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Optimizing Recursive Function in Java

I'm having problem with optimizing a function AltOper. In set {a, b, c}, there is a given multiplication (or binomial operation, whatsoever) which does not follow associative law. AltOper gets string consists of a, b, c such as "abbac", and calculates any possible answers for the operation, such as ((ab)b)(ac) = c, (a(b(ba)))c = a. AltOper counts every operation (without duplication) which ends with a, b, c, and return it as a triple tuple.
Though this code runs well for small inputs, it takes too much time for bit bulky ones. I tried memoization for some small ones, but apparently it's not enough. Struggling some hours, I finally figured out that its time complexity is basically too large. But I couldn't find any better algorithm for calculating this. Can anyone suggest idea for enhancing (significantly) or rebuilding the code? No need to be specific, but just vague idea would also be helpful.
public long[] AltOper(String str){
long[] triTuple = new long[3]; // result: {number-of-a, number-of-b, number-of-c}
if (str.length() == 1){ // Ending recursion condition
if (str.equals("a")) triTuple[0]++;
else if (str.equals("b")) triTuple[1]++;
else triTuple[2]++;
return triTuple;
}
String left = "";
String right = str;
while (right.length() > 1){
// splitting string into two, by one character each
left = left + right.substring(0, 1);
right = right.substring(1, right.length());
long[] ltemp = AltOper(left);
long[] rtemp = AltOper(right);
// calculating possible answers from left/right split strings
triTuple[0] += ((ltemp[0] + ltemp[1]) * rtemp[2] + ltemp[2] * rtemp[0]);
triTuple[1] += (ltemp[0] * rtemp[0] + (ltemp[0] + ltemp[1]) * rtemp[1]);
triTuple[2] += (ltemp[1] * rtemp[0] + ltemp[2] * (rtemp[1] + rtemp[2]));
}
return triTuple;
}

One comment ahead: I would modify the signature to allow for a binary string operation, so you can easiely modify your "input operation".
java public long[] AltOper(BiFunction<long[], long[], long[]> op, String str) {
I recommend using some sort of lookup table for subportions you have already answered. You hinted that you tried this already:
I tried memoization for some small ones, but apparently it's not enough
I wonder what went wrong, since this is a good idea, especially since your input is strings, which are both quickly hashable and comparable, so putting them in a map is cheap. You just need to ensure, that the map does not block the entire memory by ensuring, that old, unused entries are dropped. Cache-like maps can do this. I leave it to you to find one that suites your personal preferences.
From there, I would run any recursions through the cache check, to find precalculated results in the map. Small substrings that would otherwise be calculated insanely often are then looked up quickly, which cheapens your algorithm drastically.
I rewrote your code a bit, to allow for various inputs (including different operations):
import java.util.Arrays;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.function.BiFunction;
import org.junit.jupiter.api.Test;
public class StringOpExplorer {
#Test
public void test() {
BiFunction<long[], long[], long[]> op = (left, right) -> {
long[] r = new long[3];
r[0] += ((left[0] + left[1]) * right[2] + left[2] * right[0]);
r[1] += (left[0] * right[0] + (left[0] + left[1]) * right[1]);
r[2] += (left[1] * right[0] + left[2] * (right[1] + right[2]));
return r;
};
long[] result = new StringOpExplorer().opExplore(op, "abcacbabc");
System.out.println(Arrays.toString(result));
}
#SuppressWarnings("serial")
final LinkedHashMap<String, long[]> cache = new LinkedHashMap<String, long[]>() {
#Override
protected boolean removeEldestEntry(final Map.Entry<String, long[]> eldest) {
return size() > 1_000_000;
}
};
public long[] opExplore(BiFunction<long[], long[], long[]> op, String input) {
// if the input is length 1, we return.
int length = input.length();
if (length == 1) {
long[] result = new long[3];
if (input.equals("a")) {
++result[0];
} else if (input.equals("b")) {
++result[1];
} else if (input.equals("c")) {
++result[2];
}
return result;
}
// This will check, if the result is already known.
long[] result = cache.get(input);
if (result == null) {
// This will calculate the result, if it is not yet known.
result = applyOp(op, input);
cache.put(input, result);
}
return result;
}
public long[] applyOp(BiFunction<long[], long[], long[]> op, String input) {
long[] result = new long[3];
int length = input.length();
for (int i = 1; i < length; ++i) {
// This might be easier to read...
String left = input.substring(0, i);
String right = input.substring(i, length);
// Subcalculation.
long[] leftResult = opExplore(op, left);
long[] rightResult = opExplore(op, right);
// apply operation and add result.
long[] operationResult = op.apply(leftResult, rightResult);
for (int d = 0; d < 3; ++d) {
result[d] += operationResult[d];
}
}
return result;
}
}
The idea of the rewrite was to introduce caching and to isolate the operation from the exploration. After all, your algorithm is in itself an operation, but not the 'operation under test'. So now you colud (theoretically) test any operation, by changing the BiFunction parameter.
This result is extremely fast, though I really wonder about the applicability...

Print Tree with 4 nodes (simple forest) for checking a benchmark

I implemented an experimental OOP language and now benchmark garbage collection using a Storage benchmark. Now I want to check/print the following benchmark for small depths (n=2, 3, 4,..).
The tree (forest with 4 subnode) is generated by the buildTreeDepth method. The code is as follows:
import java.util.Arrays;
public final class StorageSimple {
private int count;
private int seed = 74755;
public int randomNext() {
seed = ((seed * 1309) + 13849) & 65535;
return seed;
}
private Object buildTreeDepth(final int depth) {
count++;
if (depth == 1) {
return new Object[randomNext() % 10 + 1];
} else {
Object[] arr = new Object[4];
Arrays.setAll(arr, v -> buildTreeDepth(depth - 1));
return arr;
}
}
public Object benchmark() {
count = 0;
buildTreeDepth(7);
return count;
}
public boolean verifyResult(final Object result) {
return 5461 == (int) result;
}
public static void main(String[] args) {
StorageSimple store = new StorageSimple();
System.out.println("Result: " + store.verifyResult(store.benchmark()));
}
}
Is there a somewhat simple/straight forward way to print the tree generated by buildTreeDepth? Just the short trees of n=3, 4, 5.

As other has already suggested, you may choose some lib to do so. But if you just want a simple algo to test in command line, you may do the following, which I always use when printing tree in command line (write by handle, may have some bug. Believe you can get what this BFS algo works):
queue.add(root);
queue.add(empty);
int count = 1;
while (queue.size() != 1) {
Node poll = queue.poll();
if (poll == empty) {
count = 1;
queue.add(empty);
}
for (Node n : poll.getChildNodes()) {
n.setNodeName(poll.getNodeName(), count++);
queue.add(n);
}
System.out.println(poll.getNodeName());
}
Sample output:
1
1-1 1-2 1-3 1-4
1-1-1 1-1-2 1-1-3 1-2-1 1-2-2 1-3-1 1-3-2 1-4-1
...
And in your case you use array, which seems even easier to print.

Instead of using object arrays, use a List implementation like ArrayList. For an improved better result subclass ArrayList to also hold a 'level' value and add indentation to the toString() method.

String list get an item starting without loop

I have a array list contains thousands of data.
For Example:
List<String> custNames = new ArrayList<String>();
custNames.add("John");
custNames.add("Tom");
custNames.add("Bart");
custNames.add("Tim");
custNames.add("Broad");
Now I want to get count of names only starting with 'T'. I used looping mechanism for my solution.
List<String> filterNames = new ArrayList<String>();
String nameStarts="T";
for(int i=0;i<custNames.size();i++)
{
if(custNames.get(i).toLowerCase().startsWith(nameStarts.toLowerCase()))
{
filterNames.add(custNames.get(i));
}
}
System.out.println(filterNames.size());
But I have very large collection of data in this custNames list.
Is there any different solution without using loop?
Thanks.

There is very good solution from Java 8 for your problem.
Try this,
long filterNameCount = custNames
.stream()
.parallel()
.filter((s) -> s.startsWith(nameStarts.toLowerCase()))
.count();
System.out.println(filterNameCount);

If you are open to using a third-party library, there are a few interesting options you could use with Eclipse Collections.
If you use the ArrayList as you have it above, you can use the LazyIterate utility as follows:
int count = LazyIterate.collect(custNames, String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
Assert.assertEquals(2, count);
If you use the Eclipse Collections replacement for ArrayList, you can use the rich functional protocols available directly on MutableList:
MutableList<String> custNames =
Lists.mutable.with("John", "Tom", "Bart", "Tim", "Broad");
String nameStarts= "T";
int count = custNames.asLazy()
.collect(String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
System.out.println(count);
Assert.assertEquals(2, count);
The serial API in Eclipse Collections is eager-by-default, which is why I called asLazy() first. The collect method would otherwise create another MutableList.
If you benchmark your code with your full set of data, the following parallel version of the code may be more performant:
MutableList<String> custNames =
Lists.mutable.with("John", "Tom", "Bart", "Tim", "Broad");
String nameStarts= "T";
int processors = Runtime.getRuntime().availableProcessors();
int batchSize = Math.max(1, custNames.size() / processors);
ExecutorService executor = Executors.newFixedThreadPool(processors);
int count = custNames.asParallel(executor, batchSize)
.collect(String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
executor.shutdown();
Assert.assertEquals(2, count);
The asParallel() API in Eclipse Collections is lazy-by-default. The API forces you to pass in a an ExecutorService and an int batchSize. This gives you complete control over the parallelism.
You can also use the Stream API with all MutableCollections in Eclipse Collections because they extend java.util.Collection.
Note: I am a committer for Eclipse Collections.

You could also use a tree storage : it would very efficient for this kind of search. If you are stucked with a list the previous answered is a way to do.

remove all the items which dont start with "T" like this:
custNames.removeIf(p->!p.startsWith("T"));
you can make a copy out of your list and remove items not starting with "T".

First, you can shorten your initialization with Arrays.asList(T); Second, I would use a simple loop to build a table of counts once and then use that to determine the subsequent queries. Something like,
List<String> custNames = new ArrayList<String>(Arrays.asList("John", "Tom",
"Bart", "Tim", "Broad"));
int[] counts = new int[26];
for (String name : custNames) {
char ch = Character.toLowerCase(name.charAt(0));
counts[ch - 'a']++;
}
for (int i = 0; i < counts.length; i++) {
if (counts[i] > 0) {
System.out.printf("There are %d words that start with %c%n",
counts[i], (char) ('a' + i));
}
}
Which outputs
There are 2 words that start with b
There are 1 words that start with j
There are 2 words that start with t
Or, in the specific case - counts['t' - 'a'] is the count of words starting with t.

If you have more or less static list and perform search operation often you can sort your list or use TreeMap.
Also you don't need to create new list and get its size then. You can simply create a counter variable and increment it.

You can create your own sorting and finding implementation.
Consider the following:
public class ContainingArrayList<E> extends ArrayList<E> {
private Comparator<E> comparator;
public ContainingArrayList(Comparator<E> comparator) {
this.setComparator(comparator);
}
#Override
public boolean add(E e) {
// If the collection is empty or the new element is bigger than the last one, append it to the end of the collection
if(size() == 0 || comparator.compare(e, get(size()-1)) >= 0)
return super.add(e);
else {
for (int i = 0; i < size(); i++) {
int result = comparator.compare(e, get(i));
// If the new element is bigger than the current element, continue with the next element
if (result > 0) continue;
// If the new element is equal to the current element, no need to insert (you might insert of course)
if (result == 0) return false;
// Otherwise the new element is smaller than the current element, so insert it between the previous and the current element
super.add(i, e);
return true;
}
return super.add(e);
}
}
public E get(E containingElement) {
int start = 0;
int end = size()-1;
// If the element is the first one, return the first element
if(comparator.compare(containingElement, super.get(start)) == 0)
return super.get(start);
// If the element is the last one, return the last element
if(comparator.compare(containingElement, super.get(end)) == 0)
return super.get(end);
// Otherwise do a binary search
while(start != end) {
// Get the element between start and end positions
E mid = super.get(start + (end/2));
// Compare the two elements
int result = comparator.compare(containingElement, mid);
// If the middle element compared to the containing element is equal, return the middle element
if(result == 0) {
return mid;
}
// If the containing element is smaller than the middle, halve the end position
else if(result < 0) {
end = start + (end/2);
}
// If the containing element is bigger than the middle, set the start position to the middle position
else if(result > 0) {
start = start + (end/2);
}
}
return null;
}
public Comparator<E> getComparator() {
return comparator;
}
public void setComparator(Comparator<E> comparator) {
this.comparator = comparator;
}
}
The custom comparator is used to sort the elements and to find the element that starts with a specific character. This means that you can change the comparator implementation for your needs at any time or you can create a more dynamic finding solution.
Test:
public class SortFindTest {
public SortFindTest() {
ContainingArrayList<String> t = new ContainingArrayList<String>(new MyComparator());
t.add("John");
t.add("Tom");
t.add("Bart");
t.add("Tim");
t.add("Broad");
System.out.println(t.get("T"));
}
class MyComparator implements Comparator<String> {
#Override
public int compare(String o1, String o2) {
int o1c = o1.charAt(0);
int o2c = o2.charAt(0);
if(o1c == o2c)
return 0;
if(o1c > o2c)
return 1;
return -1;
}
}
public static void main(String[] args) {
new SortFindTest();
}
}
I'm not sure if this would be faster than Java 8 Stream API but it worth a try.

If the order in which the items are stored does not matter, you could store the names in a HashMap, where the first character of each name is the key, and an ArrayList of names with that first character are the values. And then all you need to do, assuming the HashMap is named customerList, is customerList.get("T").size().
Initializing HashList and Adding Customers
HashMap<Character, ArrayList<String>> customerList = new HashMap<Character, ArrayList<String>>();
int NUM_ALPHABETS = 26;
int ascii_char = 97;
for(int i = 0; i < NUM_ALPHABETS; i++){
char c = (char) ascii_char;
customerList.add(c, new ArrayList<String>());
ascii_char++;
}
customerList.get("t").add("Tony");
customerList.get("a").add("Alice");
customerList.get("b").add("Ben");
Getting Number of Customers Starting with "t"
int num_t = customerList.get("t").size();

algorithm to find the largest area

................................
.XXXXXXXXXXXXXXX.....XXXXXXXXXX.
.X.....X.......X.....X........X.
.X.....X.......XXXXXXX........X.
.XXXXXXXXXXXX.................X.
.X....X.....X.................X.
.X....X.....XXXX..............X.
.XXXXXX........X..............X.
......X........X..............X.
......X........X..............X.
......X........X..............X.
......XXXXXXXXXXXXXXXXXXXXXXXXX.
................................
Looking for an algorithm to find the largest area. Here, "area" is defined as a number of dots (.) bounded by Xs.
private static void readFile(File inputFile) throws IOException {
Scanner fileScanner = new Scanner(inputFile);
Point previousPoint = null;
int rowCount = 0;
while(fileScanner.hasNext()){
String line = fileScanner.next();
String[] points = line.split(" ");
for(int columnCount=0;columnCount<points.length;columnCount++){
if(points[columnCount].equalsIgnoreCase("x")){
Point currentPoint = new Point();
currentPoint.setxValue(columnCount);
currentPoint.setyValue(rowCount);
}
}
rowCount++;
}
}
This is my first and struggling to move further.

This algorithm should work. You just need to implement it in Java.
Load the file into a char[][]. (1 char[] per line)
Loop through the char[][] (2 dimensionally)
upon finding a '.', perform flood fill, changing all '.' to ',', also incrementing a counter on every change.
At the end of flood fill, compare this counter with a globally set maximum. If it's higher, then set it as the new highest. (If the edges are not a proper boundary, then do not set this counter if you reached an edge during flood fill by setting a flag during 3)
Return the highest you set.
If you have any specific problems with the Java implementation, then let me know
Geobits:
Note: If you want to exclude the area "outside" any boxes, flood as
usual, but discard any area that hits the edge during the fill(skip
step 2.2 for that flood).
When doing the flood fill, you have 2 types of boundaries. A wall ('X'), and the edge of the array(which you need to explicitly check for to avoid OutOfBounds exceptions). If you hit an out of bounds, keep doing the fill, but set a flag so you know later to not consider the number you counted for the biggest box.

I was given this as assignment in an interview process and this is the compile and running code
import java.io.BufferedReader;
import java.io.FileReader;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;
public class FindArea {
public static void main(String[] args)
{
String fileName="C:\\map.txt";
FindArea area = new FindArea();
try{
FileReader inputFile = new FileReader(fileName);
BufferedReader bufferReader = new BufferedReader(inputFile);
char[][] twoArray= new char[100][100];
String line;
int i=0;
while ((line = bufferReader.readLine()) != null) {
twoArray[i] = line.toCharArray();
System.out.println(line);
i++;
}
bufferReader.close();
System.out.println("file read");
System.out.println("Max area: " + area.getMaxArea(twoArray));
} catch(Exception e) {
System.out.println("error : " + e.getMessage());
}
}
/**
* Get the maximum area from the given map
*
* #param charArray
* #return
*/
private int getMaxArea(char[][] charArray) {
HashMap<Integer, ArrayList<String>> numberOfBoxes = convertToBoxes(charArray);
numberOfBoxes = mergeOverlapAreas(numberOfBoxes);
int largeSize = 0;
for (Integer key : numberOfBoxes.keySet()) {
ArrayList<String> list = numberOfBoxes.get(key);
System.out.println("Key : " + key + " Size : " + list.size());
if (largeSize < list.size()) {
largeSize = list.size();
}
}
return largeSize;
}
/**
* Convert the 2d Array to HashMap
* Key being the count of boxes and
* Value being the list of indexes associations
*
* #param charArray
* #return
*/
private HashMap<Integer, ArrayList<String>> convertToBoxes(char[][] charArray) {
HashMap<Integer, ArrayList<String>> numberOfBoxes = new HashMap<Integer, ArrayList<String>>();
int boxes = 0;
for(int i=1; i<charArray.length; i++) {
for (int j=0; j<charArray[i].length; j++) {
if (charArray[i][j] == '.') {
boolean isExists = false;
for(Integer key : numberOfBoxes.keySet()) {
ArrayList<String> arrList = numberOfBoxes.get(key);
if(arrList != null) {
if(arrList.contains((i-1) + "-" + j) ||
arrList.contains(i + "-" + (j-1))) {
isExists = true;
arrList.add(i + "-" + j);
numberOfBoxes.put(key, arrList);
}
}
}
if (!isExists) {
ArrayList<String> list = new ArrayList<String>();
list.add(i + "-" + j);
numberOfBoxes.put(boxes, list);
boxes++;
}
}
}
}
return numberOfBoxes;
}
/**
* Check for the points exists in more than one area
* #param numberOfBoxes
* #return
*/
private HashMap<Integer, ArrayList<String>> mergeOverlapAreas( HashMap<Integer, ArrayList<String>> numberOfBoxes) {
for(Integer key : numberOfBoxes.keySet()) {
ArrayList<String> list1 = numberOfBoxes.get(key);
for (Integer key2 : numberOfBoxes.keySet()) {
if (key < key2) {
ArrayList<String> list2 = numberOfBoxes.get(key2);
Iterator<String> listIter = list2.iterator();
while(listIter.hasNext()) {
if (list1.contains(listIter.next())) {
list1.addAll(list2);
Set<String> noDuplicates = new HashSet<String>(list1);
numberOfBoxes.put(key, new ArrayList<String>(noDuplicates));
break;
}
}
}
}
}
return numberOfBoxes;
}
}

Here's an algorithm that's an alternative to flood fill. This method sweeps through the 2d array and whenever you encounter a node(pixel) that's outside to the left (right, top, bottom), it flags the current node as outside, ie if your neighbour is 'outside', you're marked 'outside' too.
The algorithm continues like this until there're no more updates. That means that all the nodes that are reachable from the 'outside' have been flagged. BTW, this is a very similar problem to level sets functions and updating them (where flood fill is also used). The nice this about this method is that it is ideal for parallelization.
1. Load 2D Symbol Array from File
2. hasupdates = false
3. Create 'isinside' bool array -> {
if(symbolarray[row][col] == '.' and row or col is at boundary)
isinside[row][col] = false
else
isinside[row][col] = true
}
4. do{
Do a sweep from left to right (for all rows) -> //This loop can be run parallely on all rows.
If (!isinside[row][col-1] and symbolarray[row][col] == '.'){
isinside[row][col] = false //mark current value as 'outside'
hasupdates = true
}
Do similar sweeps from right to left, top to bottom(all columns) and bottom to top.
}while(hasupdates)
5. Go through 'isinside' array and count the number of falses.
If you have huge files where you have to do this area calculation, you can have the sweeps along the rows and columns run parallely, because each row update (column update) is independent of the other updates.

How to leftshift an ArrayList

I'm using an ArrayList to hold a history of objects. Each new object I add using the .add method, like:
if(event.getAction() == MotionEvent.ACTION_UP)
{
if(currentWord != null)
{
wordHist.add(currentWord);
}
if(wordHist.size() > WORDHIST_MAX_COUNT)
{
wordHist.remove(0);
}
}
However I don't want this to grow indefinitely, but to be limited to a certain value. If it reaches this maximum value, I want the oldest object (index 0) to be removed, and the rest to be left shifted, so previous index 1 is now index 0, etc.
How can this be done?
Thanks

ArrayList is not really a good choice in this case, but it can by done by calling remove(0) method. But if you want to do that efficiently, a linked list is better
(edited to make it clear that LinkedList is not generally better than ArrayList, but only in this case)

If it reaches this maximum value, I want the oldest object (index 0) to be removed
Then do wordHist.remove(0). That will remove the element at index 0.
To be precise:
wordHist.add(new Word("hello"));
if (wordHist.size() > MAX_SIZE)
wordHist.remove(0);
As user658991 states however, you should be aware of that this is a linear operation, i.e., takes time proportional to the number of elements in the list.
You could do this in constant time using LinkedList methods add and removeFirst.
Another option would be to wrap an array, or ArrayList in a class called something like CircularArrayList. In circular list structures you'll override the oldest element when adding a new one.
Edit:
Your code works fine:
import java.util.*;
class Test {
static int WORDHIST_MAX_COUNT = 3;
static List<String> wordHist = new ArrayList<String>();
public static void add(String currentWord) {
// VERBATIM COPY OF YOUR CODE
if (true/*event.getAction() == MotionEvent.ACTION_UP*/)
{
if(currentWord != null)
{
wordHist.add(currentWord);
}
if(wordHist.size() > WORDHIST_MAX_COUNT)
{
wordHist.remove(0);
}
}
}
public static void main(String[] args) {
add("a");
add("b");
add("c");
for (int i = 0; i < wordHist.size(); i++)
System.out.printf("i: %d, word: %s%n", i, wordHist.get(i));
System.out.println();
add("d");
for (int i = 0; i < wordHist.size(); i++)
System.out.printf("i: %d, word: %s%n", i, wordHist.get(i));
}
}
Prints:
i: 0, word: a
i: 1, word: b
i: 2, word: c
i: 0, word: b <-- b is now at index 0.
i: 1, word: c
i: 2, word: d

Use the remove( ) method.
Using remove(0) will remove the element from the 0th index.

U can use list.remove(index)// here index being '0', this internally shifts rest of the array up. An alternative solution wud be to use a queue or dequeue.

One simple implementation of what Op De Cirkel suggested
import java.util.ArrayList;
import java.util.List;
public class SimpleCircularHistory {
private int sizeLimit, start = 0, end = 0;
boolean empty = false;
private List<String> history;
public SimpleCircularHistory(int sizeLimit) {
this.sizeLimit = sizeLimit;
history = new ArrayList<String>(sizeLimit);
}
public void add(String state){
empty = false;
end = (end + 1) % sizeLimit;
if(history.size() < sizeLimit){
history.add(state);
}else {
history.set(end, state);
start = (end + 1) % sizeLimit;
}
}
public String rollBack(){
if(empty){ // Empty
return null;
}else {
String state = history.get(end);
if(start == end){
empty = true;
}else {
end = (end + sizeLimit - 1) % sizeLimit;
}
return state;
}
}
public void print(){
if(empty){
System.out.println("Empty");
}else {
for(int i = start;; i = (i + 1) % sizeLimit){
System.out.println(history.get(i));
if(i == end) break;
}
System.out.println();
}
}
public static void main(String[] args) {
SimpleCircularHistory h = new SimpleCircularHistory(3);
h.add("a");
h.add("b");
h.add("c");
h.add("d");
h.add("e");
h.add("f");
h.print();
h.add("X");
h.add("Y");
h.rollBack();
h.rollBack();
h.print();
h.add("t");
h.add("v");
h.add("w");
h.print();
h.rollBack();
h.rollBack();
h.rollBack();
h.print();
h.rollBack();
h.print();
}
}
This would print out :
d
e
f
f
t
v
w
Empty
Empty

Yeah, I've noticed this behaviour in adroid's lists too. It's REALLY irritating.
Anyway, there is a way to get around it if I don't mind object creation/destruction and the resulting garbage collection (NEVER do this in a onDraw of a surfaceview or something).
What I do is basically have two tracking int's; one to place the new object, and one to remove it:
int trackInt = 0;
int removeInt = 0;
//and then, in the method/class you use this:
Object newobject = new Object();
//add to list
objectList.add(trackInt, newobject);
trackInt++;
if (bugList.size() > 20) //20 is the max number of object you want, ie the maximum size of the list
{
objectList.remove(removeInt);
trackInt = removeInt;
removeInt++;
if (removeInt > 19) //remember, the list is zero indexed!
{
removeInt = 0;
}
}

Commons-collections has exactly what you're looking for:
http://commons.apache.org/collections/apidocs/org/apache/commons/collections/buffer/CircularFifoBuffer.html