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Java Bucket Sort of Elements in another Array

I just recently started learning java and today I learned how I can do the so called selection sort. I have been trying for the last 3 hours to do a bucket sort, but there are some parts which I don't know how to code. Important : I am learning java completely by myself with a book. I am not a student and I am doing this as a hobby. I already googled everything I could think of and I didn't find a solution. I don't have a teacher or anybody who I can ask, so yea, any help would be appreciated!
Code:
private int[] bucketSort() {
int[]bucket=new int[maxSize+1];
int[]sortedElements = new int[elementaros.length];
for(int i=0; i<elementaros.length;i++) {
bucket[elementaros[i]]++; //it says that I can't convert from Car to int. How can I add the elements of array elemenators to bucket?
}
int outPos = 0;
for (int j = 0; j < bucket.length; j++){
for (int k = 0; k < bucket[i]; k++){
sortedElements[outPos++] = i;
}
}
return bucket;
}
The idea of the code :
I have an array elements of type Car(Car is another class of my program). It looks like this - Car[] elementaros. int maxSize shows the maximum number of administrable Car objects. What I want to do is the following - I want to sort the elements in the elementaros array alphabetically. I would really really appreciate it if somebody has the time to show me how this would function with an example code or would just give me some tips. As I said - I have nobody who I can ask.

A selection sort is a combination of searching and sorting.
The principle is quite simple but I always prefer a diagram than huge explanations.
Start a pointer at the beginning of your unsorted array. Then, for each value of the array, search for the minimum value (or search for the alphabeticaly ordered car) in your array and switch the position of the founded Car with the pointer (which is a Car too)
Then you can advance the pointer to the next element of the array.
Here is a basic implementation to do this
public static Car[] doSelectionSort(Car[] elementaros) {
for (int i = 0; i < elementaros.length - 1; i++) {
int index = i;
for (int j = i + 1; j < elementaros.length; j++) {
if (elementaros[j].getName().compareTo(elementaros[index].getName()) < 0) {
index = j;
}
}
Car nextOrderedCar = elementaros[index];
elementaros[index] = elementaros[i];
elementaros[i] = nextOrderedCar;
}
return elementaros;
}
Just for example sake, I imagine your object of type Car has a name that we could use for the comparison.
UPDATE 1:
I have read your initial question too quickly and it leds me to answer a total different sorting algorithm. My bad.
I found an implementation here that does the trick:
williamfiset bucket sort
explanation of bucket sort plus different implementation
Hope this helps.

Where is the Data Changing? - Permutations in Java

So for this extra credit problem in my calculus class, my other nerdy classmates and I decided that we would build a program to brute force a solution. One of these steps involves permutations. Through this algorithm, I managed to get it to work (I think):
public void genPermutations(int[] list, int k){
System.out.println("List: " + Arrays.toString(list));
System.out.println("----------------------");
if(k > list.length){
System.out.println("Not enough elements!");
return;
}
int[] counts = new int[list.length];
for(int i = 0; i < counts.length; i++){
counts[i] = 1;
}
int[] data = new int[k];
permutationHelper(list, counts, data, 0, k);
}
public void permutationHelper(int[] list, int[] counts, int[] data, int index, int k){
if(index == k){
//System.out.println(Arrays.toString(data));
permutations.add(data);
}else{
for(int i = 0; i < list.length; i++){
if(counts[i] == 0){
continue;
}
data[index] = list[i];
counts[i]--;
permutationHelper(list, counts, data, index + 1, k);
counts[i]++;
}
}
}
I have an ArrayList that stores all of the possible permutations (as integer arrays) that can be made from k elements of the list that I pass into the function. The problem is that if I print all of these permutations outside of the function, say after I call the genPermutations function, every permutation now is the same. But, when I print out the data where the comment is in the permutationHelper function, it correctly lists every possible permutation; I'm just unable to access them within the program later. My question is why are the values changing when I exit the function? Any help would be greatly appreciated.
Here are some pictures:
What is printed where the comment is.
What is printed later in the program.
The code used to print everything outside of the function is:
for(int i = 0; i < permutations.size(); i++){
System.out.println(Arrays.toString(permutations.get(i)));
}
I don't really know if that's necessary to know, but I just thought I'd include it just in case. Thanks in advance.

You're constantly modifying the same array object. Instead of adding different arrays to your list, you're in fact adding a reference to the same array over and over again.
To fix, instead of adding the data array to your list, you would have to add a copy of it, e.g. using Arrays.copyOf():
permutations.add(Arrays.copyOf(data, data.length));

Here the problem is that you are modifying the array after adding it to the list, you are modifying the same object again and again in different iterations. You were getting [3,2,1] in the list is because that was the outcome from last iteration. So as a fix you can use the following code. What it does is it will create a copy of data array and add that to the list.
int[] temp = Arrays.copyOf(data, data.length);
permutations.add(temp);
OR you can use clone() from array as follows.
int[] temp = data.clone();
permutations.add(temp);

Can you remove an element from an array by placing it at the end of the array and decreasing the size of the array?

for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
tmp = bag[i];
bag[i] = bag[bag.length-1];
bag[bag.length-1] = tmp;
numElements--;
break;
}
}
The goal of this is to find an object in the array and then remove it? is it possible??

Changing the length of an array is not possible. Recall that array is a static data structure whose size is determined before hand. Increasing or decreasing is not supported in this data structure. The fact that one has to increase or decrease the size depending on the usecase means that they have picked up the wrong data structure. They should perhaps go with an ArrayList.
Anyway, coming back to your question, you can simulate the 'size decrease' by maintaining a variable which you let track the array index and decrease the size of this variable. This lets you give the impression of shrinking the array.
The code you have provided does the same. Note however, that you should be using this modified index to track the contents of your array.
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
tmp = bag[i];
bag[i] = bag[bag.length-1];
bag[bag.length-1] = tmp;
numElements--;
break;
}
}
Whenever a particular bag at a given index equals to the item under question i.e., 'a', we swap elements so that the current bag element to be removed moves to the last and also we reduce the size of our new index - numElements by 1 to simulate this.
If you have the full code with you, please consider adding the following snippet at the end of that program to understand this more:
// Simulation of the array shrinking.
for(int i = 0; i < numElements; i++)
{
System.out.println( bag[i] );
}
// Movement of uninteresting elements to the end of the array.
for(int i = 0; i < bag.length; i++)
{
System.out.println( bag[i] );
}

It's not possible to change the length of an array. You can overwrite the element you wish to remove with the last element of the array and then copy the first bag.length - 1 elements of your array to a new array whose length is bag.length - 1.
for(int i = 0; i < bag.length; i++) {
if(bag[i].equals(a)) {
bag[i] = bag[bag.length-1];
bag = Arrays.copyOf (bag, bag.length - 1);
break;
}
}

public static String[] removeElements(String[] input) {
List<String> result = new ArrayList<String>();
String deleteValue = "somevalue";
for(String item : input)
if(!deleteValue .equals(item))
result.add(item);
return result.toArray(input);
}
This is one method you can fit this into your program.

You cannot decrease the size of an array. okay no problem! you can create your own data structure which supports that right?
Now, create a class named say MyArray with functions like increaseLenght(int) and decreseLength(int). Try it if you want to, will be fun for sure..

You cannot reduce the size of an array. Arrays are fixed length. What you can do is have a variable that indicates how many entries of the array you are using. This is what you are doing with numElements. The standard class ArrayList is implemented like this. The data is kept in an array and a private field size is used. With an ArrayList, when you remove an element, all the elements to the right are shifted left. However I also like your idea.
I would suggest 2 changes.
Make the last element null instead. If you are removing the element, why does it still need to be in the array?
Use numElements - 1 rather than bag.length-1 as the array could be bigger.
With these changes it becomes:
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
bag[i] = bag[numElements-1];
bag[numElements-1] = null;
numElements--;
break;
}
}

Delete some elements from the array

I need to delete elements from the array points. This is how I do this. The problem is that pts.length is always the same, and the removed elements have the value null. Therefore at some moment I receive the error message java.lang.NullPointerException.
for (int i = 0; i < points.length; i++) {
int ind = r.nextInt(pts.length);
TSPPoint pt = points[ind];
pts = removeElements(points,ind);
solPoints[i] = pt;
System.out.println(pts.length);
}
private static TSPPoint[] removeElements(TSPPoint[] input, int ind) {
List<TSPPoint> result = new LinkedList<TSPPoint>();
for(int i=0; i<input.length; i++)
if(i != ind)
result.add(input[i]);
return (TSPPoint[]) result.toArray(input);
}

What your code seems to (be supposed to) do is to remove random elements from the original array of points and append them to the pts array, i.e., create a permutation of points.
If this is the case, I suggest converting your array to a List and using Collections.shuffle.

#nrathaus has found the bug for you. It's simply that you've got your arrays confused (you're passing points into removeElements, but using pts everywhere else).
But if memory churn is being an issue, there's a much more efficient way to implement removeElements, using System.arraycopy rather than a temporary LinkedList.:
private static TSPPoint[] removeElements(TSPPoint[] input, int ind) {
TSPPoint[] rv;
if (ind >= 0 && ind < input.length) {
// New array will be one smaller
rv = new TSPPoint[input.length - 1];
if (rv.length > 0) {
// Copy the bit before the element we delete
if (ind > 0) {
System.arraycopy(input, 0, rv, 0, ind);
}
// Copy the rest
System.arraycopy(input, ind + 1, rv, ind, input.length - ind);
}
}
else {
// No change
rv = input;
}
return rv;
}
Mind you, if you're doing this a lot, creating and releasing all of these arrays may not be ideal. Using a List throughout may be better.

Wow, for deleting every element you recreate the rest of the array in a LinkedList, which you then turn into an array... The performance of this code, both in time and space, is terrible and so is readability, maintainability and testability.
Why not drop using arrays all together?.. convert points into an ArrayList and use remove(index) directly on that list and use an iterator of the llist to remove multiple elements as you iterate through?

Delete data from ArrayList with a For-loop

I got a weird problem.
I thought this would cost me few minutes, but I am struggling for few hours now...
Here is what I got:
for (int i = 0; i < size; i++){
if (data.get(i).getCaption().contains("_Hardi")){
data.remove(i);
}
}
The data is the ArrayList.
In the ArrayList I got some strings (total 14 or so), and 9 of them, got the name _Hardi in it.
And with the code above I want to remove them.
If I replace data.remove(i); with a System.out.println then it prints out something 9 times, what is good, because _Hardi is in the ArrayList 9 times.
But when I use data.remove(i); then it doesn't remove all 9, but only a few.
I did some tests and I also saw this:
When I rename the Strings to:
Hardi1
Hardi2
Hardi3
Hardi4
Hardi5
Hardi6
Then it removes only the on-even numbers (1, 3, 5 and so on).
He is skipping 1 all the time, but can't figure out why.
How to fix this? Or maybe another way to remove them?

The Problem here is you are iterating from 0 to size and inside the loop you are deleting items. Deleting the items will reduce the size of the list which will fail when you try to access the indexes which are greater than the effective size(the size after the deleted items).
There are two approaches to do this.
Delete using iterator if you do not want to deal with index.
for (Iterator<Object> it = data.iterator(); it.hasNext();) {
if (it.next().getCaption().contains("_Hardi")) {
it.remove();
}
}
Else, delete from the end.
for (int i = size-1; i >= 0; i--){
if (data.get(i).getCaption().contains("_Hardi")){
data.remove(i);
}
}

You shouldn't remove items from a List while you iterate over it. Instead, use Iterator.remove() like:
for (Iterator<Object> it = list.iterator(); it.hasNext();) {
if ( condition is true ) {
it.remove();
}
}

Every time you remove an item, you are changing the index of the one in front of it (so when you delete list[1], list[2] becomes list[1], hence the skip.
Here's a really easy way around it: (count down instead of up)
for(int i = list.size() - 1; i>=0; i--)
{
if(condition...)
list.remove(i);
}

Its because when you remove an element from a list, the list's elements move up. So if you remove first element ie at index 0 the element at index 1 will be shifted to index 0 but your loop counter will keep increasing in every iteration. so instead you of getting the updated 0th index element you get 1st index element. So just decrease the counter by one everytime you remove an element from your list.
You can use the below code to make it work fine :
for (int i = 0; i < data.size(); i++){
if (data.get(i).getCaption().contains("_Hardi")){
data.remove(i);
i--;
}
}

It makes perfect sense if you think it through. Say you have a list [A, B, C]. The first pass through the loop, i == 0. You see element A and then remove it, so the list is now [B, C], with element 0 being B. Now you increment i at the end of the loop, so you're looking at list[1] which is C.
One solution is to decrement i whenever you remove an item, so that it "canceles out" the subsequent increment. A better solution, as matt b points out above, is to use an Iterator<T> which has a built-in remove() function.
Speaking generally, it's a good idea, when facing a problem like this, to bring out a piece of paper and pretend you're the computer -- go through each step of the loop, writing down all of the variables as you go. That would have made the "skipping" clear.

I don't understand why this solution is the best for most of the people.
for (Iterator<Object> it = data.iterator(); it.hasNext();) {
if (it.next().getCaption().contains("_Hardi")) {
it.remove();
}
}
Third argument is empty, because have been moved to next line. Moreover it.next() not only increment loop's variable but also is using to get data. For me use for loop is misleading. Why you don't using while?
Iterator<Object> it = data.iterator();
while (it.hasNext()) {
Object obj = it.next();
if (obj.getCaption().contains("_Hardi")) {
it.remove();
}
}

Because your index isn't good anymore once you delete a value
Moreover you won't be able to go to size since if you remove one element, the size as changed.
You may use an iterator to achieve that.

for (Iterator<Object> it = data.iterator(); it.hasNext();) {
if ( it.getCaption().contains("_Hardi")) {
it.remove(); // performance is low O(n)
}
}
If your remove operation is required much on list. Its better you use LinkedList which gives better performance Big O(1) (roughly).
Where in ArrayList performance is O(n) (roughly) . So impact is very high on remove operation.

It is late but it might work for someone.
Iterator<YourObject> itr = yourList.iterator();
// remove the objects from list
while (itr.hasNext())
{
YourObject object = itr.next();
if (Your Statement) // id == 0
{
itr.remove();
}
}

In addition to the existing answers, you can use a regular while loop with a conditional increment:
int i = 0;
while (i < data.size()) {
if (data.get(i).getCaption().contains("_Hardi"))
data.remove(i);
else i++;
}
Note that data.size() must be called every time in the loop condition, otherwise you'll end up with an IndexOutOfBoundsException, since every item removed alters your list's original size.

This happens because by deleting the elements you modify the index of an ArrayList.

import java.util.ArrayList;
public class IteratorSample {
public static void main(String[] args) {
// TODO Auto-generated method stub
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(1);
al.add(2);
al.add(3);
al.add(4);
System.out.println("before removal!!");
displayList(al);
for(int i = al.size()-1; i >= 0; i--){
if(al.get(i)==4){
al.remove(i);
}
}
System.out.println("after removal!!");
displayList(al);
}
private static void displayList(ArrayList<Integer> al) {
for(int a:al){
System.out.println(a);
}
}
}
output:
before removal!!
1
2
3
4
after removal!!
1
2
3

There is an easier way to solve this problem without creating a new iterator object. Here is the concept. Suppose your arrayList contains a list of names:
names = [James, Marshall, Susie, Audrey, Matt, Carl];
To remove everything from Susie forward, simply get the index of Susie and assign it to a new variable:
int location = names.indexOf(Susie);//index equals 2
Now that you have the index, tell java to count the number of times you want to remove values from the arrayList:
for (int i = 0; i < 3; i++) { //remove Susie through Carl
names.remove(names.get(location));//remove the value at index 2
}
Every time the loop value runs, the arrayList is reduced in length. Since you have set an index value and are counting the number of times to remove values, you're all set. Here is an example of output after each pass through:
[2]
names = [James, Marshall, Susie, Audrey, Matt, Carl];//first pass to get index and i = 0
[2]
names = [James, Marshall, Audrey, Matt, Carl];//after first pass arrayList decreased and Audrey is now at index 2 and i = 1
[2]
names = [James, Marshall, Matt, Carl];//Matt is now at index 2 and i = 2
[2]
names = [James, Marshall, Carl];//Carl is now at index 3 and i = 3
names = [James, Marshall,]; //for loop ends
Here is a snippet of what your final method may look like:
public void remove_user(String name) {
int location = names.indexOf(name); //assign the int value of name to location
if (names.remove(name)==true) {
for (int i = 0; i < 7; i++) {
names.remove(names.get(location));
}//end if
print(name + " is no longer in the Group.");
}//end method

This is a common problem while using Arraylists and it happens due to the fact that the length (size) of an Arraylist can change. While deleting, the size changes too; so after the first iteration, your code goes haywire. Best advice is either to use Iterator or to loop from the back, I'll recommend the backword loop though because I think it's less complex and it still works fine with numerous elements:
//Let's decrement!
for(int i = size-1; i >= 0; i--){
if (data.get(i).getCaption().contains("_Hardi")){
data.remove(i);
}
}
Still your old code, only looped differently!
I hope this helps...
Merry coding!!!