new Julia();
String result=obj.decrypt(s);
System.out.println(result);
}
public String decrypt(String msg)
{
Pattern regex = Pattern.compile("((\\d\\d#)|(\\d))");
Matcher regexMatcher = regex.matcher(msg);
StringBuffer result = new StringBuffer();
while (regexMatcher.find())
regexMatcher.appendReplacement(result, getCharForNumber(Integer.parseInt(regexMatcher.group(1).replace("#",""))));
return result.toString();
}
private String getCharForNumber(int i) {
return i > 0 && i < 27 ? String.valueOf((char)(i + 96)) : null;
}
}
I hope it helps.

hs.get(s) will always return null, since s is not a String.
Try hs.get(s.toString())
hs.get(msg.charAt(i)) will also always return null, since you are passing a char to get instead of String.
There may also be logic problems in your code, but it's hard to tell.

Optimized version of your code
public class Main {
public static void main(String[] args) {
String cipher = "10#21#12#91";
System.out.print(decrypt(cipher));
//output : julia
}
static String decrypt(String cipher) {
//split with # to obtain array of code in string array
String[] cipher_char_codes = cipher.split("#");
//create empty message
StringBuilder message = new StringBuilder();
//loop for each code
for (String code : cipher_char_codes) {
//get index of character
int index = Integer.parseInt(code);
if (index > 26) {
char[] pair = code.toCharArray();
for (int i = 0; i < pair.length; i++) {
int x = Integer.parseInt("" + code.charAt(i));
message.append((char) ('a' + ((x - 1) % 26)));
}
} else {
//map index into 1 to 26
//find ascii code and cast into char
message.append((char) ('a' + ((index - 1) % 26)));
}
}
return message.toString();
}
}

Regex is indeed the way to go, and the code proposed by Pirate_Jack can be improved. It calls the expensive regex two superfluous times (replace is a regex operation).
Following is a yet improved version:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public final class Julia3 {
public static void main(final String[] args) {
final String s = "10#21#12#91";
final String result = decrypt(s);
System.out.println(result);
}
public static String decrypt(final String msg) {
final Pattern regex = Pattern.compile("((\\d\\d)(#)|(\\d))");
final Matcher regexMatcher = regex.matcher(msg);
final StringBuffer result = new StringBuffer();
String c;
while (regexMatcher.find()) {
if (regexMatcher.group(2) == null) {
c = regexMatcher.group(1);
} else {
c = regexMatcher.group(2);
}
result.append((char) ((Integer.parseInt(c) + 'a') - 1));
}
return result.toString();
}
}

This is not right :
hs.get(s)
s is a StringBuilder. It should be hs.get(Char)
Edit: an optional different solution:
public class Julia {
public static void main(String[] args) {
String s="10#21#12#91";
List<String> numbers = splitToNumbers(s);
Julia obj=new Julia();
String result=obj.decrypt(numbers);
System.out.println(result);
}
/**
*#param s
*#return
*/
private static List<String> splitToNumbers(String s) {
//add check s is not null
char[] chars = s.toCharArray();
char delimiter = '#';
List<String> numberAsStrings = new ArrayList<String>();
int charIndex = 0;
while (charIndex < (chars.length -3)) {
char theirdChar = chars[charIndex+2];
if(theirdChar == delimiter) {
numberAsStrings.add(""+chars[charIndex]+chars[charIndex+1]);
charIndex +=3;
}else {
numberAsStrings.add(""+chars[charIndex]);
charIndex ++;
}
}
//add what's left
while (charIndex < chars.length) {
numberAsStrings.add(""+chars[charIndex]);
charIndex++;
}
return numberAsStrings;
}
public String decrypt(List<String> numbersAsStings){
StringBuilder sb=new StringBuilder();
for (String number : numbersAsStings) {
int num = Integer.valueOf(number);
sb.append(intToChar(num-1));
}
return sb.toString();
}
private char intToChar(int num) {
if((num<0) || (num>25) ) {
return '?' ;
}
return (char)('a' + num);
}
}

Replacing substrings with Strings not using replace()

The problem is: mutation() is passed two Strings and returns a String. Every occurrence in the first String of fzgh is replaced with the second String.
mutation("Hello. I want an fzgh. Give me an fzgh now.", "IPhone 6")-> "Hello. I want an IPhone 6. Give me an IPhone 6 now."
Here's my attempt:
public static String mutation(String s, String t){
int f=s.indexOf("fzgh");
String w="";
if(f !=-1){
w=w+s.substring(0,f)+t;
}
return w;
}
I know there's .replace(), but we are not allowed to use it. We have to use indexOf()

You can define mutation like this
public static String mutation(String s,String t){
int f=s.indexOf("fzgh");
int l =4;//length of "fzgh"
String w = s;
while(f!=-1){
w=w.substring(0,f)+t+w.substring(f+l,w.length());
f=w.indexOf("fzgh");
}
return w;
}
This will remove all the "fzgh" from String s and will replace them with String t.

This can be easily solved if you can use split and join methods.
public static String mutation(String s, String t) {
return String.join(t, s.split("fzgh"));
}
split will split the input string around "fzgh" and return an array of String, in this case, it will return ["Hello. I want an ", ". Give me an ", " now."]. Now you can join the each element together with "IPhone 6" by using join method.
Notice that join is introduced in Java8.
But if you should only indexOf and substring to solve this problem, the code would be more complicated, but not so complex.
public static String mutation(String s, String t) {
StringBuilder sb = new StringBuilder();
int p = 0, q = 0;
while ((q = s.indexOf("fzgh", p)) >= 0) {
sb.append(s.substring(p, q));
sb.append(t);
p = q+4;
}
sb.append(s.substring(p));
return sb.toString();
}

Its kinda clumsy but there you go, i will edit meanwhile if u need explanation :)
*Edit: its not good enough yet, will try to fix it. This is kinda other way around, kinda different solution but not that good, will delete it if i fail to fix.
*Edit #2: Fixed
import java.util.Scanner;
import java.io.*;
class Mutation
{
public static String mutation(String s, String t){
String toBeReplaced = "fzgh";
int sizeOfThatString = toBeReplaced.length();
String[] array = s.split(" ");
for (int i = 0; i < array.length; i++) {
String tempString= array[i];
if(tempString.length()>sizeOfThatString) {
if(tempString.contains(toBeReplaced)) {
String other = array[i].substring(sizeOfThatString);
array[i] = t+other;
}
}
if(tempString.equals(toBeReplaced)) {
array[i] = t;
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < array.length; i++) {
if(i!=array.length) {
sb.append(array[i]+" ");
}
else {
sb.append(array[i]);
}
}
return sb.toString();
}
public static void main(String[] args) {
Arabic_Randomizer ar= new Arabic_Randomizer();
System.out.println(ar.mutation("Hello. I want an fzgh. Give me an fzgh now.", "IPhone 6"));
}
}

use this code
public class run {
public static String mutation(String w,String s, String t){
String toBeReplaced = w;
int sizeOfThatString = toBeReplaced.length();
String[] array = s.split(" ");
for (int i = 0; i < array.length; i++) {
String tempString= array[i];
if(tempString.length()>sizeOfThatString) {
tempString = array[i].substring(0,sizeOfThatString);
System.out.println("TempString " +tempString);
}
if(tempString.equals(w)) {
array[i] = t;
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < array.length; i++) {
if(i!=array.length) {
sb.append(array[i]+" ");
}
else {
sb.append(array[i]);
}
}
return sb.toString();
}
public static void main(String[] args) {
System.out.println(mutation("fzgh","Hello. I want an fzgh. Give me an fzgh now.", "IPhone 6"));
}
}
out:
TempString Hell
TempString fzgh
Hello. I want an IPhone 6 Give me an IPhone 6 now.

Replace all instances of a character in a String

I'm trying to create a method that replace all instances of a certain character in a word with a new character. This is what I have so far:
public class practice {
public static void main(String[] args) {
String test3 = updatePartialword("----", "test", 't');
System.out.println(test3); }
public static String updatePartialword(String partial, String secret, char c) {
String newPartial = "";
int len = secret.length();
for (int i=0; i<=secret.length()-1; i++){
char x = secret.charAt(i);
if (c==x) {
String first = partial.substring(0,i);
String second = partial.substring(i+1,len);
newPartial = first+x+second;
}
}
return newPartial;
}
}
I want it to return t--t, but it will only print the last t. Any help would be greatly appreciated!

Java already has a built in method in String for this. You can use the the replace() method to replace all occurrences of the given character in the String with the another character
String str = "Hello";
str.replace('l', '-'); //Returns He--o
str.replace('H', '-'); //Returns -ello

I suspect you are looking for something like
public static void main(String[] args) {
String test3 = updatePartialword("----", "test", 't');
System.out.println(test3);
}
public static String updatePartialword(String partial, String secret, char c) {
char[] tmp = partial.toCharArray();
for (int i = 0; i < secret.length(); i++) {
char x = secret.charAt(i);
if (c == x) {
tmp[i] = c;
}
}
return new String(tmp);
}

In your code you overwrite the String each time you found the character. Instead of overwriting, you should expand the string each time.
public class practice {
public static void main(String[] args) {
String test3 = updatePartialword("----", "test", 't');
System.out.println(test3);
}
public static String updatePartialword(String partial, String secret, char c) {
StringBuilder sb = new Stringbuilder();
sb.append(""); // to prevent the Stringbuilder from calculating with the chars
for (int i = 0; i < partial.lenght; i++)
if (secret.charAt(i) == c)
sb.append(c);
else
sb.append('-');
return sb.toString();
}
}