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In Java code i the method i created only put the first duplicate instance to a new array

I want to remove the duplicates by putting them in a new array but somehow I only get a first instance and a bunch of zeros.
Here is my code:
public class JavaApplication7 {
public static void main(String[] args) {
int[] arr = new int[] {1,1,2,2,2,2,3,4,5,6,7,8};
int[] res = removeD(arr);
for (int i = 0; i < res.length; i++) {
System.out.print(res[i] + " ");
}
}
public static int[] removeD(int[] ar) {
int[] tempa = new int[ar.length];
for (int i = 0; i < ar.length; i++) {
if (ar[i] == ar[i+1]) {
tempa[i] = ar[i];
return tempa;
}
}
return null;
}
}
expected: 1,2
result: 1,0,0,0,0,0,0....

why dont you make use of HashSet?
final int[] arr = new int[] { 1, 1, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8 };
final Set<Integer> set = new HashSet<>();
for (final int i : arr) {
// makes use of Integer's hashCode() and equals()
set.add(Integer.valueOf(i));
}
// primitive int array without zeros
final int[] newIntArray = new int[set.size()];
int counter = 0;
final Iterator<Integer> iterator = set.iterator();
while (iterator.hasNext()) {
newIntArray[counter] = iterator.next().intValue();
counter++;
}
for (final int i : newIntArray) {
System.out.println(i);
}
Edit
if you want your array to be ordered
final int[] arr = new int[] { 9, 9, 8, 8, 1, 1, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8 };
Set<Integer> set = new HashSet<>();
for (final int i : arr) {
// makes use of Integer's hashCode() and equals()
set.add(Integer.valueOf(i));
}
// priomitive int array without zeros
final int[] newIntArray = new int[set.size()];
int counter = 0;
// SetUtils.orderedSet(set) requires apache commons collections
set = SetUtils.orderedSet(set);
final Iterator<Integer> iterator = set.iterator();
while (iterator.hasNext()) {
newIntArray[counter] = iterator.next().intValue();
counter++;
}
for (final int i : newIntArray) {
System.out.println(i);
}

A couple of points to help you:
1) With this: for(int i =0; i<ar.length; i++){ - you will get an IndexOutOfBoundsException because you are checking [i+1]. Hint: it is only the last element that will cause this...
2) Because you're initialising the second array with the length of the original array, every non-duplicate will be a 0 in it, as each element is initialised with a 0 by default. So perhaps you need to find how many duplicates there are first, before setting the size.
3) As mentioned in the comments, you are returning the array once the first duplicate is found, so remove that and just return the array at the end of the method.
4) You will also get multiple 2s because when you check i with i+1, it will find 3 2s and update tempa with each of them, so you'll need to consider how to not to include duplicates you've already found - based on your expected result.
These points should help you get the result you desire - if I (or someone else) just handed you the answer, you wouldn't learn as much as if you researched it yourself.

Here:
int[] tempa = new int[ar.length];
That creates a new array with the same size as the incoming one. All slots in that array are initialized with 0s!
When you then put some non-0 values into the first slots, sure, those stick, but so do the 0s in all the later slots that you don't "touch".
Thus: you either have to use a data structure where you can dynamically add new elements (like List/ArrayList), or you have to first iterate the input array to determine the exact count of objects you need, to then create an appropriately sized array, to then fill that array.

Return statement
As both commenters said, you return from the method as soon as you find your first duplicate. To resolve that issue, move the return to the end of the method.
Index problems
You will then run into another issue, an ArrayIndexOutOfBoundsException because when you are checking your last item (i = ar.length - 1) which in your example would be 11 you are then comparing if ar[11] == ar[12] but ar has size 12 so index 12 is out of the bounds of the array. You could solve that by changing your exit condition of the for loop to i < ar.length - 1.
Zeros
The zeros in your current output come from the initialization. You initialize your tempa with int[ar.length] this means in the memory it will reserve space for 12 ints which are initialized with zero. You will have the same problem after resolving both issues above. Your output would look like this: 1 0 2 2 2 0 0 0 0 0 0 0. This is because you use the same index for tempa and ar. You could solve that problem in different ways. Using a List, Filtering the array afterwards, etc. It depends what you want to do exactly.
The code below has the two first issues solved:
public class JavaApplication7 {
public static void main(String[] args) {
int[] arr = new int[] { 1, 1, 2, 2, 2, 2, 3, 4, 5, 6, 7, 8 };
int[] res = removeD(arr);
for (int i = 0; i < res.length; i++) {
System.out.print(res[i] + " ");
}
}
public static int[] removeD(int[] ar) {
int[] tempa = new int[ar.length];
for (int i = 0; i < ar.length - 1; i++) {
if (ar[i] == ar[i + 1]) {
tempa[i] = ar[i];
}
}
return tempa;
}
}

There were a some error mentioned already:
return exits the method.
with arr[i+1] the for condition should bei+1 < arr.length`.
the resulting array may be smaller.
So:
public static int[] removeD(int[] ar) {
// Arrays.sort(ar);
int uniqueCount = 0;
for (int i = 0; i < ar.length; ++i) {
if (i == 0 || ar[i] != ar[i - 1]) {
++uniqueCount;
}
}
int[] uniques = new int[uniqueCount];
int uniqueI = 0;
for (int i = 0; i < ar.length; ++i) {
if (i == 0 || ar[i] != ar[i - 1]) {
uniques[uniqueI] = arr[i];
++uniqueI;
}
}
return uniques;
}

How to do a "cross-over" with two integer arrays

I'm trying to learn about genetic algorithms and am currently working on "crossing over" two "genes".
A gene is an integer array, consisting of ones and zeros. To exemplify my problem let's say we have two genes.
int[] geneA = {1,0,0,0,0};
int[] geneB = {0,1,1,1,0};
The expected result from a cross-over, for example at position 3 would be:
geneA = [1,0,0,1,0]
geneB = [0,1,1,0,0]
Meaning that every element at the index of 3 or above would be swapped with the equivalent element of the other gene. To achieve this I wrote the following method:
private void crossOver(int[] geneA, int[] geneB, int pos) {
int copyA[];
int copyB[];
copyA = geneA;
copyB = geneB;
for(int i = pos; i < geneA.length; i++) {
geneA[i] = copyB[i];
geneB[i] = copyA[i];
}
System.out.println(Arrays.toString(geneA);
System.out.println(Arrays.toString(geneB);
}
However, it seems that the elements of geneB simply get copied into geneA at the index of 3 or higher.
The console output is as following:
[1, 0, 0, 1, 0]
[0, 1, 1, 1, 0]
Any explanations or help are highly appreciated. Thanks in advance!

copyA = geneA does not create a copy. Both variables now refer to the same array.
There is no need to waste time and space on copying the entire array.
When you swap values, you just need to store one of the values in a temporary variable.
private static void crossOver(int[] geneA, int[] geneB, int pos) {
for (int i = pos; i < geneA.length; i++) {
int temp = geneA[i];
geneA[i] = geneB[i];
geneB[i] = temp;
}
}
That will update the arrays in-place, so the caller will see the change.
int[] geneA = {1,0,0,0,0};
int[] geneB = {0,1,1,1,0};
crossOver(geneA, geneB, 3);
System.out.println(Arrays.toString(geneA));
System.out.println(Arrays.toString(geneB));
Output
[1, 0, 0, 1, 0]
[0, 1, 1, 0, 0]

I think you are going wrong when you create copies of the arrays ... currently you are not making copies but copyA and copyB are just references pointing to geneA and geneB resp
Use Arrays.copy like so,
copyA = Arrays.copyOf(geneA, geneA.length);
copyB = Arrays.copyOf(geneB, geneB.length);

There is problem. you copy both array very easy to copyA and copyB which doesn't make a new copy. it just copy reference of array and both array refer to same memory. so any change in one of them will change other too. for this you can use Arrays.copyof() instead.
I think this is what you want:
private void crossOver(int[] geneA, int[] geneB, int pos) {
int[] copyA = Arrays.copyOf(geneA, geneA.length);
int[] copyB = Arrays.copyOf(geneB, geneB.length);
for(int i = pos; i < geneA.length; i++) {
geneA[i] = copyB[i];
geneB[i] = copyA[i];
}
System.out.println(Arrays.toString(geneA));
System.out.println(Arrays.toString(geneB));
}
instead of that, as I see, you change geneA and geneB after making copy and swapping, so I think this code will do same with less code but I don't know this is what you want or not:
private void crossOver(int[] geneA, int[] geneB, int pos) {
for(int i = pos; i < geneA.length; i++) {
int temp = geneA[i];
geneA[i] = geneB[i];
geneB[i] = temp;
}
System.out.println(Arrays.toString(geneA));
System.out.println(Arrays.toString(geneB));
}
I hope this help you.

This code will work:
void crossOver(int[] geneA, int[] geneB, int pos) {
int copyA[];
int copyB[];
copyA = Arrays.copyOf(geneA,geneA.length);
copyB = Arrays.copyOf(geneB,geneB.length);
for(int i = pos; i < geneA.length; i++) {
geneA[i] = copyB[i];
geneB[i] = copyA[i];
}
System.out.println(Arrays.toString(geneA));
System.out.println(Arrays.toString(geneB));
}
Problem with your code was that you was doing copy of array by:
copyA = geneA;
copyB = geneB;
which just pointed to the same reference. So in the rest of the code you was working on the same array, instead of working on a copy.
Instead proper way of copying arrays is:
copyA = Arrays.copyOf(geneA,geneA.length);
copyB = Arrays.copyOf(geneB,geneB.length);

How can I make an array without one number of the other array?

I am trying to make a code with two arrays. The second array has the same values of the first except for the smallest number. I have already made a code where z is the smallest number. Now I just want to make a new array without z, any feedback would be appreciated.
public static int Second_Tiny() {
int[] ar = {19, 1, 17, 17, -2};
int i;
int z = ar[0];
for (i = 1; i < ar.length; i++) {
if (z >ar[i]) {
z=ar[i];
}
}
}

Java 8 streams have built in functionality that can achieve what you're wanting.
public static void main(String[] args) throws Exception {
int[] ar = {19, 1, 17, 17, -2, -2, -2, -2, 5};
// Find the smallest number
int min = Arrays.stream(ar)
.min()
.getAsInt();
// Make a new array without the smallest number
int[] newAr = Arrays
.stream(ar)
.filter(a -> a > min)
.toArray();
// Display the new array
System.out.println(Arrays.toString(newAr));
}
Results:
[19, 1, 17, 17, 5]
Otherwise, you'd be looking at something like:
public static void main(String[] args) throws Exception {
int[] ar = {19, 1, 17, 17, -2, -2, -2, -2, 5};
// Find the smallest number
// Count how many times the min number appears
int min = ar[0];
int minCount = 0;
for (int a : ar) {
if (minCount == 0 || a < min) {
min = a;
minCount = 1;
} else if (a == min) {
minCount++;
}
}
// Make a new array without the smallest number
int[] newAr = new int[ar.length - minCount];
int newIndex = 0;
for (int a : ar) {
if (a != min) {
newAr[newIndex] = a;
newIndex++;
}
}
// Display the new array
System.out.println(Arrays.toString(newAr));
}
Results:
[19, 1, 17, 17, 5]

I think the OP is on wrong track seeing his this comment:
"I am trying to find out the second smallest integer in array ar[]. I
should get an output of 1 once I am done. The way I want to achieve
that is by making a new array called newar[] and make it include all
the indexes of ar[], except without -2."
This is a very inefficient way to approach this problem. You'll have to do 3 passes, Once to find to smallest indexed element, another pass to remove the element (this is an array so removing an element will require a full pass), and another one to find smallest one again.
You should just do a single pass algorithm and keep track of the smallest two integers,
or even better use a tree for efficiency. Here are the best answers of this problem:
Find the 2nd largest element in an array with minimum number of comparisons
Algorithm: Find index of 2nd smallest element from an unknown array
UPDATE: Here is the algorithm with OP's requirements,
3 passes, and no external libraries:
public static int Second_Tiny() {
int[] ar = {19, 1, 17, 17, -2};
//1st pass - find the smallest item on original array
int i;
int z = ar[0];
for (i = 1; i < ar.length; i++) {
if (z >ar[i]){
z=ar[i];
}
}
//2nd pass copy all items except smallest one to 2nd array
int[] ar2 = new int[ar.length-1];
int curIndex = 0;
for (i=0; i<ar.length; i++) {
if (ar[i]==z)
continue;
ar2[curIndex++] = ar[i];
}
//3rd pass - find the smallest item again
z = ar2[0];
for (i = 1; i < ar2.length; i++) {
if (z >ar2[i]){
z=ar2[i];
}
}
return z;
}

This grabs the index of the element specified in variable z and then sets a second array to the first array minus that one element.
Essentially this gives ar2 = ar1 minus element z
public static int Second_Tiny() {
int[] ar = {19, 1, 17, 17, -2};
int[] ar2;
int i;
int z = ar[0];
int x = 0;
for (i = 1; i < ar.length; i++) {
if (z >ar[i]){
z=ar[i];
x=i;
}
}
ar2 = ArrayUtils.remove(ar, x);
return(z);
}

Removing duplicates from array without using Util classes

Please read the question before marking it as duplicate
I have written following code to remove duplicates from array without using Util classes but now I am stuck
public class RemoveDups{
public static void main(String[] args) {
int[] a = { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 3, 1, 4, 52, 1, 45, };
int temp;
for (int i : a) {
for (int j = 0; j < a.length - 1; j++) {
if (a[j] > a[j + 1]) {
temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
a = removeDups(a);
for (int i : a) {
System.out.println(i);
}
}
private static int[] removeDups(int[] a) {
int[] result = new int[a.length];
int j = 0;
for (int i : a) {
if (!isExist(result, i)) {
result[j++] = i;
}
}
return result;
}
private static boolean isExist(int[] result, int i) {
for (int j : result) {
if (j == i) {
return true;
}
}
return false;
}
}
and now the output is
1
2
3
4
5
6
45
52
0
0
0
0
0
0
0
0
0
0
Here my problem is
My code is not working in case of 0s
I am not able to understand how sorting an array can reduce time of execution
Is there any way to remove elements from array without using Util classes I know one way to remove convert array into list and then remove but for that also we need Util classes is there any way to implement by myself.

Since the numbers you deal with are limited to a small range you can remove duplicates by a simple "counting sort": mark the numbers you have found in a set-like data structure and then go over the data structure. An array of boolean works just fine, for less memory usage you could create a basic bitset or hash table. If n is the number of elements in the array and m is the size of the range, this algorithm will have O(n+m) complexity.
private static int[] removeDups(int[] a, int maxA) {
boolean[] present = new boolean[maxA+1];
int countUnique = 0;
for (int i : a) {
if (!present[i]) {
countUnique++;
present[i] = true;
}
}
int[] result = new int[countUnique];
int j = 0;
for (int i=0; i<present.length; i++) {
if (present[i]) result[j++] = i;
}
return result;
}
I am not able to understand how sorting an array can reduce time of execution
In a sorted array you can detect duplicates in a single scan, taking O(n) time. Since sorting is faster than checking each pair - O(n log n) compared to O(n²) time complexity - it would be faster to sort the array instead of using the naive algorithm.

As you are making the result array of the same length as array a
so even if you put only unique items in it, rest of the blank items will have the duplicate values in them which is 0 for int array.
Sorting will not help you much, as you code is searching the whole array again and again for the duplicates. You need to change your logic for it.
You can put some negative value like -1 for all the array items first in result array and then you can easily create a new result array say finalResult array from it by removing all the negative values from it, It will also help you to remove all the zeroes.

In java , arrays are of fixed length. Once created, their size can't be changed.
So you created an array of size18.
Then after you applied your logic , some elements got deleted. But array size won't change. So even though there are only 8 elements after the duplicate removal, the rest 10 elements will be auto-filled with 0 to keep the size at 18.
Solution ?
Store the new list in another array whose size is 8 ( or whatever, calculate how big the new array should be)
Keep a new variable to point to the end of the last valid element, in this case the index of 52. Mind you the array will still have the 0 values, you just won't use them.
I am not able to understand how sorting an array can reduce time of execution
What ? You sort an array if you need it to be sorted. Nothing else. Some algorithm may require the array to be sorted or may work better if the array is sorted. Depends on where you are using the array. In your case, the sorting will not help.
As for your final question , you can definitely implement your own duplicate removal by searching if an element exists more than once and then deleting all the duplicates.

My code is not working in case of 0
There were no zeroes to begin with in your array. But because its an int[], after the duplicates are removed the remaining of the indexes are filled with 0. That's why you can see a lot of zeroes in your array. To get rid of those 0s, you need to create another array with a lesser size(size should be equal to the no. of unique numbers you've in your array, excluding 0).
If you can sort your array(I see that its already sorted), then you could either bring all the zeroes to the front or push them to the last. Based on that, you can iterate the array and get the index from where the actual values start in the array. And, then you could use Arrays.copyOfRange(array, from, to) to create a copy of the array only with the required elements.

try this
package naveed.workingfiles;
public class RemoveDups {
public static void main(String[] args) {
int[] a = { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 3, 1, 4, 52, 1, 45, };
removeDups(a);
}
private static void removeDups(int[] a) {
int[] result = new int[a.length];
int j = 0;
int count = 0;
for (int i : a) {
if (!isExist(result, i)) {
result[j++] = i;
count++;
}
}
System.out.println(count + "_____________");
for (int i=0;i<count;i++) {
System.out.println(result[i]);
}
// return result;
}
private static boolean isExist(int[] result, int i) {
for (int j : result) {
if (j == i) {
return true;
}
}
return false;
}
}

public class RemoveDups {
public static void main(String[] args) {
int[] a = { 1, 2, 0, 3, 1,0, 3, 6, 2};
removeDups(a);
}
private static void removeDups(int[] a) {
int[] result = new int[a.length];
int j = 0;
int count = 0;
boolean zeroExist = false;
for (int i : a) {
if(i==0 && !zeroExist){
result[j++] = i;
zeroExist = true;
count++;
}
if (!isExist(result, i)) {
result[j++] = i;
count++;
}
}
System.out.println(count + "_____________");
for (int i=0;i<count;i++) {
System.out.println(result[i]);
}
// return result;
}
private static boolean isExist(int[] result, int i) {
for (int j : result) {
if (j == i) {
return true;
}
}
return false;
}
}
// It works even Array contains 'Zero'

class Lab2 {
public static void main(String[] args) {
int[] a = { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 3, 1, 4, 52, 1, 45 };
removeDups(a);
}
private static void removeDups(int[] a) {
int[] result = new int[a.length];
int j = 0;
int count = 0;
for (int i : a) {
if (!isExist(result, i)) {
result[j++] = i;
count++;
}
}
System.out.println(count + "_____________");
for (int i = 0; i < count; i++) {
System.out.println(result[i]);
}
}
private static boolean isExist(int[] result, int i) {
for (int j : result) {
if (j == i) {
return true;
}
}
return false;
}
}

How do I divide an ordered list of integers into evenly sized sublists?

Does anyone have a good algorithm for taking an ordered list of integers, i.e.:
[1, 3, 6, 7, 8, 10, 11, 13, 14, 17, 19, 23, 25, 27, 28]
into a given number of evenly sized ordered sublists, i.e. for 4 it will be:
[1, 3, 6] [7, 8, 10, 11] [13, 14, 17, 19] [23, 25, 27, 28]
The requirement being that each of the sublists are ordered and as similar in size as possible.

Splitting the lists evenly means you will have two sizes of lists - size S and S+1.
With N sublists, and X elements in the original, you would get:
floor(X/N) number of elements in the smaller sublists (S), and X % N is the number of larger sublists (S+1).
Then iterate over the original array, and (looking at your example) creating small lists firsts.
Something like this maybe:
private static List<Integer[]> splitOrderedDurationsIntoIntervals(Integer[] durations, int numberOfIntervals) {
int sizeOfSmallSublists = durations.length / numberOfIntervals;
int sizeOfLargeSublists = sizeOfSmallSublists + 1;
int numberOfLargeSublists = durations.length % numberOfIntervals;
int numberOfSmallSublists = numberOfIntervals - numberOfLargeSublists;
List<Integer[]> sublists = new ArrayList(numberOfIntervals);
int numberOfElementsHandled = 0;
for (int i = 0; i < numberOfIntervals; i++) {
int size = i < numberOfSmallSublists ? sizeOfSmallSublists : sizeOfLargeSublists;
Integer[] sublist = new Integer[size];
System.arraycopy(durations, numberOfElementsHandled, sublist, 0, size);
sublists.add(sublist);
numberOfElementsHandled += size;
}
return sublists;
}

Here is my own recursive solution, inspired by merge sort and breadth first tree traversal:
private static void splitOrderedDurationsIntoIntervals(Integer[] durations, List<Integer[]> intervals, int numberOfInterals) {
int middle = durations.length / 2;
Integer[] lowerHalf = Arrays.copyOfRange(durations, 0, middle);
Integer[] upperHalf = Arrays.copyOfRange(durations, middle, durations.length);
if (lowerHalf.length > upperHalf.length) {
intervals.add(lowerHalf);
intervals.add(upperHalf);
} else {
intervals.add(upperHalf);
intervals.add(lowerHalf);
}
if (intervals.size() < numberOfIntervals) {
int largestElementLength = intervals.get(0).length;
if (largestElementLength > 1) {
Integer[] duration = intervals.remove(0);
splitOrderedDurationsIntoIntervals(duration, intervals);
}
}
}
I was hoping someone might have a suggestion for an iterative solution.

Here's a solution for Python. You can translate it to Java, you need a way to get a piece of of a list and then to return it. You cannot use the generator approach though, but you can append each sublist to a new list.

pseudocode...
private static void splitOrderedDurationsIntoIntervals(Integer[] durations, List<Integer[]> intervals, int numberOfInterals) {
int num_per_interval = Math.floor(durations.length / numberOfInterals);
int i;
int idx;
// make sure you have somewhere to put the results
for (i = 0; i < numberOfInterals; i++) intervals[i] = new Integer[];
// run once through the list and put them in the right sub-list
for (i = 0; i < durations.length; i++)
{
idx = Math.floor(i / num_per_interval);
intervals[idx].add(durations[i]);
}
}
That code will need a bit of tidying up, but I'm sure you get the point. Also I suspect that the uneven sized interval list will be at the end rather than at the beginning. If you really want it that way round you can probably do that by reversing the order of the loop.

That should be an Answer in a more iterative fashion.
public static void splitList(List<Integer> startList, List<List<Integer>> resultList,
int subListNumber) {
final int subListSize = startList.size() / subListNumber;
int index = 0;
int stopIndex = subListSize;
for (int i = subListNumber; i > 0; i--) {
resultList.add(new ArrayList<Integer>(startList.subList(index, stopIndex)));
index = stopIndex;
stopIndex =
(index + subListSize > startList.size()) ? startList.size() : index + subListSize;
}
}

You might consider something like this:
public static int[][] divide(int[] initialList, int sublistCount)
{
if (initialList == null)
throw new NullPointerException("initialList");
if (sublistCount < 1)
throw new IllegalArgumentException("sublistCount must be greater than 0.");
// without remainder, length / # lists will always be the minimum
// number of items in a given subset
int min = initialList.length / sublistCount;
// without remainer, this algorithm determines the maximum number
// of items in a given subset. example: in a 15-item sample,
// with 4 subsets, we get a min of 3 (15 / 4 = 3r3), and
// 15 + 3 - 1 = 17. 17 / 4 = 4r1.
// in a 16-item sample, min = 4, and 16 + 4 - 1 = 19. 19 / 4 = 4r3.
// The -1 is required in samples in which the max and min are the same.
int max = (initialList.length + min - 1) / sublistCount;
// this is the meat and potatoes of the algorithm. here we determine
// how many lists have the min count and the max count. we start out
// with all at max and work our way down.
int sublistsHandledByMax = sublistCount;
int sublistsHandledByMin = 0;
while ((sublistsHandledByMax * max) + (sublistsHandledByMin * min)
!= initialList.length)
{
sublistsHandledByMax--;
sublistsHandledByMin++;
}
// now we copy the items into their new sublists.
int[][] items = new int[sublistCount][];
int currentInputIndex = 0;
for (int listIndex = 0; listIndex < sublistCount; listIndex++)
{
if (listIndex < sublistsHandledByMin)
items[listIndex] = new int[min];
else
items[listIndex] = new int[max];
// there's probably a better way to do array copies now.
// it's been a while since I did Java :)
System.arraycopy(initialList, currentInputIndex, items[listIndex], 0, items[listIndex].length);
currentInputIndex += items[listIndex].length;
}
return items;
}
This isn't quite polished - I got into an infinite loop (I think) when I tried to pass an 18-item array in with 10 sublists. I think the algorithm breaks down when min == 1.
This should be fairly fast. Good luck :)