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algorithm visualization: implementing insertion sort without loops but with variable increments everytime a function is called

I'm building a sorting algorithm visualizer in processing (extension of java with extra libraries for visualization) and i'm very much stuck on this problem which I think others will be able to help me solve.
In processing there is a function called draw() that is being called 60 times each second. It's here that I want to execute, each time draw() is called, one step of the insertion algorithm. I already implemented it with a bubble sort. (see code below).
updateBubble() is being called in draw() and 'colors' is the name of the arraylist I use to keep the different values of colors to sort.
picture to get a better understanding:
[![visualisation algorithm preview][1]][1]
...
int j = 0
...
void updateBubble() {
bubble.sort(j);
j++;
if (i<bubble.colors.size()) {
if (j >= bubble.colors.size()-i-1) {
j = 0;
i++;
}
} else {
bubble.sorted = true;
}
}
and here is the function in the class BubbleSort (bubble is an object of this class)
void sort(int j) {
if (j<colors.size()-1) {
if (colors.get(j) > colors.get(j+1))
{
int temp = colors.get(j);
colors.set(j, colors.get(j+1));
colors.set((j+1), temp);
}
}
}
This way I was able to slow down the visualization process to the the pace of the framerate which I can control myself without using loops which would execute the sorting algorithm immediately. Now I also wanted to make a similar implementation for the insertion sort algorithm but i feel like i'm stuck because I don't seem to be able to use a similar implementation that works or there might be a better way to do this?
What I have at the moment executes it immediately as expected, without being able to see the process.
void updateInsertion() {
insertion.sort();
}
void sort() {
int n = colors.size();
for (int i = 1; i < n; ++i) {
int key = colors.get(i);
int j = i - 1;
while (j >= 0 && colors.get(j) > key) {
colors.set(j+1, colors.get(j));
j = j - 1;
}
colors.set(j+1, key);
}
}
this is what i got now: which is still wrong but is getting closer and clearifies what i'm trying to reach, making a function that only works with increments and if statements instead of whiles and fors so each different step is being executed with each call of the method.
// i resembles for loop variable
if (i<insertion.colors.size()) {
if (j<0 || insertion.colors.get(j) <= insertion.colors.get(i)) { // negative check to go out of while loop
insertion.colors.set(j+1, keap);
if(notSortedYet()){
i++;
keap = insertion.colors.get(i);
j = i - 1;
}
} else { // resembles being in the while loop
insertion.colors.set((j+1), insertion.colors.get(j));
j = j - 1;
}
}
}
EDIT: I fixed it and you can find my solution beneath :) everytime updateInsertion() is called, my code will execute exact one step in the algorithm! thanks to everyone who put effort into commenting, I dont know if this is best practise, so keep me updated on that if you want!
void updateInsertion() {
// i resembles for loop variable
if (i<insertion.colors.size()) {
if (j>=0 && insertion.colors.get(j) > firstUnsorted) {
int temp = insertion.colors.get(j+1);
insertion.colors.set((j+1), insertion.colors.get(j));
insertion.colors.set(j,temp);
j = j - 1;
} else {
insertion.colors.set(j+1, firstUnsorted);
if (i<insertion.colors.size()-1) {
i++;
}
firstUnsorted = insertion.colors.get(i);
j = i - 1;
}
}
}

I love this project.
Processing also have a millis() method which returns how many milliseconds were spent since you've started your sketch. I sometimes use it to time my animations, which could come in handy right here. Here's an implementation of a timer class:
class Delay {
int limit;
Delay (int l) {
limit = millis() + l;
}
boolean expired () {
return (millis() > limit);
}
}
I suggest that you use this class instead of tweaking the FPS. By using the Delay to slow down your implementation of the sort, you're letting the computer work at it's own rhythm and only draw a new frame when you need it. Like this (excuse the parts where I say "do stuff"):
Delay holdTheFrame = new Delay(1);
void draw() {
if(holdTheFrame.expired()) {
holdTheFrame = new Delay(500); // half a second before the next frame
// Advance one step forward in your sorting
// Draw the visualization of the data
}
}
You can fine tune at what pace your data is sorted and only paint it when it changes. It's win-win!
Have fun!
EDIT
To help you with the implementation, here's an example one. You can copy and paste this code in an empty Processing sketch and it'll run as-is. To make things easier on my side I print to console instead of using the graphical display, but you should be able to get what I'm doing.
The secret here is that my sorting algorithm have been subtly modified so they instead always run only ONE sorting step when I call them. See for yourself:
int _numberOfItems = 10;
int _sortingStep = 0;
IntList _bubbleList = new IntList();
boolean _bubbleListSorted = false;
IntList _selectionList = new IntList();
IntList _insertionList = new IntList();
Delay _delay = new Delay(1);
void setup() {
for (int i=0; i<_numberOfItems; i++) {
_bubbleList.append((int)random(10, 99));
}
for (int i=0; i<_numberOfItems; i++) {
_selectionList.append((int)random(10, 99));
}
for (int i=0; i<_numberOfItems; i++) {
_insertionList.append((int)random(10, 99));
}
}
void draw() {
if (_delay.expired()) {
_delay = new Delay(500);
// sort one step with every algo you want to display
if (!_bubbleListSorted) {
singleStepBubbleSort(_bubbleList);
}
if (_sortingStep < _numberOfItems) {
singleStepSelectionSort(_selectionList, _sortingStep);
singleStepInsertionSort(_insertionList, _sortingStep);
}
_sortingStep++;
// update the display (I'm printing to console instead for simplicity)
for (int i : _bubbleList) {
print(i + " ");
}
print(" | ");
for (int i : _selectionList) {
print(i + " ");
}
print(" | ");
for (int i : _insertionList) {
print(i + " ");
}
print("\n");
}
}
// An "single-step" implementation of Insertion Sort
void singleStepInsertionSort(IntList list, int step) {
int k = list.get(step);
int j = step - 1;
while (j >= 0 && list.get(j) > k) {
list.set(j+1, list.get(j));
j = j - 1;
}
list.set(j+1, k);
}
// An "single-step" implementation of Bubble Sort
void singleStepBubbleSort(IntList list) {
int temp;
boolean swapped = false;
for (int i=0; i<list.size()-1; i++)
{
if (list.get(i) > list.get(i + 1))
{
// swap arr[j] and arr[j+1]
temp = list.get(i);
list.set(i, list.get(i+1));
list.set(i+1, temp);
swapped = true;
}
}
if (!swapped) {
_bubbleListSorted = true;
}
}
// An "single-step" implementation of Selection Sort
void singleStepSelectionSort(IntList list, int step)
{
int min_idx = step;
for (int j = step+1; j < list.size(); j++) {
if (list.get(j) < list.get(min_idx)) {
min_idx = j;
}
}
int temp = list.get(min_idx);
list.set(min_idx, list.get(step));
list.set(step, temp);
}
class Delay {
int limit;
Delay (int l) {
limit = millis() + l;
}
boolean expired () {
return (millis() > limit);
}
}
Let me know if you have questions.
MORE EDITS:
Every swap of an insertion sort means many, many swaps. It's a real pain because this algorithm is kinda complicated to stop in it's tracks.
Luckily, I don't care. Thinking outside the box, I opted instead to create a class dedicated to sort an array while recording how to sort it, then be able to play it back "as if it was happening in real time". take a look:
int numberOfItems = 10;
int sortingStep = 0;
Delay delay = new Delay(1);
ManagedSelectionSort managedSelectionSort; // I created a class just to manage this madness
void setup() {
IntList list = new IntList();
for (int i=0; i<numberOfItems; i++) {
list.append((int)random(10, 99)); // some random numbers to sort later
}
managedSelectionSort = new ManagedSelectionSort(list); // take a look at the instantiation of this class
print("Step " + String.format("%02d", sortingStep) + ": ");
printArray(managedSelectionSort.list);
print("\n");
}
void draw() {
if (delay.expired()) {
delay = new Delay(100); // i put a very short delay, you'll probably want to tweak this
managedSelectionSort.sortOneStep(); // this is not what it seems
sortingStep++;
print("Step " + String.format("%02d", sortingStep) + ": ");
printArray(managedSelectionSort.list);
print("\n");
}
}
// this class is where the magic happens
// we'll sort the array all at once while recording every move
// then we'll play back those moves on a copy of the array
class ManagedSelectionSort {
IntList list, hiddenList; // list is the "official" list, while hiddenList is where the heavy lifting happens
ArrayList<SwapIndex> swapList; // this is where I record how to sort the array
ManagedSelectionSort(IntList baseList) { // this way I can instantiate several similar objects with the same list
list = new IntList();
hiddenList = new IntList();
swapList = new ArrayList<SwapIndex>();
for (int i : baseList) {
// both lists have the same initial numbers
list.append(i);
hiddenList.append(i);
}
// as soon as this object is instantiated, it knows how it'll sort the array
// because it already did...
hiddenSort();
}
// this method plays the moves which were recorded earlier according to the current sortingStep
// the swapList array was filled with every swap needed to sort the array, one by one
// now it's just a matter of playing them back on a copy of the initial array
void sortOneStep() {
if (sortingStep < swapList.size()) {
swap(list, swapList.get(sortingStep).index1, swapList.get(sortingStep).index2);
}
}
// this is the real implementation of the insertion sort
void hiddenSort()
{
for (int i=1; i<hiddenList.size(); i++) {
int j = i;
while (j>0 && hiddenList.get(j) < hiddenList.get(j-1)) {
swap(hiddenList, j, j-1, true); // swap is a class specific helper method, it swaps the numbers and also records the move
j--;
}
}
}
// this is an overload, i could have done without but it's confortable
void swap(IntList list, int index1, int index2) {
swap(list, index1, index2, false);
}
void swap(IntList list, int index1, int index2, boolean recordMove) {
// the swap first
int temp = list.get(index1);
list.set(index1, list.get(index2));
list.set(index2, temp);
// if the method is set on 'record', it adds this move to the swapList array
if (recordMove) {
swapList.add(new SwapIndex(index1, index2));
}
}
}
// this class could have been a struct, but I like to start in OOP right from the bat in case things gets complicated
class SwapIndex {
int index1;
int index2;
SwapIndex(int index1, int index2) {
this.index1 = index1;
this.index2 = index2;
}
}
// this method is just an helper method to print to console
void printArray(IntList list) {
for (int i : list) {
print(i + " ");
}
}
class Delay {
int limit;
Delay (int l) {
limit = millis() + l;
}
boolean expired () {
return millis() > limit;
}
}
This should solve your initial problem, if I understood it right this time!

One way to achieve this is via a some sort of stored state. Below is at a high level what I'm talking about.
// Starts the procedure. Must be called before draw().
void init() {
state = "forLoop";
i = 1;
n = colors.size();
}
// Single iteration of a loop.
void draw(){
switch(state) {
case "forLoop":
doForBody();
break;
case "whileLoop":
doWhileLoopBody();
break;
...
}
}
// Executes everything in the while loop and the one or two things
// just after it.
void doWhileLoopBody() {
if (isThisIterationOfWhileDone()) {
// Get out of the while loop and prepare for the next iteration of for.
// A better way to what I'm doing on the next couple lines here would
// be to introduce an additional state (ex: "postWhile") that would
// execute just after this method and would handle the colors.set(),
// incrementing i, etc.
state = "forLoop";
colors.set(j+1, key);
i++;
return;
}
// update colors, value of j, etc...
}
// Executes everything before the while loop.
void doForLoopBody() {
if (isThisIterationOfForDone()) {
state = "END";
return;
}
// update colors, get values of key and j initialized, etc
// switch to processing the body of the while loop
state = "whileLoop";
}

CoinChange Problem with DP in Java using 2D array

I am implementing an algorithm to solve the Coin Change problem, where given an array that indicates types of coins (i.e. int[] coinValues = {1,4,6};) and a value to achieve (i.e. int totalAmount=8;), an array is returned where the value at position 0 indicates the minimum number of coins needed to achieve totalAmount. The rest of the array will keep a track of how many coins are needed to achieve the total sum.
An example input of coins = {1,4,6} and total = 8 should return the array [3,2,0,1]. However, my code is returning [1,2,0,1].
Another example would be coins = {2,4,8,16,34,40,64} and total = 50 should return the array [2, 0, 0, 0, 1, 1, 0, 0]. My code is not returning that result.
The algorithm is implemented with 2 methods: CoinChange and CoinCount. CoinChange creates the coin matrix and CoinCount keeps track of the coins required to achieve the total sum.
package P5;
import java.util.Arrays;
public class CoinChange7 {
public static int[] CoinChange(int[] v, int sum) {
int[][] aux = new int[v.length + 1][sum + 1];
// Initialising first column with 0
for(int i = 1; i <= v.length; i++) {
aux[i][0] = 0;
}
// Implementing the recursive solution
for(int i = 1; i <= v.length-1; i++) {
for(int j = 1; j <= sum; j++) {
if(i == 1) {
if(v[1] > j) {
aux[i][0]=999999999; //instead of Integer.MAX_VALUE
} else {
aux[i][j]=1 + aux[1][j-v[1]];
}
} else {
if(v[i] > j) { //choose best option ,discard this coin or use it.
aux[i][j] = aux[i - 1][j];
} else
aux[i][j] = Math.min(aux[i-1][j],1 + aux[i][j-v[i]]);
}
}
}
int []z=CoinCount(sum,aux,v);
return z; // Return solution to the initial problem
}
public static int[] CoinCount(int A, int[][] aux, int[] d){
int coin = d.length-1;
int limit=A;
int [] typo=new int[d.length+1]; //We create solution array, that will count no of coins
for (int k=0;k<typo.length;k++) {
typo[k]=0;
} while (coin>0 || limit>0){
if(limit-d[coin]>=0 && coin-1>=0){
if(1+aux[coin][limit-d[coin]]<aux[coin-1][limit]){
typo[coin+1]=typo[coin+1]+1;
limit=limit-d[coin];
} else {
coin=coin-1;
}
} else if(limit-d[coin]>=0) {
typo[coin+1]=typo[coin+1]+1;
limit=limit-d[coin];
} else if(coin-1>=0) {
coin=coin-1;
}
}
typo[0]= aux[d.length-1][A];
return typo; //return the final array with solutions of each coin
}
public static void main(String[] args) {
int[] coins = {1,4,6};
int sum = 8;
int[] x=CoinChange(coins,sum);
System.out.println("At least " + Arrays.toString(x) +" from set "+ Arrays.toString(coins)
+ " coins are required to make a value of " + sum);
}
}

Clarification
I don't know if you still need the answer to this question but I will try to answer it anyway.
First, there are a few things I would like to clarify. The coin change problem does not have a unique solution. If you want both the minimum of coins used to make the change and frequencies of coins usage, I think that depends on the approach used to solve the program and the arrangement of the coins.
For example: Take the coins to be [4,6,8] and amount = 12. You'll quickly see that the minimum coins required to make this change is 2. Going by your choice of output, the following are all correct: [2,0,2,0] and [2,1,0,1].
By the way, the Coin change problem can be solved in many ways. A simple recursive DP approach in Java is here. It only returns the min coins needed to make the change at O(nlog(n)) time and O(n) space.
Another approach is by using a 2D DP matrix (same with the approach you tried using) at both O(n^2) time and space. Explanation on how to use this approach is here. Please be careful with the explanation because it is not generally correct. I noticed it's almost the same as the one you used.
Your solution
I will mention a few things about your solution that may have affected the result.
The number of rows of the DP matrix is v.length not v.length + 1.
Based on your solution, this should not affect the result because I noticed you don't seem comfortable with zero indexes.
I think it is not necessary to initialize the first column of the DB matrix since the data type you used is int, which is 0 by default. Again, this does not affect the answer, though.
The way you filled row 1 (supposed to be the first row, but you ignored row 0) is not good and may affect the result of some solutions.
The only mistake I see there is that there is no uniform value to specify amounts (i.e. j) that cannot be solved using the single coin (i.e. v[0]). Negative numbers could have been better because any positive integer is a potential valid solution for the cell. You could use -1 (if you're going by the Leetcode instruction). This way, you'll easily know cells that contain invalid values while filling the rest of the matrix.
The way you compute aux[i][j] is wrong because you are using the wrong coins. you are using v[i] instead of v[i-1] since you aux.length is one bigger than the v.length.
I did not look at the countCoint method. It looks complex for a seemingly simple problem. Please see my solution.
My Solution
public int[] change(int[] coins, int amount){
int[][] DP = new int[coins.length][amount+1];
//fill the first column with 0
//int array contains 0 by default, so this part is not necessary
/*
for (int i = 0; i < coins.length; i++) {
DP[i][0] =0;
}
*/
//fill the first row.
//At 0th row, we are trying to find the min number of ways to change j amount using only
//one coin i.e. coins[0] (that is the meaning of DP[0][j];
for (int j = 1; j <= amount; j++) {
if(coins[0] > j || j % coins[0] != 0){
DP[0][j] = -1;
}else{
DP[0][j] = j /coins[0];
}
}
//iterate the rest of the unfilled DP
for (int i = 1; i < coins.length; i++) {
for (int j = 1; j < amount+1; j++) {
if(coins[i] > j){
DP[i][j] = DP[i-1][j];
}else {
int prev = DP[i-1][j];
int cur = 1+DP[i][j-coins[i]];
if(cur == 0){
DP[i][j] = DP[i-1][j];
} else if(prev == -1) {
DP[i][j] = 1 + DP[i][j - coins[i]];
}else{
DP[i][j] = Math.min(DP[i-1][j],1+DP[i][j-coins[i]]);
}
}
}
}
return countCoin(coins,amount,DP);
}
public int[] countCoin(int[] coins, int amount, int[][] DP){
int[] result = new int[coins.length+1];//The 1 added is to hold result.
int i = coins.length -1;
int j = amount;
//while the rest will contain counter for coins used.
result[0] = DP[i][j];
if(result[0] ==0 || result[0] ==-1)return result;
while (j > 0 ){
if(i-1 >= 0 && DP[i][j] == DP[i-1][j]){
i = i-1;
}else{
j = j - coins[i];
result[i+1] += 1;
}
}
return result;
}

GC overhead limit exceeded - Arrays

i get this error after waiting long time for my code to execute and its pointing me to this method
public Iterable<Board> neighbors() {
Queue<Board> q = new LinkedList<>();
int n = dimension();
int x = 0, y = 0;
outer:
// do some stuff to get the x and y
if (y+1 < n) {
the line where i get the error -> int [][]arr = new int[n][n];
for (int i = 0; i < tiles.length; i++) {
arr[i] = Arrays.copyOf(tiles[i], n);
}
// do some stuff
Board br = new Board(arr);
if(!this.equals(br)) {
q.add(new Board(arr));
}
}
if (y-1 >= 0) {
int [][]arr = new int[n][n];
for (int i = 0; i < tiles.length; i++) {
arr[i] = Arrays.copyOf(tiles[i], n);
}
// do some stuff
Board br = new Board(arr);
if(!this.equals(br)) {
q.add(new Board(arr));
}
}
if (x-1 >= 0) {
int [][]arr = new int[n][n];
for (int i = 0; i < tiles.length; i++) {
arr[i] = Arrays.copyOf(tiles[i], n);
}
// do some stuff
Board br = new Board(arr);
if(!this.equals(br)) {
q.add(new Board(arr));
}
}
if (x+1 < n) {
int [][]arr = new int[n][n];
for (int i = 0; i < tiles.length; i++) {
arr[i] = Arrays.copyOf(tiles[i], n);
}
// do some stuff
Board br = new Board(arr);
if(!this.equals(br)) {
q.add(new Board(arr));
}
}
return q;
}
i basically need to copy tiles array and make changes to the copy "arr" but keep the tiles array without changing to use it later..i really don't like the way i'm doing it copying and pasting code i think its inefficient but no other way comes to my mind so i would like to know why i get this error "i know its because GC taking more time and not doing alot" but i want to know why its happening in this case also if there is better way to copy the array.
also i increased the heap memory to -Xmx1600m
Thanks for your time.

The Problem
It is likely that the problem arises from creating a lot of objects in a short period of time. See this answer for more information.
At the moment, one Board consist of at least four objects:
The Board itself
The array arr inside the board
The three arrays inside arr
Creating Less Objects
Our goal is to create fewer objects (arrays). Since you want to deal with small boards only, we could use one long to store the complete 3×3 board. A long has 64 bit. We use 64 / 9 = 7 bits per field to store the value on that field:
state = ... 0000100 0000011 0000010 0000001 0000000
4th field ↑ 2nd field ↑ 0th field
3rd field 1st field
The following class handles the bit operations.
class Board {
private final static int SIDE_LENGTH = 3;
private final static int FIELDS = SIDE_LENGTH * SIDE_LENGTH;
private final static int BITS_PER_FIELD = 64 / FIELDS;
private final static long FIELD_MASK = (1 << BITS_PER_FIELD) - 1;
private long state;
public Board() {
for (int field = 0; field < FIELDS; ++field) {
set(field, field);
}
}
/** Copy constructor. */
public Board(Board other) {
this.state = other.state;
}
public int get(int x, int y) {
return get(coordinatesToField(x, y));
}
public void set(int x, int y, int value) {
set(coordinatesToField(x, y), value);
}
private int coordinatesToField(int x, int y) {
return SIDE_LENGTH * y + x;
}
private int get(int field) {
return (int) ((state >>> (field * BITS_PER_FIELD)) & FIELD_MASK);
}
private void set(int field, int value) {
int shift = field * BITS_PER_FIELD;
state &= ~(FIELD_MASK << shift);
state |= (long) value << shift;
}
public String toString() {
StringBuilder sb = new StringBuilder();
for (int field = 0; field < FIELDS; ++field) {
sb.append(get(field));
sb.append((field + 1) % SIDE_LENGTH == 0 ? "\n" : "\t");
}
return sb.toString();
}
// TODO implement equals and hashCode
}
When using this class, you don't have to deal with arrays anymore, which saves not only a lot of objects, but also the copy code in your prorgram.
The class also works for 1×1, 2×2, and 4×4 boards, but not for larger ones due to the 64 bit limit.
Usage Examples
public static void main(String[] args) {
// Create and print the initial board
// 0 1 2
// 3 4 5
// 6 7 8
Board b = new Board();
System.out.println(b);
// Copy an existing board
Bord copy = new Board(b);
// Set the upper right field to value 8
copy.set(2, 0, 8);
// Print the center field
// 4
Syste.out.println(copy.get(1, 1));
}
Additional Ideas
You even could avoid creating Board objects at all, and just store the long values. But that doesn't help when you are using generics (such as LinkedList) because of Java's auto boxing.
Also note that LinkedList wraps each entry in an additional node object. Maybe you can use a more efficient DataStructure like a circular buffer.
Depending on what you are doing, you might as well have a look at the Flyweight design pattern.

Need help developing a proper print method for this Java program

This program takes integers from user input and puts them in a collection. It then prints the positive values first, then the negative values, and doesn't print repeated numbers. It stops asking for input once the user enters 0. Here is the code:
public class Intcoll2
{
private int[] c;
private int[] d;
private int howmany = 0;
public Intcoll2()
{
c = new int[500];
}
public Intcoll2(int i)
{
c = new int[i]
}
public void insert(int i)
{
if (i > 0)
{
int j = 0;
while ((j <= howmany) && (c[j] != i)) j++;
if (j == howmany)
{
if (j == c.length - 1)
{
d = new int[2*c.length];
for (int k = 0; k<c.length; i++){
d[k] = c[k];
}
c = d;
}
c[j] = i; c[j + 1] = 0;
}
howmany++;
}
}
public int get_howmany()
{
int j=0, howmany=0;
while (c[j]!=0) {howmany++; j++;}
return howmany;
}
Now my current print method looks like this:
public void print()
{
int j = 0;
System.out.println();
while (j <= howmany)
{
System.out.println(c[j]); j++;
}
}
But when I try to use that in my client, it only prints out zeros. Any help with what I'm doing wrong would be greatly appreciated.

An answer that you were probably not looking for, but still on the only real answer you should care about.
Your problem is not that somewhere in that code a bug is hiding. The problem is that your code is confusing beyond limits:
Dont use single-character variable names.
The constructor that takes an int ... creates an empty array!
Dont say "collection" when you are using arrays.
Dont give fields and local variables the same name.
Seriously: understanding this mess is mainly complicated and hard because you wrote code that is hard to read.
Now you are asking other people to debug such complicated code that you (the author who created it!) do not understand in the first place.
Instead, you might throw this whole thing away. And slowly write it again; but in a way that isn't at all confusing to the reader.

I took a look at your class and rewrote it in a more legible manner. I didn't test it but I'm confident it works. You can check it out and hopefully understand what's happening. Hope this helps!
public class IntCollection2 {
private int[] collection; // A large allocation, not neccessarily filled up.
private int currentSize; // The number of spots currently filled in the collection.
public IntCollection2() {
collection = new int[500];
currentSize = 0;
}
public IntCollection2(int size) {
collection = new int[size];
currentSize = 0;
}
/**
* Inserts a new element into the internal array. If the current array is filled up,
* a new array double the size of the current one is allocated.
* #param element An int to insert into the collection. Must not be '0'.
* #return True if the element was successfully inserted, false if the element was
* equal to '0' and was ignored.
*/
public boolean insert(int element) {
if (element != 0) {
if (currentSize < collection.length - 1) {
collection[currentSize] = element;
} else {
int[] newCollection = new int[collection.length * 2];
for (int i = 0; i < currentSize; i++) {
newCollection[i] = collection[i];
}
newCollection[currentSize] = element;
collection = newCollection;
}
currentSize++;
return true;
} else {
return false;
}
}
/**
* Not actually necessary because the class automatically updates its currentSize
* every time a new element is inserted.
* #return The current number of filled spots in the internal array.
*/
public int getCurrentSize() {
int size = 0;
for (int i = 0; i < collection.length && collection[i] != 0; i++) {
size++;
}
return size;
}
/**
* Prints out all the elements currently in the collection, one on each line.
*/
public void print() {
System.out.println();
for (int i = 0; i < currentSize; i++) {
System.out.println(collection[i]);
}
}
}
FYI: this class just prints out every element in the collection, in order. You mentioned something about printing positive then negative values, but I leave that to you.
EDIT: I'm guessing you're brand new to programming, so I just want to clarify exactly what a collection is. An array is an ordered list of elements. When you create an array, the computer sets aside a bit of memory to hold exactly the number of elements you told it to. You cannot change the size of an existing array. A collection is basically a wrapper around an array. It makes a bigger array than it needs to hold its elements, and when its array becomes full, it allocates a new, bigger one that can hold more elements.

Resolving method calls from generic object in enhanced for loop

This is more than likely a simple question for someone who is more familiar with Java than I am. Here's the gist of my issue:
I have a function that basically generates the possible combinations of the objects contained within an ArrayList. Being that I have multiple objects that need to use this function, the function is screaming at me to be made generic. The issue I'm encountering, though, is that an enhanced for-loop is unable to resolve method calls from the generic iterator. I understand why this happening, but I'm not familiar enough with Java to know how to resolve this issue. In any case, here is my code:
private <T> ArrayList<T> determineIdealOrderCombination(ArrayList<T> orders, int position){
// Local Variable Declarations
List<ArrayList<T>> subsets = new ArrayList<>();
int k = orders.size()+1; // Add one due to the do-while loop
int theoreticalQuantity;
int indexOfMaxProfit;
double maxProfit;
int[] s; // Here we'll keep indices pointing to elements in input array
double[] profits; // Here we'll keep track of the profit of each combination
// Begin searching for valid combinations
do {
// Setup
k--;
s = new int[k];
profits = new double[k];
// Generate combinations
if ( (k <= orders.size()) && (k > 0) ) {
// Set the first index sequence: 0, 1, 2,...
for (int i = 0; (s[i] = i) < k - 1; i++) ;
subsets.add(getSubset(orders, s));
for (; ; ) {
int i;
// Find position of item that can be incremented
for (i = k - 1; i >= 0 && s[i] == orders.size() - k + i; i--) ;
if (i < 0) {
break;
} else {
s[i]++; // increment this item
for (++i; i < k; i++) { // fill up remaining items
s[i] = s[i - 1] + 1;
}
subsets.add(getSubset(orders, s));
}
}
// All combinations have been evaluated, now throw away invalid combinations that violate the upper limit
// and calculate the valid combinations profits.
for (int i = 0; i < subsets.size(); i++) {
// Calculate the final position
theoreticalQuantity = position;
profits[i] = 0;
for (T t : subsets.get(i)) {
theoreticalQuantity += t.getQuantity(); // <-- THE PROBLEM
profits[i] += calculateProjectedProfit(t.getSecurity(), t.getQuantity(), t.getPrice()); // <-- THE PROBLEM
}
if(theoreticalQuantity > _MAX_POSITION_PER_ASSET){
// Negate profits if final position violates the position limit on an asset
profits[i] = Double.MIN_VALUE;
}
}
}
else{
break;
}
}
while( (subsets.size() == 0) );
// Verify that the subset array is not zero - it should never be zero
if(subsets.size() == 0){
return new ArrayList<>();
}
// Return the most profitable combination, if any.
indexOfMaxProfit = -1;
maxProfit = Double.MIN_VALUE;
for(int i = 0; i < profits.length; i++){
if(profits[i] != Double.MIN_VALUE){
if(profits[i] > maxProfit){
maxProfit = profits[i];
indexOfMaxProfit = i;
}
}
}
if( (maxProfit > 0) && (indexOfMaxProfit != -1) ){
return subsets.get(indexOfMaxProfit);
}
else{
return new ArrayList<>();
}
}
Any help would be appreciated.

This is how you tell the compiler that the incoming objects have the relevant methods:
public interface MyCommonInterface {
public int getQuantity();
}
private <T extends MyCommonInterface> ArrayList<T> determineIdealOrderCombination(ArrayList<T> orders, int position) {
As an additional note, i would read some tutorials on generics before attempting to use them. they are a little tricky to get the hang of initially. however, once you put out a little effort to learn the basics, you should be in a much better place to actually utilize them.