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Make The All of Combination from Palindrome

I have a problem for my task. I must make a program that the input is a palindrome / not a palindrome, and the output is return the substring of the string that can be a palindrome in recursive. Example :
"marah" , the output should be, ("m","a","r","a","h") , ("m","ara","h") . I dont know to implement this in recursive. Please anyone who can help me, i'm very need that code. I worked it in java. Thank you, i hope there is a help coming :D .
public static String palindrome(String s) {
String s, sub;
int i, c, length;
Scanner in = new Scanner(System.in);
System.out.println("Enter a string to print it's all substrings");
s = in.nextLine();
length = string.length();
System.out.println("Substrings of \"" + string + "\" are :-");
for (c = 0; c < length; c++) {
for (i = 1; i <= length - c; i++) {
sub = string.substring(c, c + i);
System.out.println(sub);
}
}
}

public static String longestPalindrome(String s) {
if (s.isEmpty()) {
return null;
}
if (s.length() == 1) {
return s;
}
String longest = s.substring(0, 1);
for (int i = 0; i < s.length(); i++) {
// get longest palindrome with center of i
String tmp = helper(s, i, i);
if (tmp.length() > longest.length()) {
longest = tmp;
}
// get longest palindrome with center of i, i+1
tmp = helper(s, i, i + 1);
if (tmp.length() > longest.length()) {
longest = tmp;
}
}
return longest;
}
// Given a center, either one letter or two letter,
// Find longest palindrome
public static String helper(String s, int begin, int end) {
while (begin >= 0 && end <= s.length() - 1 && s.charAt(begin) == s.charAt(end)) {
begin--;
end++;
}
return s.substring(begin + 1, end);
}
if the input is "mama", the output is only "ama", the expected is, "m","a","m","a" , "mam","a" , and "m","ama" . Anybody can help?

This is called palindrom partition, you can find it here http://www.programcreek.com/2013/03/leetcode-palindrome-partitioning-java/

implement basic string compression

I am working on question 1.5 from the book Cracking The Coding interview. The problem is to take a string "aabcccccaaa" and turn it into a2b1c5a3.
If the compressed string is not smaller than the original string, then return the original string.
My code is below. I used an ArrayList because I would not know how long the compressed string would be.
My output is [a, 2, b, 1, c, 5], aabc, []. When the program gets to the end of string, it doesn't have a character to compare the last character too.
import java.util.*;
import java.io.*;
public class stringCompression {
public static void main(String[] args) {
String a = "aabcccccaaa";
String b = "aabc";
String v = "aaaa";
check(a);
System.out.println("");
check(b);
System.out.println("");
check(v);
}
public static void check(String g){
ArrayList<Character> c = new ArrayList<Character>();
int count = 1;
int i = 0;
int h = g.length();
for(int j = i + 1; j < g.length(); j++)
{
if(g.charAt(i) == g.charAt(j)){
count++;
}
else {
c.add(g.charAt(i));
c.add((char)( '0' + count));
i = j;
count = 1;
}
}
if(c.size() == g.length()){
System.out.print(g);
}
else{
System.out.print(c);
}
}
}

In the last loop you're not adding the result to the array. When j = g.length() still needs to add the current char and count to the array. So you could check the next value of j before increment it:
for(int j = i + 1; j < g.length(); j++)
{
if(g.charAt(i) == g.charAt(j)){
count++;
}
else {
c.add(g.charAt(i));
c.add((char)( '0' + count));
i = j;
count = 1;
}
if((j + 1) = g.length()){
c.add(g.charAt(i));
c.add((char)( '0' + count));
}
}

I would use a StringBuilder rather than an ArrayList to build your compressed String. When you start compressing, the first character should already be added to the result. The count of the character will be added once you've encountered a different character. When you've reached the end of the String you should just be appending the remaining count to the result for the last letter.
public static void main(String[] args) throws Exception {
String[] data = new String[] {
"aabcccccaaa",
"aabc",
"aaaa"
};
for (String d : data) {
System.out.println(compress(d));
}
}
public static String compress(String str) {
StringBuilder compressed = new StringBuilder();
// Add first character to compressed result
char currentChar = str.charAt(0);
compressed.append(currentChar);
// Always have a count of 1
int count = 1;
for (int i = 1; i < str.length(); i++) {
char nextChar = str.charAt(i);
if (currentChar == nextChar) {
count++;
} else {
// Append the count of the current character
compressed.append(count);
// Set the current character and count
currentChar = nextChar;
count = 1;
// Append the new current character
compressed.append(currentChar);
}
}
// Append the count of the last character
compressed.append(count);
// If the compressed string is not smaller than the original string, then return the original string
return (compressed.length() < str.length() ? compressed.toString() : str);
}
Results:
a2b1c5a3
aabc
a4

You have two errors:
one that Typo just mentioned, because your last character was not added;
and another one, if the original string is shorter like "abc" with only three chars: "a1b1c1" has six chars (the task is "If the compressed string is not smaller than the original string, then return the original string.")
You have to change your if statement, ask for >= instead of ==
if(c.size() >= g.length()){
System.out.print(g);
} else {
System.out.print(c);
}

Use StringBuilder and then iterate on the input string.
private static string CompressString(string inputString)
{
var count = 1;
var compressedSb = new StringBuilder();
for (var i = 0; i < inputString.Length; i++)
{
// Check if we are at the end
if(i == inputString.Length - 1)
{
compressedSb.Append(inputString[i] + count.ToString());
break;
}
if (inputString[i] == inputString[i + 1])
count++;
else
{
compressedSb.Append(inputString[i] + count.ToString());
count = 1;
}
}
var compressedString = compressedSb.ToString();
return compressedString.Length > inputString.Length ? inputString : compressedString;
}

Reverse every 2nd word of a sentence

I am trying to reverse every 2nd words of every single sentence like
If a given string is :
My name is xyz
The desired output should be :
My eman is zyx
My current output is:
Ym eman s1 zyx
I am not able to achieve my desired output.Don't know what I am doing wrong here
Here is my code
char[] sentence = " Hi my name is person!".toCharArray();
System.out.println(ReverseSentence(sentence));
}
private static char[] ReverseSentence(char[] sentence)
{
//Given: "Hi my name is person!"
//produce: "iH ym eman si !nosrep"
if(sentence == null) return null;
if(sentence.length == 1) return sentence;
int startPosition=0;
int counter = 0;
int sentenceLength = sentence.length-1;
//Solution handles any amount of spaces before, between words etc...
while(counter <= sentenceLength)
{
if(sentence[counter] == ' ' && startPosition != -1 || sentenceLength == counter) //Have passed over a word so upon encountering a space or end of string reverse word
{
//swap from startPos to counter - 1
//set start position to -1 and increment counter
int begin = startPosition;
int end;
if(sentenceLength == counter)
{
end = counter;
}
else
end = counter -1;
char tmp;
//Reverse characters
while(end >= begin){
tmp = sentence[begin];
sentence[begin] = sentence[end];
sentence[end] = tmp;
end--; begin++;
}
startPosition = -1; //flag used to indicate we have no encountered a character of a string
}
else if(sentence[counter] !=' ' && startPosition == -1) //first time you encounter a letter in a word set the start position
{
startPosition = counter;
}
counter++;
}
return sentence;
}

If you want to reverse the alternate word you can try something like splitting the whole String into words delimited by whitespaces and apply StringBuilder reverse() on every second word like :-
String s = "My name is xyz";
String[] wordsArr = s.split(" "); // broke string into array delimited by " " whitespace
StringBuilder sb = new StringBuilder();
for(int i = 0 ; i< wordsArr.length; i++){ // loop over array length
if(i%2 == 0) // if 1st word, 3rd word, 5th word..and so on words
sb.append(wordsArr[i]); // add the word as it is
else sb.append(new StringBuilder(wordsArr[i]).reverse()); // else use StringBuilder revrese() to reverse it
sb.append(" ");// add a whitespace in between words
}
System.out.println(sb.toString().trim()); //remove extra whitespace from the end and convert StringBuilder to String
Output :- My eman is zyx

You can solve your problem vary easy way! Just use a flag variable which will indicate the even or odd position, more precisely whether any word will gonna be reversed or not!
Look at the following modification I made in your code, just added three extra line:
private static boolean flag = true;// added a variable flag to check if we reverse the word or not.
private static char[] ReverseSentence(char[] sentence)
{
//Given: "Hi my name is person!"
//produce: "iH ym eman si !nosrep"
if(sentence == null) return null;
if(sentence.length == 1) return sentence;
int startPosition=0;
int counter = 0;
int sentenceLength = sentence.length-1;
//Solution handles any amount of spaces before, between words etc...
while(counter <= sentenceLength)
{
if(sentence[counter] == ' ' && startPosition != -1 || sentenceLength == counter) //Have passed over a word so upon encountering a space or end of string reverse word
{
flag = !flag; // first time (odd position) we are not going to reverse!
//swap from startPos to counter - 1
//set start position to -1 and increment counter
int begin = startPosition;
int end;
if(sentenceLength == counter)
{
end = counter;
}
else
end = counter -1;
char tmp;
//Reverse characters
while(end >= begin & flag){ //lets see whether we are going to reverse or not
tmp = sentence[begin];
sentence[begin] = sentence[end];
sentence[end] = tmp;
end--; begin++;
}
startPosition = -1; //flag used to indicate we have no encountered a character of a string
}
else if(sentence[counter] !=' ' && startPosition == -1) //first time you encounter a letter in a word set the start position
{
startPosition = counter;
}
counter++;
}
return sentence;
}
Input
My name is xyz
Output:
My eman is zyx

The following code does this "special reverse" which reverses any other word in the sentence:
public static void main(String[] args) {
String sentence = "My name is xyz";
System.out.println(specialReverse(sentence)); // My eman is zyx
}
private static String specialReverse(String sentence) {
String result = "";
String[] words = sentence.split(" ");
// we'll reverse only every second word according to even/odd index
for (int i = 0; i < words.length; i++) {
if (i % 2 == 1) {
result += " " + reverse(words[i]);
} else {
result += " " + words[i];
}
}
return result;
}
// easiest way to reverse a string in Java in a "one-liner"
private static String reverse(String word) {
return new StringBuilder(word).reverse().toString();
}

Just for completeness here's Java-8 solution:
public static String reverseSentence(String input) {
String[] words = input.split(" ");
return IntStream.range(0, words.length)
.mapToObj( i -> i % 2 == 0 ? words[i] :
new StringBuilder(words[i]).reverse().toString())
.collect(Collectors.joining(" "));
}
reverseSentence("My name is xyz"); // -> My eman is zyx

package com.eg.str;
// Without using StringBuilder
// INPUT: "Java is very cool prog lang"
// OUTPUT: Java si very looc prog gnal
public class StrRev {
public void reverse(String str) {
String[] tokens = str.split(" ");
String result = "";
String k = "";
for(int i=0; i<tokens.length; i++) {
if(i%2 == 0)
System.out.print(" " + tokens[i] + " ");
else
result = tokens[i];
for (int j = result.length()-1; j >= 0; j--) {
k = "" + result.charAt(j);
System.out.print(k);
}
result = "";
}
}
public static void main(String[] args) {
StrRev obj = new StrRev();
obj.reverse("Java is very cool prog lang");
}
}

//reverse second word of sentence in java
public class ReverseSecondWord {
public static void main(String[] args) {
String s="hello how are you?";
String str[]=s.split(" ");
String rev="";
for(int i=0;i<str[1].length();i++)
{
char ch=str[1].charAt(i);
rev=ch+rev;
}
str[1]=rev;
for(int i=0;i<str.length;i++)
{
System.out.print(str[i]+" ");
}
}
}

Counting Letters Between the First and Last - Java

OK, so I'm doing this project that requires that I have the first and last setters of a string appear with the number of letters in between them counted, and output. I've tried repurposing some reverse a string code I had handy, but I cannot get the output to appear in my IDE.
Can anyone look over my code, and make some suggestions?
public static void main(String[] args) {
String countWord;
countWord = JOptionPane.showInputDialog(null,
"Enter the string you wish to have formatted:");
}
static String countMe(String countWord) {
int count = 1;
char first = countWord.charAt (0);
char last = countWord.charAt(-1);
StringBuilder word = new StringBuilder();
for(int i = countWord.length() - 1; i >= 0; --i)
if (countWord.charAt(i) != first ) {
if (countWord.charAt(i) != last) {
count++;
}
}
return countWord + first + count + last;
}
}

Just build it using charAt():
return "" + str.charAt(0) + (str.length() - 2) + str.charAt(str.length() - 1);
The "" at the front causes the numeric values that follow to be concatenated as Strings (instead of added arithmetically).
A slightly more terse alternative is:
return countWord.replaceAll("(.).*(.)", "$1" + (str.length() - 2) + "$2")

Once you determined the first and last chars, it is no need for unnecessary conditions. Just try this:
static String countMe(String countWord) {
char first = countWord.charAt(0);
char last = countWord.charAt(countWord.length()-1);
int count=0;
for (int i = 1; i < countWord.length()-1; i++)
{
count++;
}
return first + String.valueOf(count) + last;
}
Or, if it is not mandatory to use for loop, you can make it simple as this
static String countMe(String countWord) {
char first = countWord.charAt(0);
char last = countWord.charAt(countWord.length()-1);
int count = countWord.substring(1, countWord.length()-1).length();
return first + String.valueOf(count) + last;
}

You could use the string.length() method to obtain the total length of the string. Your code would be something like:
int totalLength = countWord.length();
int betweenLength = totalLength - 2; // This gives the count of characters between first and last letters
char first = countWord.charAt(0);
char last = countWord.charAt(str.length() - 1);
String answer = first + betweenLength + last;

import javax.swing.JOptionPane;
public class Main{
public static void main(String[] args) {
String countWord;
countWord = JOptionPane.showInputDialog(null,
"Enter the word you wish to have formatted:");
JOptionPane.showMessageDialog(null, countMe(countWord));
}
static String countMe(String countWord) {
int count = 0;
String first = String.valueOf(countWord.charAt(0));
String last = String.valueOf(countWord.charAt(countWord.length() - 1));
for(int i = 1; i < countWord.length() - 1; i++) {
if (String.valueOf(countWord.charAt(i)) != first ) {
count++;
}
}
return first + count + last;
}
}

Java: method to get position of a match in a String?

String match = "hello";
String text = "0123456789hello0123456789";
int position = getPosition(match, text); // should be 10, is there such a method?

The family of methods that does this are:
int indexOf(String str)
indexOf(String str, int fromIndex)
int lastIndexOf(String str)
lastIndexOf(String str, int fromIndex)
Returns the index within this string of the first (or last) occurrence of the specified substring [searching forward (or backward) starting at the specified index].
String text = "0123hello9012hello8901hello7890";
String word = "hello";
System.out.println(text.indexOf(word)); // prints "4"
System.out.println(text.lastIndexOf(word)); // prints "22"
// find all occurrences forward
for (int i = -1; (i = text.indexOf(word, i + 1)) != -1; i++) {
System.out.println(i);
} // prints "4", "13", "22"
// find all occurrences backward
for (int i = text.length(); (i = text.lastIndexOf(word, i - 1)) != -1; i++) {
System.out.println(i);
} // prints "22", "13", "4"

This works using regex.
String text = "I love you so much";
String wordToFind = "love";
Pattern word = Pattern.compile(wordToFind);
Matcher match = word.matcher(text);
while (match.find()) {
System.out.println("Found love at index "+ match.start() +" - "+ (match.end()-1));
}
Output :
Found 'love' at index 2 - 5
General Rule :
Regex search left to right, and once the match characters has been used, it cannot be reused.

text.indexOf(match);
See the String javadoc

Finding a single index
As others have said, use text.indexOf(match) to find a single match.
String text = "0123456789hello0123456789";
String match = "hello";
int position = text.indexOf(match); // position = 10
Finding multiple indexes
Because of #StephenC's comment about code maintainability and my own difficulty in understanding #polygenelubricants' answer, I wanted to find another way to get all the indexes of a match in a text string. The following code (which is modified from this answer) does so:
String text = "0123hello9012hello8901hello7890";
String match = "hello";
int index = text.indexOf(match);
int matchLength = match.length();
while (index >= 0) { // indexOf returns -1 if no match found
System.out.println(index);
index = text.indexOf(match, index + matchLength);
}

You can get all matches in a file simply by assigning inside while-loop, cool:
$ javac MatchTest.java
$ java MatchTest
1
16
31
46
$ cat MatchTest.java
import java.util.*;
import java.io.*;
public class MatchTest {
public static void main(String[] args){
String match = "hello";
String text = "hello0123456789hello0123456789hello1234567890hello3423243423232";
int i =0;
while((i=(text.indexOf(match,i)+1))>0)
System.out.println(i);
}
}

int match_position=text.indexOf(match);

import java.util.StringTokenizer;
public class Occourence {
public static void main(String[] args) {
String key=null,str ="my name noorus my name noorus";
int i=0,tot=0;
StringTokenizer st=new StringTokenizer(str," ");
while(st.hasMoreTokens())
{
tot=tot+1;
key = st.nextToken();
while((i=(str.indexOf(key,i)+1))>0)
{
System.out.println("position of "+key+" "+"is "+(i-1));
}
}
System.out.println("total words present in string "+tot);
}
}

I have some big code but working nicely....
class strDemo
{
public static void main(String args[])
{
String s1=new String("The Ghost of The Arabean Sea");
String s2=new String ("The");
String s6=new String ("ehT");
StringBuffer s3;
StringBuffer s4=new StringBuffer(s1);
StringBuffer s5=new StringBuffer(s2);
char c1[]=new char[30];
char c2[]=new char[5];
char c3[]=new char[5];
s1.getChars(0,28,c1,0);
s2.getChars(0,3,c2,0);
s6.getChars(0,3,c3,0); s3=s4.reverse();
int pf=0,pl=0;
char c5[]=new char[30];
s3.getChars(0,28,c5,0);
for(int i=0;i<(s1.length()-s2.length());i++)
{
int j=0;
if(pf<=1)
{
while (c1[i+j]==c2[j] && j<=s2.length())
{
j++;
System.out.println(s2.length()+" "+j);
if(j>=s2.length())
{
System.out.println("first match of(The) :->"+i);
}
pf=pf+1;
}
}
}
for(int i=0;i<(s3.length()-s6.length()+1);i++)
{
int j=0;
if(pl<=1)
{
while (c5[i+j]==c3[j] && j<=s6.length())
{
j++;
System.out.println(s6.length()+" "+j);
if(j>=s6.length())
{
System.out.println((s3.length()-i-3));
pl=pl+1;
}
}
}
}
}
}

//finding a particular word any where inthe string and printing its index and occurence
class IndOc
{
public static void main(String[] args)
{
String s="this is hyderabad city and this is";
System.out.println("the given string is ");
System.out.println("----------"+s);
char ch[]=s.toCharArray();
System.out.println(" ----word is found at ");
int j=0,noc=0;
for(int i=0;i<ch.length;i++)
{
j=i;
if(ch[i]=='i' && ch[j+1]=='s')
{
System.out.println(" index "+i);
noc++;
}
}
System.out.println("----- no of occurences are "+noc);
}
}

String match = "hello";
String text = "0123456789hello0123456789hello";
int j = 0;
String indxOfmatch = "";
for (int i = -1; i < text.length()+1; i++) {
j = text.indexOf("hello", i);
if (i>=j && j > -1) {
indxOfmatch += text.indexOf("hello", i)+" ";
}
}
System.out.println(indxOfmatch);

If you're going to scan for 'n' matches of the search string, I'd recommend using regular expressions.
They have a steep learning curve, but they'll save you hours when it comes to complex searches.

for multiple occurrence and the character found in string??yes or no
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class SubStringtest {
public static void main(String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter the string");
String str=br.readLine();
System.out.println("enter the character which you want");
CharSequence ch=br.readLine();
boolean bool=str.contains(ch);
System.out.println("the character found is " +bool);
int position=str.indexOf(ch.toString());
while(position>=0){
System.out.println("the index no of character is " +position);
position=str.indexOf(ch.toString(),position+1);
}
}
}

public int NumberWordsInText(String FullText_, String WordToFind_, int[] positions_)
{
int iii1=0;
int iii2=0;
int iii3=0;
while((iii1=(FullText_.indexOf(WordToFind_,iii1)+1))>0){iii2=iii2+1;}
// iii2 is the number of the occurences
if(iii2>0) {
positions_ = new int[iii2];
while ((iii1 = (FullText_.indexOf(WordToFind_, iii1) + 1)) > 0) {
positions_[iii3] = iii1-1;
iii3 = iii3 + 1;
System.out.println("position=" + positions_[iii3 - 1]);
}
}
return iii2;
}

class Main{
public static int string(String str, String str1){
for (int i = 0; i <= str.length() - str1.length(); i++){
int j;
for (j = 0; j < str1.length(); j++) {
if (str1.charAt(j) != str.charAt(i + j)) {
break;
}
}
if (j == str1.length()) {
return i;
}}
return -1;
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter the string");
String str=sc.nextLine();
System.out.println("Enter the Substring");
String str1=sc.nextLine();
System.out.println("The position of the Substring is "+string(str, str1));
}
}