Develop Reference

how to split a string by using charAt and string.length()

only allow charAt method and length method . Thank you so much!
void runApp() {
String str = "345, 688"; //->"345" "688"
String value = strCut(str);
}
String strCut(String str) {
int result = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(3) == ',') {
what should i write here ? ?
}

Your code needs some refactoring, try this:
void runApp() {
String str = "345, 688"; //->"345" "688"
String value = strCut(str);
}
String strCut(String str) {
int result = 0;
for (int i = 0; i < str.length(); i++) {
int cutStringIndex;
if (str.charAt(i) == ',') {
cutStringIndex = i;
}
for (int i = 0; i < cutStringIndex(); i++) {
String cutStringOne = "";
cutStringOne = cutStringOne + str.charAt(i);
}
for (int i = cutStringIndex() + 1; i < str.length(); i++) {
String cutStringTwo = "";
cutStringTwo = cutStringTwo + str.charAt(i);
}
cutString = cutStringOne + " " + cutStringTwo;
return cutString;
}
This will take out the comma which appears to be what you were looking for. I only used the two methods you asked for. Essentially this code gets the index of the comma, then reconstructs the two parts of the strings until it reaches the point of the comma, and skips over it. It may need some minor tweaks for your situation but this should be what you're looking for.

It can be done like this, Suppose String s="200,300,450,600" and you have to split given string using charAt() and string.length() method then first add ',' at the end of the string as given in the code below.
String s="200,300,450,600,",str="";
for(int i=0;i<s.length();i++){
char ch=s.charAt(i);
if(ch!=','){ //checking if particular character is not ','
str+=ch; //storing it in str string
}
else{
System.out.println(str); //printing each string before ',' is found
str="";
}
}
The output of above code will be:200
300
450
600(all the numbers will be printed on next line)

If you want to use only charAt and string.length() then you should try this
void runApp{
String str = "345, 688, 123";
String values[] = strCut(str); //values[0] = 345, values[1] = 688, values[2] = 123
for(String value : values){
System.out.print(value + " ");
}
}
String[] strCut(String str) {
int elements = 1;
int index = 0;
for(int i = 0; i < str.length(); i++){
if(str.charAt(i) == ',')
elements++;
}
String result[] = new String[elements];
for(int i = 0; i < result.length; i++)
result[i] = "";
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) != ',') {
if(str.charAt(i) != ' ')
result[index] = result[index] + str.charAt(i);
}
else index++;
}
return result;
}

You can do it as follows:
public class Main {
public static void main(String[] args) {
// Test
runApp();
}
static void runApp() {
String str = "345, 688"; // Expected->"345" "688"
String value = strCut(str);
System.out.println(value);// Display the result
}
static String strCut(String str) {
// Initialise result with a "
String result = "\"";
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == ',') {// Check char at the index, i
// Add " at the end of one number and again " at the start of the next
result += "\" \"";
} else if (str.charAt(i) != ' ') {
result += str.charAt(i);
}
}
// Add " at the end
result += "\"";
// Finally, return result
return result;
}
}
Output:
"345" "688"

if you must want to make use of charAt() then do like below..
ArrayList<String> stringArr= new ArrayList<String>();
int startindex=0;
for (int i = 0; i < str.length(); i++)
{
if (str.charAt(i) == ',')
{
String partString = str.substring(startindex, i) ;
startindex=i+1;
stringArr.add(partString);
}
}
String lastString = str.substring(startindex, str.length()) ;
stringArr.add(lastString);
OR
You can simply use split method like below
String[] parts = string.split(",");
String part1 = parts[0]; // 345
String part2 = parts[1]; // 688

You can achieve it by simply doing this,
This will give you the desired result.
String str = "345,688";
ArrayList<String> stringArray = new ArrayList<>();
int startindex=0;
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == ',') {
String subStr = str.substring(startindex, i);
startindex = i+1;
stringArray.add(subStr);
}
}
stringArray.add(str.substring(startindex));

Reverse encoded string

i am asking for help on how i would go about reversing my code so that the input 'A2B5C2' will give me the output 'AABBBBBCC', any suggestions?
Thanks
public static void printRLE(String str) {
int n = str.length();
for (int i = 0; i < n; i++) {
// Count occurrences of current character
int count = 1;
while (i < n - 1 && str.charAt(i) == str.charAt(i + 1)) {
count++;
i++;
}
// Print character and its count
System.out.print(str.charAt(i));
System.out.print(count);
}
}
public static void main(String[] args) {
String str = "AABBBBBCC";
printRLE(str);
}

To get the case, the number will more than 9, I'd suggest a simple regex to match letter+number, then just repeat the letter the number of times you need :
static String getRevRLE(String str) {
StringBuilder res = new StringBuilder();
Matcher m = Pattern.compile("([a-zA-Z][0-9]+)").matcher(str);
while (m.find()) {
String g = m.group();
res.append(g.substring(0, 1).repeat(Integer.parseInt(g.substring(1))));
}
return res.toString();
}
Using the Streams API you can reduce to
static String getRevRLE(String str) {
return Pattern.compile("([a-zA-Z][0-9]+)").matcher(str).results()
.map(MatchResult::group)
.map(g -> g.substring(0, 1).repeat(Integer.parseInt(g.substring(1))))
.collect(Collectors.joining());
}
Testing
public static void main(String[] args) {
String str = "AABBBBBCCCCCCCCCCCCCCCCCCCC";
String rle = getRLE(str);
String res = getRevRLE(rle);
System.out.println(res + " " + res.equals(str)); // AABBBBBCCCCCCCCCCCCCCCCCCCC true
}

Here you go:
public static String encode(String input) {
String output = "";
while (true) {
char c = input.charAt(0);
String countStr = "";
char current;
for (int i = 1; i < input.length() && Character.isDigit(current = input.charAt(i)); i++)
countStr += current;
int count = Integer.parseInt(countStr);
for (int i = 0; i < count; i++)
output += c;
int trimLength = 1 + countStr.length();
if (trimLength >= input.length())
return output;
else
input = input.substring(trimLength);
}
}

You can do this task like this:
public static String printRLE(String str) {
int n = str.length();
String result = "";
for (int i = 0; i < n - 1; i++) {
char ch = str.charAt(i);
if (!Character.isDigit(ch)) {
if (Character.isDigit(str.charAt(i + 1))) {
int fi = i + 1;
i += 2;
while (i < n && Character.isDigit(str.charAt(i))) i++;
int repeat = Integer.parseInt(str.substring(fi, i));
result += String.valueOf(ch).repeat(repeat);
i--;
} else result += ch;
}
}
return result;
}
public static void main(String[] args) {
String str = "10A10B32C1";
System.out.println(printRLE(str));
}
, output
AAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBC

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

Finding patterns in strings using basic Java methods

say we have a string with these characters
"ABGCCFFGTBG"
then we have another string with characters "GECCCDOABG"
So the pattern is the prefix and suffix but if your given strings larger then this but have common prefix and suffix patterns how to pull those out into a substring in java. Keep in mind we dont always know the characters in the string were getting we just know there is a pattern in it.
my start is something like this
for(int i = 0. i < strA.length(); i++)
{
for(int j = 0; j < strB.length(); j++)
{
if(strA.charAt(i) == strB.charAt(j))
{
String subPattern = strA.substring(0,i);
String subPattern2 = strB.substring(0,j);
}
}
}
but this doesn't work. Any ideas?

Try to select best-matched pattern at first:
public static void main(String[] args) {
String strA = "ABGCCFFGTBG";
String strB = "GECCCDOABG";
System.out.println("Pattern: " + findPattern(strA, strB));
}
public static String findPattern(String strA, String strB) {
for (int length = Math.min(strA.length(), strB.length()); length > 0; length--) {
for (int i = 0; i <= strA.length() - length; i++) {
String pattern = strA.substring(i, i + length);
if (strB.contains(pattern)) {
return pattern;
}
}
}
throw new NoSuchElementException("No common pattern between " + strA + " and " + strB);
}
Output:
Pattern: ABG

this solution will find a pattern, no matter where it is in the string:
public static void main(String[] args) {
String strA = "uioABCDqwert";
String strB = "yxcvABCDwrk";
StringBuilder sb = new StringBuilder();
for (int i = 0; i < strA.length(); i++) {
for (int j = 0; j < strB.length(); j++) {
if (strA.charAt(i) == strB.charAt(j)) {
sb.append(strB.charAt(j));
i++;
}
}
if (sb.length() > 0)
break;
}
System.out.println(sb.toString());
}
this should give you an idea how it could be done

How to fill the null array with a specific number in Java ?

I'm trying to solve a palindrome problem that the input consists of Strings , if the concatenation of two strings represent a palindrome word(A palindrome is a word which can be read the same way in either direction. For example, the following
words are palindromes: civic, radar, rotor, and madam)
then save it into array to print it latter otherwise print "0"
but I'm having a problem in filling the null index with zeros , here I get Exception
for (int re = 0; re < result.length; re++) {
if (result[re].equals(null)) {
result[re] = "0";
}
}
"Exception in thread "main" java.lang.NullPointerException"
here is my full code
import java.util.Scanner;
public class Palindrome {
public static String reverse(String R2) {
String Reverse = "";
String word_two = R2;
int ln = word_two.length();
for (int i = ln - 1; i >= 0; i--) {
Reverse = Reverse + word_two.charAt(i);
}
return Reverse;
}
public static void main(String[] args) {
Scanner inpoot = new Scanner(System.in);
int stop = 0;
String pal1;
int Case = inpoot.nextInt();
String result[] = new String[Case];
String Final;
int NumberofWords;
for (int i = 0; i < Case; i++) {
NumberofWords = inpoot.nextInt();
String words[] = new String[NumberofWords];
for (int array = 0; array < words.length; array++) {
words[array] = inpoot.next();
}
for (int word1 = 0; word1 < NumberofWords; word1++) {
if (stop > Case) {
break;
}
for (int word2 = 0; word2 < NumberofWords; word2++) {
if (word1 == word2) {
continue;
}
Final = "" + words[word1].charAt(0);
if (words[word2].endsWith(Final)) {
pal1 = words[word1].concat(words[word2]);
} else {
continue;
}
if (pal1.equals(reverse(pal1))) {
result[i] = pal1;
stop++;
break;
} else {
pal1 = "";
}
}
}
}
// HERE IS THE PROBLEM
for (int re = 0; re < result.length; re++) {
if (result[re].equals(null)) {
result[re] = "0";
}
}
for (int x = 0; x < result.length; x++) {
System.out.println("" + result[x]);
}
}
}

A test such as anObject.equals(null) makes no sense. Indeed, if anObject is null, it will throw a NullPointerException (NPE), and if it is not, it will always return false.
To test if a reference is null, just use anObject == null.

If you want to check whether result[re] is null, you cannot use equals. Use the identity comparison:
if (result[re] == null) {
result[re] = "0";
}