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Need to encode repetitive pattern in String with * , such that * means "repeat from beginning"

Encoding format: introduce * to indicate "repeat from beginning". Example. Input-{a,b,a,b,c,a,b,a,b,c,d} can be written as {a , b, * ,c, * , d}. Output:5; E.g 2: ABCABCE, output- 5.
Here * means repeat from beginning. For example if given String is ABCABCABCABC , it will return ABC**, another example is if String is ABCABCABC, it will return ABC*ABC.
I have the below code but this code assumes that the string will contain the repetitive pattern only and no other characters, I want to modify it to check :
1. Which pattern is repeating
2. Ignore non repeating patterns
2. encode that pattern according to the problem statement
import java.util.Scanner;
public class Magicpotion {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter the string:");
String str = sc.nextLine();
int len = str.length();
if (len != 0) {
int lenby3 = len / 3;
int starcount = ( int).(Math.log(lenby3) / Math.log(2));
int leftstring = (lenby3 - (int) Math.pow(2, starcount));
int resultlen = (1 * 3) + starcount + (leftstring * 3);
System.out.println("ResultLength: " + resultlen);
System.out.print("ABC");
for (int i = 0; i < starcount; i++) {
System.out.print("*");
}
for (int i = 0; i < leftstring; i++) {
System.out.print("ABC");
}
} else
System.out.println("ResultLength: " + 0);
}
}
Here my assumption is that ABC will always be repeating pattern , hence I have divided the length by 3. I want to generalise it such that I find the repeating pattern which can be a AB or BC or ABCD and proceed accordingly.

This looks like homework. So instead of a full solution just some hints:
You can process the input string character by character and encode as you go. If you have at some point already read k characters and the next k characters are exactly the same, output a * and advance to position 2k.
Otherwise, output the next input character and advance position to k+1.
As mentioned by dyukha this algorithm does not always result in the shortest possible encoding. If this is required some more effort has to be put into the search.

This problem can be solved using dynamic programming.
Assume that you processed your stay at some position i. You want to understand what it the minimal length of encoding of str[0..i]. Let's call it ans[i]. You have two options:
Just add i-th character to the encoding. So the length is ans[i-1] + 1.
You may write *, when possible. In this case the length is ans[i / 2] + 1 or something like this.
The final length is in ans[n-1]. You can store how you obtained ans[i] to recover the encoding itself.
Checking whether you can write * can be optimized, using some hashing (to obtain O(n) solution instead of O(n^2)).
The difference with Henry's solution is that he always applies * when it's possible. It's not clear to me that it results into the minimal length (if I understood correctly, aaaaaa is a counterexample), so I'm giving a solution I'm sure about.

/**
* #author mohamed ali
* https://www.linkedin.com/in/oo0shaheen0oo/
*/
public class Magic_potion_encoding
{
private static int minimalSteps( String ingredients )
{
StringBuilder sb = new StringBuilder(ingredients);
for(int i =0;i<sb.length();i++)
{
char startChar = sb.charAt(i);
int walkingIndex1=i;
int startIndex2 =sb.toString().indexOf(startChar,i+1);
int walkingIndex2=startIndex2;
while(walkingIndex2 !=-1 && walkingIndex2<sb.length() && sb.charAt(walkingIndex1) == sb.charAt(walkingIndex2) )
{
if(walkingIndex1+1==startIndex2)
{
String subStringToBeEncoded = sb.substring(i,walkingIndex2+1);//substring the string found and the original "substring does not include the last index hence the +1
int matchStartIndex = sb.indexOf(subStringToBeEncoded,walkingIndex2+1);// look for first match for the whole string matched
int matchEndeIndex= matchStartIndex+subStringToBeEncoded.length();
int origStartIndex=i;
int origEndIndex = i+subStringToBeEncoded.length();
if (matchStartIndex!=-1 )
{
if(origEndIndex==matchStartIndex)
{
sb.replace(matchStartIndex,matchEndeIndex,"*");
}
else
{
while(matchStartIndex!=-1 && matchEndeIndex<sb.length() && sb.charAt(origEndIndex) == sb.charAt(matchEndeIndex) )
{
if(origEndIndex==matchStartIndex-1)// if the index of the 2 strings are right behind one another
{
sb.replace(matchStartIndex,matchEndeIndex+1,"*");
}
else
{
origEndIndex++;
matchEndeIndex++;
}
}
}
}
sb.replace(startIndex2,walkingIndex2+1,"*");
break;
}
walkingIndex1++;
walkingIndex2++;
}
}
System.out.println("orig= " + ingredients + " encoded = " + sb);
return sb.length();
}
public static void main( String[] args )
{
if ( minimalSteps("ABCABCE") == 5 &&
minimalSteps("ABCABCEA") == 6 &&
minimalSteps("abbbbabbbb") == 5 &&
minimalSteps("abcde") == 5 &&
minimalSteps("abcbcbcbcd") == 6 &&
minimalSteps("ababcababce") == 6 &&
minimalSteps("ababababxx") == 6 &&
minimalSteps("aabbccbbccaabbccbbcc") == 8)
{
System.out.println( "Pass" );
}
else
{
System.out.println( "Fail" );
}
}
}

Given that the repetitions are from the beginning, every such repeating substring will have the very first character of the given string. [Every repetition needs to be represented by a "star". (i.e ABCABCABC ans = ABC** ) . If all sequential repetitions are to be represented with one "star". (i.e ABCABCABC and = ABC* ), a slight modification to (2) will do the thing (i.e remove the if case where the just a star is added)]
Divide the given string to substrings based on the first character.
Eg. Given String = "ABABCABD"
Sub Strings = {"AB", "ABC", "AB", "ABD"}
Just traverse through the list of substrings and get the required result. I've used a map here, to make the search easy.
Just a rough write up.
SS = {"AB", "ABC", "AB", "ABD"};
result = SS[0];
Map<string, bool> map;
map.put(SS[0],true);
for (i = 1; i < SS.length; i++){
if (map.hasKey(SS[i])){
result += "*";
}
else {
res = nonRepeatingPart(SS[i], map);
result += "*" + res;
map.put(SS[i], true);
}
}
String nonRepeatingPart(str, map){
for (j = str.length-1; j >= 0; j--){
if (map.hasKey(str.subString(0, j))){
return str.subString(j, str.length-1);
}
}
return throwException("Wrong Input");
}

string getCompressed(string str){
string res;
res += str[0];
int i=1;
while(i<str.size()){
//check if current char is the first char in res
char curr = str[i];
if(res[0]==curr){
if(str.substr(0,i)==str.substr(i,i)){
res += '*';
i+=i; continue;
}else{
res += curr;
i++; continue;
}
}else {
res += curr;
i++; continue;
}
}
return res;
}
int main()
{
string s = "ABCABCABC";
string res = getCompressed(s);
cout<<res.size();
return 0;
}

Abbreviation expander for a given lexicon

I am trying to write a program that will allows users to make short blog entries by typing abbreviations for common words. On completion of the input, Program will expand the abbreviations according to the lexicon defined.
Conditions
A substituted word must be the shortest word that can be formed by adding zero or more letters (or punctuation symbols) to the abbreviation.
If two or more unique words can be formed by adding the same number of letters, then the abbreviation should be printed as it is.
Input
The input is divided into two sections.
The first section is the lexicon itself, and the second section is a user's blog entry that needs to be expanded. The sections are divided by a single | character.
For example:-
cream chocolate every ever does do ice is fried friend friends lick like floor favor flavor flower best but probably poorly say says that what white our you your strawberry storyboard the | wht flvr ic crm ds yr bst fnd lke? ur frds lk stbry, bt choc s prly th bs flr vr!
Output
what flavor ice cream does your best friend like? our friends lk strawberry, but chocolate is poorly the best floor ever!
I have written the program for this and tested it locally with many different test cases with success but it fails on submission to test server.
An automated Test suit runs to validate the program’s output on its submission to test server. In case of failure, details of the failing test case/cases are not visible.
Below is the program
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class BlogEntry {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String[][] info = readInput();
String[] output = inputExpander(info[0],info[1]);
//System.out.println();
for(int i = 0; i < output.length; ++i) {
if(i!=0)
System.out.print(" ");
System.out.print(output[i]);
}
}
public static String[][] readInput() {
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(
System.in));
String input = null;
String[][] info = new String[2][];
String[] text;
String[] abbr;
try {
input = bufferReader.readLine();
StringTokenizer st1 = new StringTokenizer(input, "|");
String first = "", second = "";
int count = 0;
while (st1.hasMoreTokens()) {
++count;
if(count == 1)
first = st1.nextToken();
if(count == 2)
second = st1.nextToken();
}
st1 = new StringTokenizer(first, " ");
count = st1.countTokens();
text = new String[count];
count = 0;
while (st1.hasMoreTokens()) {
text[count] = st1.nextToken();
count++;
}
st1 = new StringTokenizer(second, " ");
count = st1.countTokens();
abbr = new String[count];
count = 0;
while (st1.hasMoreTokens()) {
abbr[count] = st1.nextToken();
count++;
}
info[0] = text;
info[1] = abbr;
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return info;
}
public static String[] inputExpander(String[] text, String[] abbr) {
String[] output = new String[abbr.length];
boolean result;
for (int i = 0; i < abbr.length; ++i) {
String abbrToken = abbr[i];
char[] char_abbr_token = abbrToken.toCharArray();
for (int j = 0; j < text.length; ++j) {
String textToken = text[j];
boolean flag2 = false;
if ((char_abbr_token[char_abbr_token.length - 1] == '!')
|| (char_abbr_token[char_abbr_token.length - 1] == '?')
|| (char_abbr_token[char_abbr_token.length - 1] == ',')
|| (char_abbr_token[char_abbr_token.length - 1] == ';')) {
flag2 = true;
}
char[] char_text_token = textToken.toCharArray();
result = ifcontains(char_text_token, char_abbr_token);
if (result) {
int currentCount = textToken.length();
int alreadyStoredCount = 0;
if (flag2)
textToken = textToken
+ char_abbr_token[char_abbr_token.length - 1];
if (output[i] == null)
output[i] = textToken;
else {
alreadyStoredCount = output[i].length();
char[] char_stored_token = output[i].toCharArray();
if ((char_stored_token[char_stored_token.length - 1] == '!')
|| (char_stored_token[char_stored_token.length - 1] == '?')
|| (char_stored_token[char_stored_token.length - 1] == ',')
|| (char_stored_token[char_stored_token.length - 1] == ';')) {
alreadyStoredCount -= 1;
}
if (alreadyStoredCount > currentCount) {
output[i] = textToken;
} else if (alreadyStoredCount == currentCount) {
output[i] = abbrToken;
}
}
}
}
if(output[i] == null)
output[i] = abbrToken;
}
return output;
}
public static boolean ifcontains(char[] char_text_token,
char[] char_abbr_token) {
int j = 0;
boolean flag = false;
for (int i = 0; i < char_abbr_token.length; ++i) {
flag = false;
for (; j < char_text_token.length; ++j) {
if ((char_abbr_token[i] == '!') || (char_abbr_token[i] == '?')
|| (char_abbr_token[i] == ',')
|| (char_abbr_token[i] == ';')) {
flag = true;
break;
}
if (char_abbr_token[i] == char_text_token[j]) {
flag = true;
break;
}
}
if (!flag)
return flag;
}
//System.out.println("match found" + flag);
return flag;
}
}
Can someone direct/hint me to/about the possible use case which I may have missed in the implementation? Thanks in advance.

Ran your program with duplicate word in input (lexicon). When a word is repeated in the lexicon, it is not getting expanded because the check is only on the length(line no. 112) of the stored word not its content.
I think you need to check:-
If same word appears more than once then expand.
If 2 or more unique words of same length appear then keep it short.

How would I approach solving this:
Parse the input, tokenize the lexicon and the text.
For each (possibly abbreviated) token like choc convert it to a regular expression like .*c.*h.*o.*c.*.
Search for shortest lexicon words matching this regular expression. Replace the text token if exactly one is found, otherwise leave it alone.
It is quite hard to say what's wrong with your code without careful debugging. It is hard to understand what one or the other part of the code does, it's not quite self-evident.

Applying Linear and Binary Searches to Arrays

I have to create a program that takes a user input (a number) and then the program should have that number and apply a search to the array and output the corresponding title by matching the index and the number the user inputted. However during run time, nothing happens. I have set breakers in my code and noticed a problem with the for loop (search algorithm). Please help me and let me know what is wrong is my search algorithm. What I am trying to do is use the number of that the user inputs to match a index and then output the book title that is stored in the index.
private void btnFindActionPerformed(java.awt.event.ActionEvent evt) {
// TODO add your handling code here:
// declares an array
String[] listOfBooks = new String [101];
// assigns index in array to book title
listOfBooks[1] = "The Adventures of Tom Sawyer";
listOfBooks[2] = "Huckleberry Finn";
listOfBooks[4] = "The Sword in the Stone";
listOfBooks[6] = "Stuart Little";
listOfBooks[10] = "Treasure Island";
listOfBooks[12] = "Test";
listOfBooks[14] = "Alice's Adventures in Wonderland";
listOfBooks[20] = "Twenty Thousand Leagues Under the Sea";
listOfBooks[24] = "Peter Pan";
listOfBooks[26] = "Charlotte's Web";
listOfBooks[31] = "A Little Princess";
listOfBooks[32] = "Little Women";
listOfBooks[33] = "Black Beauty";
listOfBooks[35] = "The Merry Adventures of Robin Hood";
listOfBooks[40] = "Robinson Crusoe";
listOfBooks[46] = "Anne of Green Gables";
listOfBooks[50] = "Little House in the Big Woods";
listOfBooks[52] = "Swiss Family Robinson";
listOfBooks[54] = "The Lion, the Witch and the Wardrobe";
listOfBooks[54] = "Heidi";
listOfBooks[66] = "A Winkle in Time";
listOfBooks[100] = "Mary Poppins";
// gets user input
String numberInput = txtNumberInput.getText();
int number = Integer.parseInt(numberInput);
// Linear search to match index number and user input number
for(int i = 0; i < listOfBooks.length - 1; i++) {
if (listOfBooks.get(i) == number) {
txtLinearOutput.setText(listOfBooks[i]);
break;
}
}
*There is a problem with the listOfBooks.get in the if statement. Also I need to apply a binary search that would search the same array just using the binary method. Need help to apply this type of binary search.
How could I make a statement that checks if the int number is equal to an index?
Note that the following code is just an example of what I have to apply. Variables are all for example purposes:
public static Boolean binarySearch(String [ ] A, int left, int right, String V){
int middle;
if (left > right) {
return false;
}
middle = (left + right)/2;
int compare = V.compareTo(A[middle]);
if (compare == 0) {
return true;
}
if (compare < 0) {
return binarySearch(A, left, middle-1, V);
} else {
return binarySearch(A, middle + 1, right, V);
}
}

you can avoid for loop and check condition by just giving number like this: txtLinearOutput.setText(listOfBooks[number-1]);
remove your code
// Linear search to match index number and user input number
for(int i = 0; i < listOfBooks.length - 1; i++) {
if (listOfBooks.get(i) == number) {
txtLinearOutput.setText(listOfBooks[i]);
break;
}
with
try{
int number = Integer.parseInt(numberInput);
if(number>0 && number<101){
txtLinearOutput.setText(listOfBooks[number-1]);
}else{
// out of range
}
}catch(Exception e){
// handle exception here
}

You are comparing if (listOfBooks.get(i) == number) it is wrong, you should compare: if (i == number), becouse you need compare element position.

This isn't a binary search answer. Just an implementation of HashMap. Have a look at it.
HashMap<String, Integer> books = new HashMap();
books.put("abc", 1);
books.put("xyz", 2);
books.put("pqr", 3);
books.put("lmo", 4);
System.out.println(books.getValue("abc");
Using the inbuilt BinarySearch.
String []arr = new String[15];
arr[0] = "abc";
arr[5] = "prq";
arr[7] = "lmo";
arr[10] = "xyz";
System.out.println(Arrays.binarySearch(arr, "lmo"));
How to compare Strings using binary search.
String[] array = new String[4];
array[0] = "abc";
array[1] = "lmo";
array[2] = "pqr";
array[3] = "xyz";
int first, last, middle;
first = 0;
last = array.length - 1;
middle = (first + last) / 2;
String key = "abc";
while (first <= last) {
if (compare(array[middle], key))
first = middle + 1;
else if (array[middle].equals(key)) {
System.out.println(key + " found at location " + (middle) + ".");
break;
} else {
last = middle - 1;
}
middle = (first + last) / 2;
}
if (first > last)
System.out.println(key + " is not found.\n");
}
private static boolean compare(String string, String key) {
// TODO Auto-generated method stub
for (int i = 0; i < Math.min(string.length(), key.length()); ++i)
if (string.charAt(i) < key.charAt(i))
return true;
return false;
}

Your linear search code looks something like this
try{
txtLinearOutput.setText(listOfBooks[yourNumber]);
}
catch(IndexOutOfBoundsException ie){
// prompt that number is not an index
}
catch(Exception e){
// if any other exception is caught
}

What you are doing here:
if (listOfBooks.get(i) == number) {
is that you are matching the content of the array with the input number, which is irrelevant.
You can directly use the input number to fetch the value stored at the index.
For example:
txtLinearOutput.setText(listOfBooks[number-1]);
Additionally, int number = Integer.parseInt(numberInput); should be placed within try-catch block to validate input number parsing. And you can check if the input number is within the range of the array to avoid exceptions like:
try{
int number = Integer.parseInt(numberInput);
// Linear search to match index number and user input number
if (number > 0 && number <=100) {
txtLinearOutput.setText(listOfBooks[number-1]);
} else {
// Display error message
}
} catch(Exception e) {
// Handle exception and display error message
}
And for using binary search, the string array need to be sorted. You can use Arrays.sort() method for sorting it.
And regarding using binary search, you can use Java Arrays Binary Search method

Issues with While loop condition

I am writing a program that reads a text file which contains match results, and then should output them in a table. I have a While loop within a While loop:
Scanner fileread1 = new Scanner(new File("demo.txt"));
int x = 0;
int y = 22;
int i = 0;
while (x <= y) {
while (fileread1.hasNext()) {
fileinput = fileread1.nextLine(); // this reads the next line of
// from the file
String line = fileinput;
String[] split = line.split(":");
boolean result = false;
int homescore1 = 0;
int awayscore1 = 0;
int goalsscored = 0;
boolean att = false;
boolean htt = false;
int atscore = 0;
int htscore = 0;
// When the text line is read, it is then split into four sections.
if (split.length == 4) {
// String text = line.trim();
String userteam = userteaminput;
String hometeam = split[0].trim();
String awayteam = split[1].trim();
String home_score = split[2].trim();
String away_score = split[3].trim();
// this is a test to try convert the goals string into a
// integer. If this fails, the result is not
// not valid and does not get outputted to the console.
try {
homescore1 = Integer.parseInt(home_score);
awayscore1 = Integer.parseInt(away_score);
result = true;
}
catch (NumberFormatException exception) {
// if the try is not able to convert, this will run
errors++;
}
if (userteam.equals(Teams.get(i))) {
if (awayteam.equalsIgnoreCase(userteam)) {
att = true;
games++;
goalsfor = goalsfor + awayscore1;
goalsagainst = goalsagainst + homescore1;
}
if (att == true && awayscore1 > homescore1) {
atwc++;
gameswon++;
}
else if (att == true && awayscore1 < homescore1) {
htwc++;
gameslost++;
}
else if (att == true && awayscore1 == homescore1) {
gamesdrawn++;
}
if (hometeam.equalsIgnoreCase(userteam)) {
htt = true;
totaluser++;
games++;
goalsfor = goalsfor + homescore1;
goalsagainst = goalsagainst + awayscore1;
}
if (htt == true && homescore1 > awayscore1) {
atwc++;
gameswon++;
}
else if (htt == true && homescore1 < awayscore1) {
htwc++;
gameslost++;
}
else if (htt == true && awayscore1 == homescore1) {
gamesdrawn++;
}
}
}
else {
errors++;
}
}
// ********************************************************************
// Leeds IF Statement
// ********************************************************************
if (Rhinos.equals(Teams.get(i)) {
Rhinos.goalsfor = Rhinos.goalsfor + goalsfor;
Rhinos.gameswon = Rhinos.gameswon + gameswon;
Rhinos.gameslost = Rhinos.gameslost + gameslost;
Rhinos.goalsagainst = Rhinos.goalsagainst;
Rhinos.gamesplayed = Rhinos.gamesplayed + games;
}
else if (Bulls.equals(Teams.get(i)) {
Bulls.goalsfor = Bulls.goalsfor + goalsfor;
Bulls.gameswon = Bulls.gameswon + gameswon;
Bulls.gameslost = Bulls.gameslost + gameslost;
Bulls.goalsagainst = Bulls.goalsagainst;
Bulls.gamesplayed = Bulls.gamesplayed + games;
}
x++;
i++;
goalsfor = 0;
gameswon = 0;
gameslost = 0;
gamesagainst = 0;
}
I know that there are only ever going to be 22 teams that have results in the text file supplied, so the first loop should run for 22 times.
The inner loop, will continue whilst the file provided has a next line. The text file may sometimes have more lines of results then other files. Within this loop, I have a reference to an Array item:
if (userteam.equals(Teams.get(i)))
In the first run, this will refer to 0 in my Array which, for the record, is Leeds Rhinos. Once the inner loop has completed, it then moves onto the outer loop - this deals with the results just recorded. If the current team is Leeds Rhinos, it should then add the values. The i should then have 1 added, so for the next loop, it refers to the index of 1 of the array, not 0. (I have more IF statements here, all identical but refer to other teams) Variables get set back to 0, ready for the next run.
The issue I have, is that i does not seem to have 1 added each time it runs through, so I am only getting results passed through for one team. If I manually specify which array index to look (say 3) it will run through, and the team will have their results successfully recorded.
Is there a way I can get 1 added to i every time it loops? I'm not sure if this is the correct java loop to use, but to me, seemed the most logical. There are some objects not declared here - this is just a snippet of the code, left out the declarations as I know they work, and there's a lot declared.

If you're worried about failed incrementation, it would be better to use a For loop.
Instead of having a while (x < y) and sticking an increment statement somewhere in your code,
a
for (i = 0; i < y; i++) { // do tests here }
loop will guarantee that you always increment and run the test for the next team.
For future reference, when using while loops and incrementing, the incrementation is almost always done at the END of the while loop, and not somewhere in between. The incrementation statement should also almost never be in a conditional statement (which might cause an infinite loop).

Your question is not clear. But let's point out something in the code you provided
1) What is the difference between your if and else if statement? They are checking exact same thing
if (userteam.equals(Teams.get(i)) {
Rhinos.goalsfor = Rhinos.goalsfor + goalsfor;
Rhinos.gameswon = Rhinos.gameswon + gameswon;
Rhinos.gameslost = Rhinos.gameslost + gameslost;
Rhinos.goalsagainst = Rhinos.goalsagainst;
Rhinos.gamesplayed = Rhinos.gamesplayed + games;
}
else if (userteam.equals(Teams.get(i)) {
Bulls.goalsfor = Bulls.goalsfor + goalsfor;
Bulls.gameswon = Bulls.gameswon + gameswon;
Bulls.gameslost = Bulls.gameslost + gameslost;
Bulls.goalsagainst = Bulls.goalsagainst;
Bulls.gamesplayed = Bulls.gamesplayed + games;
}
2) What are you doing with variable x, I don't see anywhere you are increasing it.
3) On very first run, when x<=y, the inner loop will finish reading all lines, so even if you increase the X some point, from second run the inner loop will not execute. As it already finished reading all lines. So no point doing this
Again if you provide some more inside on what you want to accomplish, may be with the sample text file data, that would probably help answering your question.
Thank you.

Your formatting is working against you here; properly indented, your code structure is something like this (note, I had to add in missing closing braces, }, at the end of the code you provided as I assume you just missed them when you copied your code over):
Scanner fileread1 = new Scanner(new File("demo.txt"));
int x = 0;
int y = 22;
int i = 0;
while (x <= y) {
while (fileread1.hasNext()) {
fileinput = fileread1.nextLine(); // this reads the next line of
/* stuff */
if (split.length == 4) {
/* stuff */
x++;
i++;
}
}
}
Your incrementation of x and i is nested within if (split.length == 4) {, meaning that x and i will only be incremented in that specific case and not at the end of each iteration of the inner while loop.

List collections interface in java

Please find below a function in my code:
private static List<String> formCrfLinesWithMentionClass(int begin, int end, String id,
List<String> mList, int mListPos, List<String> crf) {
List<String> crfLines = crf;
int yes = 0;
mListPosChanged = mListPos;
//--------------------------------------------------------------------------
for (int crfLinesMainIter = begin; crfLinesMainIter < end; ) {
System.out.println(crfLines.get(crfLinesMainIter));
//---------------------------------------------------------------------------
//the total number of attributes without orthographic features
//in a crfLine excluding the class attribute is 98
if (!crfLines.get(crfLinesMainIter).equals("") && crfLines.get(crfLinesMainIter).split("\\s").length == 98) {
//in mList parenthesis are represented by the symbol
//in crfLines parenthesis are represented by -LRB- or -RRB-
//we make a check to ensure the equality is preserved
if(val.equals(crfLines.get(crfLinesMainIter).split("\\s")[0])) {
yes = checkForConsecutivePresence(crfLinesMainIter, mList, mListPos, id, crfLines);
if (yes > 0) {
mListPosChanged += yes;
System.out.println("formCrfLinesWithMentionClass: "+mListPosChanged);
for (int crfLinesMentionIter = crfLinesMainIter;
crfLinesMentionIter < crfLinesMainIter + yes;
crfLinesMentionIter++) {
String valString = "";
if (crfLinesMentionIter == crfLinesMainIter) {
valString += crfLines.get(crfLinesMentionIter);
valString += " B";
crfLines.add(crfLinesMentionIter, valString);
}
else {
valString += crfLines.get(crfLinesMentionIter);
valString += " I";
crfLines.add(crfLinesMentionIter, valString);
}
}
crfLinesMainIter += yes;
}
else {
++crfLinesMainIter;
}
}
else {
++crfLinesMainIter;
}
}
else {
++crfLinesMainIter;
}
}
return crfLines;
}
The problem I face is as follows:
crfLines is a List collections interface.
When the for loop (between //-----) starts out, the crfLines.get(crfLinesMainIter) works fine. But once, it enters into the if and other processing is carried out on it, even though "crfLinesMainIter" changes the crfLines.get(crfLinesMainIter) seems to get a certain previous value. It does not retrieve the actual value at the index. Has anyone faced such a scenario? Would anyone be able to tell me why this occurs?
My actual question is, when does it occur that even though the indexes might be different a list.get() function still retrieves a value from before which was at another index?
For example:
List crfLines = new LinkedList<>();
if crfLinesMainIter = 2
crfLines.get(crfLinesMainIter) brings me a value say 20 and this value 20 satisfies the if loop condition. So then further processing happens. Now when the for loop executes the values of crfLinesMainIter changes to say 5. In this case, crfLines.get(5) should actually bring me a different value, but it still brings me the previous value 20.

(Not an answer.)
Reworked (more or less) for some modicum of readability:
private static List<String> formCrfLinesWithMentionClass(int begin, int end, String id, List<String> mList, int mListPos, List<String> crf) {
List<String> crfLines = crf;
mListPosChanged = mListPos;
int i = begin;
while (i < end) {
if (crfLines.get(i).equals("") || (crfLines.get(i).split("\\s").length != 98)) {
++i;
continue;
}
if (!val.equals(crfLines.get(i).split("\\s")[0])) {
++i;
continue;
}
int yes = checkForConsecutivePresence(i, mList, mListPos, id, crfLines);
if (yes <= 0) {
++i;
continue;
}
mListPosChanged += yes;
for (int j = i; j < i + yes; j++) {
String valString = crfLines.get(j);
valString += (j == i) ? " B" : " I";
crfLines.add(j, valString);
}
i += yes;
}
return crfLines;
}
What is mListPostChanged? I find it confusing that it's being set to the value of a parameter named mListPos--it makes me think the m prefix is meaningless.
What is val in the line containing the split?