https://leetcode.com/discuss/interview-question/373202/amazon-oa-2019-optimal-utilization Given 2 lists a and b. Each element is a pair of integers where the first integer represents the unique id and the second integer represents a value. Your task is to find an element from a and an element form b such that the sum of their values is less or equal to target and as close to target as possible. Return a list of ids of selected elements. If no pair is possible, return an empty list.
question was this but I had to use Lists (like <<1,1>, <2,2> <3,3>>)
my solution was something like below. I kept failing some test cases with a NullPointerException. I am trying to find out WHAT SPECIFIC INPUT(foreground, background) COULD HAVE CAUSED THIS. (the assessment website HID the error line). the problem did not have any specifications or guarantees like 0 < deviceCapacity < 1000000, so I do NOT know what was passed in
Things I checked for:
foreground and background were not null. they did not have null values (how i checked is below)
I put a System.out.println(foregroundApplications.get(i)); System.out.println(backgroundApplications.get(j)) just before initializing "sums", the goal being to see if any values were something like null or <null, null>, but they were all valid number pairs like <1,8>. (if it was null, it would have printed null right? unsure about this). example of what I saw (there were no nulls): <1, 14> <2, 14> <3, 14> <4, 14>
i checked in the beginning if the lists (foreground and background) were null with an if(foregroundApps == null), they weren't.
I can't change my code anymore, this was an assessment with obfuscated test cases which I am trying to figure out.
P.S. If there is a better approach than O(M*N) time, I would like to know
public List<List<Integer>> optimize(int deviceCapacity, List<List<Integer> foregroundApplications, List<List<Integer>> backgroundApplications)
{
TreeMap<Integer, List<List<Integer>>> map = new TreeMap<>();
for(int i = 0; i < foregroundApplications.size(); i++)
{
for(int j = 0; j < backgroundApplications.size(); j++)
{
int sum = foregroundApplications.get(i).get(1) + backgroundApplications.get(j).get(1);
if(sum<=deviceCapacity)
{
List<List<Integer>> list= new ArrayList<>();
if(map.containsKey(sum))
{
list = map.get(sum);
}
List<Integer> pair = new ArrayList<>();
pair.add(foregroundApplications.get(i).get(0));
pair.add(backgroundApplications.get(j).get(0));
list.add(pair);
map.put(sum, list);
}
}
}
if(map.size() == 0)
{
List<List<Integer>> list= new ArrayList<>();
List<Integer> emptyPair = new ArrayList<>();
emptyPair.add(null);
emptyPair.add(null);
list.add(emptyPair);
return list;
}
return map.get(map.lastKey());
}
Given an array of n Objects, let's say it is an array of strings, and it has the following values:
foo[0] = "a";
foo[1] = "cc";
foo[2] = "a";
foo[3] = "dd";
What do I have to do to delete/remove all the strings/objects equal to "a" in the array?
[If you want some ready-to-use code, please scroll to my "Edit3" (after the cut). The rest is here for posterity.]
To flesh out Dustman's idea:
List<String> list = new ArrayList<String>(Arrays.asList(array));
list.removeAll(Arrays.asList("a"));
array = list.toArray(array);
Edit: I'm now using Arrays.asList instead of Collections.singleton: singleton is limited to one entry, whereas the asList approach allows you to add other strings to filter out later: Arrays.asList("a", "b", "c").
Edit2: The above approach retains the same array (so the array is still the same length); the element after the last is set to null. If you want a new array sized exactly as required, use this instead:
array = list.toArray(new String[0]);
Edit3: If you use this code on a frequent basis in the same class, you may wish to consider adding this to your class:
private static final String[] EMPTY_STRING_ARRAY = new String[0];
Then the function becomes:
List<String> list = new ArrayList<>();
Collections.addAll(list, array);
list.removeAll(Arrays.asList("a"));
array = list.toArray(EMPTY_STRING_ARRAY);
This will then stop littering your heap with useless empty string arrays that would otherwise be newed each time your function is called.
cynicalman's suggestion (see comments) will also help with the heap littering, and for fairness I should mention it:
array = list.toArray(new String[list.size()]);
I prefer my approach, because it may be easier to get the explicit size wrong (e.g., calling size() on the wrong list).
An alternative in Java 8:
String[] filteredArray = Arrays.stream(array)
.filter(e -> !e.equals(foo)).toArray(String[]::new);
Make a List out of the array with Arrays.asList(), and call remove() on all the appropriate elements. Then call toArray() on the 'List' to make back into an array again.
Not terribly performant, but if you encapsulate it properly, you can always do something quicker later on.
You can always do:
int i, j;
for (i = j = 0; j < foo.length; ++j)
if (!"a".equals(foo[j])) foo[i++] = foo[j];
foo = Arrays.copyOf(foo, i);
You can use external library:
org.apache.commons.lang.ArrayUtils.remove(java.lang.Object[] array, int index)
It is in project Apache Commons Lang http://commons.apache.org/lang/
See code below
ArrayList<String> a = new ArrayList<>(Arrays.asList(strings));
a.remove(i);
strings = new String[a.size()];
a.toArray(strings);
If you need to remove multiple elements from array without converting it to List nor creating additional array, you may do it in O(n) not dependent on count of items to remove.
Here, a is initial array, int... r are distinct ordered indices (positions) of elements to remove:
public int removeItems(Object[] a, int... r) {
int shift = 0;
for (int i = 0; i < a.length; i++) {
if (shift < r.length && i == r[shift]) // i-th item needs to be removed
shift++; // increment `shift`
else
a[i - shift] = a[i]; // move i-th item `shift` positions left
}
for (int i = a.length - shift; i < a.length; i++)
a[i] = null; // replace remaining items by nulls
return a.length - shift; // return new "length"
}
Small testing:
String[] a = {"0", "1", "2", "3", "4"};
removeItems(a, 0, 3, 4); // remove 0-th, 3-rd and 4-th items
System.out.println(Arrays.asList(a)); // [1, 2, null, null, null]
In your task, you can first scan array to collect positions of "a", then call removeItems().
There are a lot of answers here--the problem as I see it is that you didn't say WHY you are using an array instead of a collection, so let me suggest a couple reasons and which solutions would apply (Most of the solutions have already been answered in other questions here, so I won't go into too much detail):
reason: You didn't know the collection package existed or didn't trust it
solution: Use a collection.
If you plan on adding/deleting from the middle, use a LinkedList. If you are really worried about size or often index right into the middle of the collection use an ArrayList. Both of these should have delete operations.
reason: You are concerned about size or want control over memory allocation
solution: Use an ArrayList with a specific initial size.
An ArrayList is simply an array that can expand itself, but it doesn't always need to do so. It will be very smart about adding/removing items, but again if you are inserting/removing a LOT from the middle, use a LinkedList.
reason: You have an array coming in and an array going out--so you want to operate on an array
solution: Convert it to an ArrayList, delete the item and convert it back
reason: You think you can write better code if you do it yourself
solution: you can't, use an Array or Linked list.
reason: this is a class assignment and you are not allowed or you do not have access to the collection apis for some reason
assumption: You need the new array to be the correct "size"
solution:
Scan the array for matching items and count them. Create a new array of the correct size (original size - number of matches). use System.arraycopy repeatedly to copy each group of items you wish to retain into your new Array. If this is a class assignment and you can't use System.arraycopy, just copy them one at a time by hand in a loop but don't ever do this in production code because it's much slower. (These solutions are both detailed in other answers)
reason: you need to run bare metal
assumption: you MUST not allocate space unnecessarily or take too long
assumption: You are tracking the size used in the array (length) separately because otherwise you'd have to reallocate your array for deletes/inserts.
An example of why you might want to do this: a single array of primitives (Let's say int values) is taking a significant chunk of your ram--like 50%! An ArrayList would force these into a list of pointers to Integer objects which would use a few times that amount of memory.
solution: Iterate over your array and whenever you find an element to remove (let's call it element n), use System.arraycopy to copy the tail of the array over the "deleted" element (Source and Destination are same array)--it is smart enough to do the copy in the correct direction so the memory doesn't overwrite itself:
System.arraycopy(ary, n+1, ary, n, length-n)
length--;
You'll probably want to be smarter than this if you are deleting more than one element at a time. You would only move the area between one "match" and the next rather than the entire tail and as always, avoid moving any chunk twice.
In this last case, you absolutely must do the work yourself, and using System.arraycopy is really the only way to do it since it's going to choose the best possibly way to move memory for your computer architecture--it should be many times faster than any code you could reasonably write yourself.
Something about the make a list of it then remove then back to an array strikes me as wrong. Haven't tested, but I think the following will perform better. Yes I'm probably unduly pre-optimizing.
boolean [] deleteItem = new boolean[arr.length];
int size=0;
for(int i=0;i<arr.length;i==){
if(arr[i].equals("a")){
deleteItem[i]=true;
}
else{
deleteItem[i]=false;
size++;
}
}
String[] newArr=new String[size];
int index=0;
for(int i=0;i<arr.length;i++){
if(!deleteItem[i]){
newArr[index++]=arr[i];
}
}
I realise this is a very old post, but some of the answers here helped me out, so here's my tuppence' ha'penny's worth!
I struggled getting this to work for quite a while before before twigging that the array that I'm writing back into needed to be resized, unless the changes made to the ArrayList leave the list size unchanged.
If the ArrayList that you're modifying ends up with greater or fewer elements than it started with, the line List.toArray() will cause an exception, so you need something like List.toArray(new String[] {}) or List.toArray(new String[0]) in order to create an array with the new (correct) size.
Sounds obvious now that I know it. Not so obvious to an Android/Java newbie who's getting to grips with new and unfamiliar code constructs and not obvious from some of the earlier posts here, so just wanted to make this point really clear for anybody else scratching their heads for hours like I was!
Initial array
int[] array = {5,6,51,4,3,2};
if you want remove 51 that is index 2, use following
for(int i = 2; i < array.length -1; i++){
array[i] = array[i + 1];
}
EDIT:
The point with the nulls in the array has been cleared. Sorry for my comments.
Original:
Ehm... the line
array = list.toArray(array);
replaces all gaps in the array where the removed element has been with null. This might be dangerous, because the elements are removed, but the length of the array remains the same!
If you want to avoid this, use a new Array as parameter for toArray(). If you don`t want to use removeAll, a Set would be an alternative:
String[] array = new String[] { "a", "bc" ,"dc" ,"a", "ef" };
System.out.println(Arrays.toString(array));
Set<String> asSet = new HashSet<String>(Arrays.asList(array));
asSet.remove("a");
array = asSet.toArray(new String[] {});
System.out.println(Arrays.toString(array));
Gives:
[a, bc, dc, a, ef]
[dc, ef, bc]
Where as the current accepted answer from Chris Yester Young outputs:
[a, bc, dc, a, ef]
[bc, dc, ef, null, ef]
with the code
String[] array = new String[] { "a", "bc" ,"dc" ,"a", "ef" };
System.out.println(Arrays.toString(array));
List<String> list = new ArrayList<String>(Arrays.asList(array));
list.removeAll(Arrays.asList("a"));
array = list.toArray(array);
System.out.println(Arrays.toString(array));
without any null values left behind.
My little contribution to this problem.
public class DeleteElementFromArray {
public static String foo[] = {"a","cc","a","dd"};
public static String search = "a";
public static void main(String[] args) {
long stop = 0;
long time = 0;
long start = 0;
System.out.println("Searched value in Array is: "+search);
System.out.println("foo length before is: "+foo.length);
for(int i=0;i<foo.length;i++){ System.out.println("foo["+i+"] = "+foo[i]);}
System.out.println("==============================================================");
start = System.nanoTime();
foo = removeElementfromArray(search, foo);
stop = System.nanoTime();
time = stop - start;
System.out.println("Equal search took in nano seconds = "+time);
System.out.println("==========================================================");
for(int i=0;i<foo.length;i++){ System.out.println("foo["+i+"] = "+foo[i]);}
}
public static String[] removeElementfromArray( String toSearchfor, String arr[] ){
int i = 0;
int t = 0;
String tmp1[] = new String[arr.length];
for(;i<arr.length;i++){
if(arr[i] == toSearchfor){
i++;
}
tmp1[t] = arr[i];
t++;
}
String tmp2[] = new String[arr.length-t];
System.arraycopy(tmp1, 0, tmp2, 0, tmp2.length);
arr = tmp2; tmp1 = null; tmp2 = null;
return arr;
}
}
It depends on what you mean by "remove"? An array is a fixed size construct - you can't change the number of elements in it. So you can either a) create a new, shorter, array without the elements you don't want or b) assign the entries you don't want to something that indicates their 'empty' status; usually null if you are not working with primitives.
In the first case create a List from the array, remove the elements, and create a new array from the list. If performance is important iterate over the array assigning any elements that shouldn't be removed to a list, and then create a new array from the list. In the second case simply go through and assign null to the array entries.
Arrgh, I can't get the code to show up correctly. Sorry, I got it working. Sorry again, I don't think I read the question properly.
String foo[] = {"a","cc","a","dd"},
remove = "a";
boolean gaps[] = new boolean[foo.length];
int newlength = 0;
for (int c = 0; c<foo.length; c++)
{
if (foo[c].equals(remove))
{
gaps[c] = true;
newlength++;
}
else
gaps[c] = false;
System.out.println(foo[c]);
}
String newString[] = new String[newlength];
System.out.println("");
for (int c1=0, c2=0; c1<foo.length; c1++)
{
if (!gaps[c1])
{
newString[c2] = foo[c1];
System.out.println(newString[c2]);
c2++;
}
}
Will copy all elements except the one with index i:
if(i == 0){
System.arraycopy(edges, 1, copyEdge, 0, edges.length -1 );
}else{
System.arraycopy(edges, 0, copyEdge, 0, i );
System.arraycopy(edges, i+1, copyEdge, i, edges.length - (i+1) );
}
If it doesn't matter the order of the elements. you can swap between the elements foo[x] and foo[0], then call foo.drop(1).
foo.drop(n) removes (n) first elements from the array.
I guess this is the simplest and resource efficient way to do.
PS: indexOf can be implemented in many ways, this is my version.
Integer indexOf(String[] arr, String value){
for(Integer i = 0 ; i < arr.length; i++ )
if(arr[i] == value)
return i; // return the index of the element
return -1 // otherwise -1
}
while (true) {
Integer i;
i = indexOf(foo,"a")
if (i == -1) break;
foo[i] = foo[0]; // preserve foo[0]
foo.drop(1);
}
to remove only the first of several equal entries
with a lambda
boolean[] done = {false};
String[] arr = Arrays.stream( foo ).filter( e ->
! (! done[0] && Objects.equals( e, item ) && (done[0] = true) ))
.toArray(String[]::new);
can remove null entries
In an array of Strings like
String name = 'a b c d e a f b d e' // could be like String name = 'aa bb c d e aa f bb d e'
I build the following class
class clearname{
def parts
def tv
public def str = ''
String name
clearname(String name){
this.name = name
this.parts = this.name.split(" ")
this.tv = this.parts.size()
}
public String cleared(){
int i
int k
int j=0
for(i=0;i<tv;i++){
for(k=0;k<tv;k++){
if(this.parts[k] == this.parts[i] && k!=i){
this.parts[k] = '';
j++
}
}
}
def str = ''
for(i=0;i<tv;i++){
if(this.parts[i]!='')
this.str += this.parts[i].trim()+' '
}
return this.str
}}
return new clearname(name).cleared()
getting this result
a b c d e f
hope this code help anyone
Regards
Assign null to the array locations.
Right now I have an array of "Dragon"s. Each item has two values. An ID and a Count. So my array would look something like this:
Dragon[] dragons = { new Dragon(2, 4),
new Dragon(83, 199),
new Dragon(492, 239),
new Dragon(2, 93),
new Dragon(24, 5)
};
As you can see, I have two Dragons with the ID of 2 in the array. What I would like to accomplish is, when a duplicate is found, just add the count of the duplicate to the count of the first one, and then remove the duplicate Dragon.
I've done this sort of successfully, but I would end up with a null in the middle of the array, and I don't know how to remove the null and then shuffle them.
This is what I have so far but it really doesn't work properly:
public static void dupeCheck(Dragon[] dragons) {
int end = dragons.length;
for (int i = 0; i < end; i++) {
for (int j = i + 1; j < end; j++) {
if (dragons[i] != null && dragons[j] != null) {
if (dragons[i].getId() == dragons[j].getId()) {
dragons[i] = new Item(dragons[i].getId(), dragons[i].getCount() + dragons[j].getCount());
dragons[j] = null;
end--;
j--;
}
}
}
}
}
You should most probably not maintain the dragon count for each dragon in the dragon class itself.
That aside, even if you are forced to use an array, you should create an intermeditate map to store your dragons.
Map<Integer, Dragon> idToDragon = new HashMap<>();
for (Dragon d : yourArray) {
// fetch existing dragon with that id or create one if none present
Dragon t = idToDragon.computeIfAbsent(d.getId(), i -> new Dragon(i, 0));
// add counts
t.setCount(t.getCount() + d.getCount());
// store in map
idToDragon.put(d.getId(), t);
}
Now the map contains a mapping between the dragons' ids and the dragons, with the correct counts.
To create an array out of this map, you can just
Dragon[] newArray = idToDragon.values().toArray(new Dragon[idToDragon.size()]);
You may be force to store the result in an array but that doesn't mean that you're force to always use an array
One solution could be using the Stream API, group the items adding the count and save the result into an array again. You can get an example of how to use the Stream API to sum values here. Converting a List<T> into a T[] is quite straightforward but anyways, you have an example here
The size of an array cannot be changed after it's created.
So you need to return either a new array or list containing the merged dragons.
public static Dragon[] merge(Dragon[] dragonArr) {
return Arrays.stream(dragonArr)
// 1. obtain a map of dragon IDs and their combined counts
.collect(groupingBy(Dragon::getId, summingInt(Dragon::getCount)))
// 2. transform the map entries to dragons
.entrySet().stream().map(entry -> new Dragon(entry.getKey(), entry.getValue()))
// 3. collect the result as an array
.toArray(Dragon[]::new);
}
So I'm trying to go through an arraylist of objects that all have a certain strength value and depending on their strength value, they go into the bigger 2d array based on that. So if their strength value is 0 then they go in the 0th array of the bigger one and this is what my code looks like so far
private ArrayList<Battleable> arr;
public BattleDeck() {
arr = new ArrayList<Battleable>();
for (Battleable creature: arr){
arr.add(creature);
}
}
public Battleable[][] export2Darray() {
//returns a two-dimensional ragged array where each row
// contains a deep copy of all of the Battleable objects
// in the BattleStack with the corresponding Level value
Battleable[][] retVal = new Battleable[10][];
int k = 0;
for (int i = 0; i<arr.size(); i++){
int levelOfObj = arr.get(i).getLevel();
if(levelOfObj == k) {
//insert it into retVal[0][0]
}
}
}
return retVal;
}
and I was wondering how I would do that? How do i syntax-tically say "get the obj that has strength 0 and put it in position 0 0 of my 2d array
A solution using Java 8 streams:
// group Battleables ArrayList by strength
Map<Integer, List<Battleable>> map =
arr.stream().collect(Collectors.groupingBy(Battleable::getStrength));
The result is a Map containing the Battleables as Lists with their strength as their key.
If you need the result as a jagged 2D array, sort the entries like this:
final Battleable[][] arrays = new Battleable[10][];
map.entrySet().forEach(entry -> {
arrays[entry.getKey()] = entry.getValue().toArray(new Battleable[entry.getValue().size()]);
});
Since arrays are of fixed size in Java, there is no clean way to add items to an array. You can resize the array each time by creating a new array each time, one larger than the last, and copying the data from the old array to the new array, but that would be messy and you would be reinventing a wheel called ArrayList. Modus Tollens has a good answer, but it uses some slightly advanced Java 8 concepts. Here's one way to write it without them:
public Battleable[][] export2Darray() {
Battleable[][] retVal = new Battleable[10][];
// create a map that will hold the items, arranged by level
Map<Integer, List<Battleable>> byLevel = new HashMap<>();
for (int i = 0; i < 10; i++) {
// initialize all levels with empty lists
byLevel.put(i, new ArrayList<>());
}
for (Battleable battleable : arr) {
int level = battleable.getLevel();
// get the list for this level and add to it
byLevel.get(level).add(battleable);
}
// Now we have a map from levels to lists of battleables;
// we need to turn each list into an array in our retVal
for (int level = 0; level < 10; level++) {
// get each list, convert it toArray and assign to slot in retVal
retVal[level] = byLevel.get(level).toArray(new Battleable[0]);
}
return retVal;
}
Here's a solution using ArrayLists, I am creating an ArrayList which will be referenced by strength, then inside of this I have another ArrayListwhich will have all of the Battleable objects of that strength level.
public ArrayList<ArrayList<Battleable>> exportBattleable() {
ArrayList<ArrayList<Battleable>> retVal = new ArrayList<ArrayList<Battleable>>();
for (int i = 0; i < arr.size(); i++){
retVal.get(arr.getLevel()).add(arr.get(i));
}
return retVal;
}
Now if you want to print all Battleable objects of strength = 3, you would do:
ArrayList<Battleable> strength3 = retVal.get(3);
for(Battleable battleable : strength3) {
System.out.println(battleable.toString());
}
This way you don't have to worry about re-sizing your arrays depending on how many Battleable objects you are adding in, same with strength levels, if you decide that instead of using strength levels from 0-9 that you wanted to use 0-20 you already have the ability to scale up or down.
I have 3 arraylist each have size = 3 and 3 arrays also have length = 3 of each. I want to copy data from arraylists to arrays in following way but using any loop (i.e for OR for each).
myArray1[1] = arraylist1.get(1);
myArray1[2] = arraylist2.get(1);
myArray1[3] = arraylist3.get(1);
I have done it manually one by one without using any loop, but code appears to be massive because in future I'm sure that number of my arraylists and arrays will increase up to 15.
I want to copy the data from arraylists to arrays as shown in the image but using the loops not manually one by one?
How about this?
List<Integer> arraylist0 = Arrays.asList(2,4,3);
List<Integer> arraylist1 = Arrays.asList(2,5,7);
List<Integer> arraylist2 = Arrays.asList(6,3,7);
List<List<Integer>> arraylistList = Arrays.asList(arraylist0, arraylist1, arraylist2);
int size = 3;
int[] myArray0 = new int[size];
int[] myArray1 = new int[size];
int[] myArray2 = new int[size];
int[][] myBigArray = new int[][] {myArray0, myArray1, myArray2};
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
myBigArray[i][j] = arraylistList.get(j).get(i);
}
}
To explain, since we want to be able to work with an arbitrary size (3, 15, or more), we are dealing with 2-dimensional data.
We are also dealing with array and List, which are slightly different in their use.
The input to your problem is List<Integer>, and so we make a List<List<Integer>> in order to deal with all the input data easily.
Similarly, the output will be arrays, so we make a 2-dimensional array (int[][]) in order to write the data easily.
Then it's simply a matter of iterating over the data in 2 nested for loops. Notice that this line reverses the order of i and j in order to splice the data the way you intend.
myBigArray[i][j] = arraylistList.get(j).get(i);
And then you can print your answer like this:
System.out.println(Arrays.toString(myArray0));
System.out.println(Arrays.toString(myArray1));
System.out.println(Arrays.toString(myArray2));
You need to have two additional structures:
int[][] destination = new int [][] {myArray1, myArray2,myArray3 }
List<Integer>[] source;
source = new List<Integer>[] {arraylist1,arraylist2,arraylist3}
myArray1[1] = arraylist1.get(1);
myArray1[2] = arraylist2.get(1);
myArray1[3] = arraylist3.get(1);
for (int i=0;i<destination.length;i++) {
for (int j=0;j<source.length;j++) {
destination[i][j] = source[j].get(i);
}
}
If you cannot find a ready made API or function for this, I would suggest trivializing the conversion from List to Array using the List.toArray() method and focus on converting/transforming the given set of lists to a another bunch of lists which contain the desired output. Following is a code sample which I would think achieves this. It does assume the input lists are NOT of fixed/same sizes. Assuming this would only make the logic easier.
On return of this function, all you need to do is to iterate over the TreeMap and convert the values to arrays using List.toArray().
public static TreeMap<Integer, List<Integer>> transorm(
List<Integer>... lists) {
// Return a blank TreeMap if not input. TreeMap explanation below.
if (lists == null || lists.length == 0)
return new TreeMap<>();
// Get Iterators for the input lists
List<Iterator<Integer>> iterators = new ArrayList<>();
for (List<Integer> list : lists) {
iterators.add(list.iterator());
}
// Initialize Return. We return a TreeMap, where the key indicates which
// position's integer values are present in the list which is the value
// of this key. Converting the lists to arrays is trivial using the
// List.toArray() method.
TreeMap<Integer, List<Integer>> transformedLists = new TreeMap<>();
// Variable maintaining the position for which values are being
// collected. See below.
int currPosition = 0;
// Variable which keeps track of the index of the iterator currently
// driving the iteration and the driving iterator.
int driverItrIndex = 0;
Iterator<Integer> driverItr = lists[driverItrIndex].iterator();
// Actual code that does the transformation.
while (driverItrIndex < iterators.size()) {
// Move to next driving iterator
if (!driverItr.hasNext()) {
driverItrIndex++;
driverItr = iterators.get(driverItrIndex);
continue;
}
// Construct Transformed List
ArrayList<Integer> transformedList = new ArrayList<>();
for (Iterator<Integer> iterator : iterators) {
if (iterator.hasNext()) {
transformedList.add(iterator.next());
}
}
// Add to return
transformedLists.put(currPosition, transformedList);
}
// Return Value
return transformedLists;
}