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isPalindrome - Collection and List Reversal

This is a homework lab for school. I am trying to reverse a LinkedList, and check if it is a palindrome (the same backwards and forwards). I saw similar questions online, but not many that help me with this. I have made programs that check for palindromes before, but none that check an array or list. So, first, here is my isPalindrome method:
public static <E> boolean isPalindrome(Collection<E> c) {
Collection<E> tmp = c;
System.out.println(tmp);
Collections.reverse((List<E>) c);
System.out.println(c);
if(tmp == c) { return true; } else { return false; }
}
My professor wants us to set the method up to accept all collections which is why I used Collection and cast it as a list for the reverse method, but I'm not sure if that is done correctly. I know that it does reverse the list. Here is my main method:
public static void main(String...strings) {
Integer[] arr2 = {1,3,1,1,2};
LinkedList<Integer> ll2 = new LinkedList<Integer>(Arrays.asList(arr2));
if(isPalindrome(ll2)) { System.out.println("Successful!"); }
}
The problem is, I am testing this with an array that is not a palindrome, meaning it is not the same backwards as it is forwards. I already tested it using the array {1,3,1} and it works fine because that is a palindrome. Using {1,3,1,1,2} still returns true for palindrome, though it is clearly not. Here is my output using the {1,3,1,1,2} array:
[1, 3, 1, 1, 2]
[2, 1, 1, 3, 1]
Successful!
So, it seems to be properly reversing the List, but when it compares them, it assumes they are equal? I believe there is an issue with the tmp == c and how it checks whether they are equal. I assume it just checks if it contains the same elements, but I'm not sure. I also tried tmp.equals(c), but it returned the same results. I'm just curious is there is another method that I can use or do I have to write a method to compare tmp and c?
Thank you in advance!
Tommy

In your code c and tmp are links to same collection and tmp == c will be always true. Your must clone your collection to new instance, for example: List<E> tmp = new ArrayList(c);.

Many small points
public static <E> boolean isPalindrome(Collection<E> c) {
List<E> list = new ArrayList<>(c);
System.out.println(list);
Collections.reverse(list);
System.out.println(list);
return list.equals(new ArrayList<E>(c));
}
Reverse only works on an ordered list.
One makes a copy of the collection.
One uses equals to compare collections.
public static void main(String...strings) {
int[] arr2 = {1, 3, 1, 1, 2};
//List<Integer> ll2 = new LinkedList<>(Arrays.asList(arr2));
List<Integer> ll2 = Arrays.asList(arr2);
if (isPalindrome(ll2)) { System.out.println("Successful!"); }
}

You need to copy the Collection to a List / array. This has to be done, since the only ordering defined for a Collection is the one of the iterator.
Object[] asArray = c.toArray();
You can apply the algorithm of your choice for checking if this array is a palindrom to check, if the Collection is a palindrom.
Alternatively using LinkedList it would be more efficient to check, if the list is a palindrom without creating a new List to reverse:
public static <E> boolean isPalindrome(Collection<E> c) {
List<E> list = new LinkedList<>(c);
Iterator<E> startIterator = list.iterator();
ListIterator<E> endIterator = list.listIterator(list.size());
for (int i = list.size() / 2; i > 0; i--) {
if (!Objects.equals(startIterator.next(), endIterator.previous())) {
return false;
}
}
return true;
}

Reusing input String[] with values from arrayList

I have the following code which sorts a mixed array of items while maintaining the position of types:
For example:
[20, "abc", "moose", 2,1] turns into [1, "abc", "moose", 2, 20]
Algorithm:
public class Algorithm {
public static String[] sortMixedArray(String[] input){
if (input.length == 0){
return input;
}
// make new arraylist for strings and numbers respectively
List<String> strs = new ArrayList<String>();
List<Integer> numbers = new ArrayList<Integer>();
// add values to the arraylist they belong to
for (String item : input){
if (NumberUtils.isNumber(item)){
numbers.add(Integer.valueOf(item));
} else {
strs.add(item);
}
}
// sort for O(nlogn)
Collections.sort(strs);
Collections.sort(numbers);
// reuse original array
for (int i = 0; i < input.length; i++){
if (NumberUtils.isNumber(input[i])) {
input[i] = String.valueOf(numbers.remove(0));
} else {
input[i] = strs.remove(0);
}
}
return input;
}
public static void main(String[] args) {
String[] test = new String[] {"moo", "boo"};
System.out.println(Arrays.toString(sortMixedArray(test)));
}
I have a two-part question:
1. Is switching between array and arraylist efficient? That is, should I have used arrays everywhere instead of arraylist if my input MUST be an array.
2. What is the best way to place arraylist items back into a array? I am checking for type, is there a better way?

1.If you do it the way you have it in your code then it's perfectly fine. If you know beforehand how many elements you will have it's better to use arrays but thats not the case in your example.
2.The best and easiest way is to use the toArray() function of the List interface.
ArrayList<String> list = ...;
String[] array = list.toArray(new String[list.size()]);
But this won't work for your code since you are merging two lists into one array. You can still improve your code a bit because you do not actually have to remove the items from the lists when putting them back in the array. This safes some computation since removing the first element from an ArrayList is very inefficient (O(N) runtime per remove operation).
for (int i = 0, s = 0, n = 0; i < input.length; i++) {
if (NumberUtils.isNumber(input[i])) {
input[i] = Integer.toString(numbers.get(n++));
} else {
input[i] = strs.get(s++);
}
}

No but it highly unlikely to matter unless you have a million of elements.
Do whatever you believe is simplest and most efficient for you, the developer.
BTW the least efficient operations is remove(0) which is O(N) so you might change that.

Removing Duplicate Values from ArrayList

I have one Arraylist of String and I have added Some Duplicate Value in that. and i just wanna remove that Duplicate value So how to remove it.
Here Example I got one Idea.
List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
System.out.println("List"+list);
for (int i = 1; i < list.size(); i++) {
String a1 = list.get(i);
String a2 = list.get(i-1);
if (a1.equals(a2)) {
list.remove(a1);
}
}
System.out.println("List after short"+list);
But is there any Sufficient way remove that Duplicate form list. with out using For loop ?
And ya i can do it by using HashSet or some other way but using array list only.
would like to have your suggestion for that. thank you for your answer in advance.

You can create a LinkedHashSet from the list. The LinkedHashSet will contain each element only once, and in the same order as the List. Then create a new List from this LinkedHashSet. So effectively, it's a one-liner:
list = new ArrayList<String>(new LinkedHashSet<String>(list))
Any approach that involves List#contains or List#remove will probably decrease the asymptotic running time from O(n) (as in the above example) to O(n^2).
EDIT For the requirement mentioned in the comment: If you want to remove duplicate elements, but consider the Strings as equal ignoring the case, then you could do something like this:
Set<String> toRetain = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
toRetain.addAll(list);
Set<String> set = new LinkedHashSet<String>(list);
set.retainAll(new LinkedHashSet<String>(toRetain));
list = new ArrayList<String>(set);
It will have a running time of O(n*logn), which is still better than many other options. Note that this looks a little bit more complicated than it might have to be: I assumed that the order of the elements in the list may not be changed. If the order of the elements in the list does not matter, you can simply do
Set<String> set = new TreeSet<String>(String.CASE_INSENSITIVE_ORDER);
set.addAll(list);
list = new ArrayList<String>(set);

if you want to use only arraylist then I am worried there is no better way which will create a huge performance benefit. But by only using arraylist i would check before adding into the list like following
void addToList(String s){
if(!yourList.contains(s))
yourList.add(s);
}
In this cases using a Set is suitable.

You can make use of Google Guava utilities, as shown below
list = ImmutableSet.copyOf(list).asList();
This is probably the most efficient way of eliminating the duplicates from the list and interestingly, it preserves the iteration order as well.
UPDATE
But, in case, you don't want to involve Guava then duplicates can be removed as shown below.
ArrayList<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
System.out.println("List"+list);
HashSet hs = new HashSet();
hs.addAll(list);
list.clear();
list.addAll(hs);
But, of course, this will destroys the iteration order of the elements in the ArrayList.
Shishir

Java 8 stream function
You could use the distinct function like above to get the distinct elements of the list,
stringList.stream().distinct();
From the documentation,
Returns a stream consisting of the distinct elements (according to Object.equals(Object)) of this stream.
Another way, if you do not wish to use the equals method is by using the collect function like this,
stringList.stream()
.collect(Collectors.toCollection(() ->
new TreeSet<String>((p1, p2) -> p1.compareTo(p2))
));
From the documentation,
Performs a mutable reduction operation on the elements of this stream using a Collector.
Hope that helps.

Simple function for removing duplicates from list
private void removeDuplicates(List<?> list)
{
int count = list.size();
for (int i = 0; i < count; i++)
{
for (int j = i + 1; j < count; j++)
{
if (list.get(i).equals(list.get(j)))
{
list.remove(j--);
count--;
}
}
}
}
Example:
Input: [1, 2, 2, 3, 1, 3, 3, 2, 3, 1, 2, 3, 3, 4, 4, 4, 1]
Output: [1, 2, 3, 4]

List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
HashSet<String> hs=new HashSet<>(list);
System.out.println("=========With Duplicate Element========");
System.out.println(list);
System.out.println("=========Removed Duplicate Element========");
System.out.println(hs);

I don't think the list = new ArrayList<String>(new LinkedHashSet<String>(list)) is not the best way , since we are using the LinkedHashset(We could use directly LinkedHashset instead of ArrayList),
Solution:
import java.util.ArrayList;
public class Arrays extends ArrayList{
#Override
public boolean add(Object e) {
if(!contains(e)){
return super.add(e);
}else{
return false;
}
}
public static void main(String[] args) {
Arrays element=new Arrays();
element.add(1);
element.add(2);
element.add(2);
element.add(3);
System.out.println(element);
}
}
Output:
[1, 2, 3]
Here I am extending the ArrayList , as I am using the it with some changes by overriding the add method.

public List<Contact> removeDuplicates(List<Contact> list) {
// Set set1 = new LinkedHashSet(list);
Set set = new TreeSet(new Comparator() {
#Override
public int compare(Object o1, Object o2) {
if(((Contact)o1).getId().equalsIgnoreCase(((Contact)2).getId()) ) {
return 0;
}
return 1;
}
});
set.addAll(list);
final List newList = new ArrayList(set);
return newList;
}

This will be the best way
List<String> list = new ArrayList<String>();
list.add("Krishna");
list.add("Krishna");
list.add("Kishan");
list.add("Krishn");
list.add("Aryan");
list.add("Harm");
Set<String> set=new HashSet<>(list);

It is better to use HastSet
1-a) A HashSet holds a set of objects, but in a way that it allows you to easily and quickly determine whether an object is already in the set or not. It does so by internally managing an array and storing the object using an index which is calculated from the hashcode of the object. Take a look here
1-b) HashSet is an unordered collection containing unique elements. It has the standard collection operations Add, Remove, Contains, but since it uses a hash-based implementation, these operation are O(1). (As opposed to List for example, which is O(n) for Contains and Remove.) HashSet also provides standard set operations such as union, intersection, and symmetric difference.Take a look here
2) There are different implementations of Sets. Some make insertion and lookup operations super fast by hashing elements. However that means that the order in which the elements were added is lost. Other implementations preserve the added order at the cost of slower running times.
The HashSet class in C# goes for the first approach, thus not preserving the order of elements. It is much faster than a regular List. Some basic benchmarks showed that HashSet is decently faster when dealing with primary types (int, double, bool, etc.). It is a lot faster when working with class objects. So that point is that HashSet is fast.
The only catch of HashSet is that there is no access by indices. To access elements you can either use an enumerator or use the built-in function to convert the HashSet into a List and iterate through that.Take a look here

Without a loop, No! Since ArrayList is indexed by order rather than by key, you can not found the target element without iterate the whole list.
A good practice of programming is to choose proper data structure to suit your scenario. So if Set suits your scenario the most, the discussion of implementing it with List and trying to find the fastest way of using an improper data structure makes no sense.

public static void main(String[] args) {
#SuppressWarnings("serial")
List<Object> lst = new ArrayList<Object>() {
#Override
public boolean add(Object e) {
if(!contains(e))
return super.add(e);
else
return false;
}
};
lst.add("ABC");
lst.add("ABC");
lst.add("ABCD");
lst.add("ABCD");
lst.add("ABCE");
System.out.println(lst);
}
This is the better way

list = list.stream().distinct().collect(Collectors.toList());
This could be one of the solutions using Java8 Stream API. Hope this helps.

public void removeDuplicates() {
ArrayList<Object> al = new ArrayList<Object>();
al.add("java");
al.add('a');
al.add('b');
al.add('a');
al.add("java");
al.add(10.3);
al.add('c');
al.add(14);
al.add("java");
al.add(12);
System.out.println("Before Remove Duplicate elements:" + al);
for (int i = 0; i < al.size(); i++) {
for (int j = i + 1; j < al.size(); j++) {
if (al.get(i).equals(al.get(j))) {
al.remove(j);
j--;
}
}
}
System.out.println("After Removing duplicate elements:" + al);
}
Before Remove Duplicate elements:
[java, a, b, a, java, 10.3, c, 14, java, 12]
After Removing duplicate elements:
[java, a, b, 10.3, c, 14, 12]

Using java 8:
public static <T> List<T> removeDuplicates(List<T> list) {
return list.stream().collect(Collectors.toSet()).stream().collect(Collectors.toList());
}

In case you just need to remove the duplicates using only ArrayList, no other Collection classes, then:-
//list is the original arraylist containing the duplicates as well
List<String> uniqueList = new ArrayList<String>();
for(int i=0;i<list.size();i++) {
if(!uniqueList.contains(list.get(i)))
uniqueList.add(list.get(i));
}
Hope this helps!

private static void removeDuplicates(List<Integer> list)
{
Collections.sort(list);
int count = list.size();
for (int i = 0; i < count; i++)
{
if(i+1<count && list.get(i)==list.get(i+1)){
list.remove(i);
i--;
count--;
}
}
}

public static List<String> removeDuplicateElements(List<String> array){
List<String> temp = new ArrayList<String>();
List<Integer> count = new ArrayList<Integer>();
for (int i=0; i<array.size()-2; i++){
for (int j=i+1;j<array.size()-1;j++)
{
if (array.get(i).compareTo(array.get(j))==0) {
count.add(i);
int kk = i;
}
}
}
for (int i = count.size()+1;i>0;i--) {
array.remove(i);
}
return array;
}
}

How to move data from multiple Arraylist to multiple Arrays (in Java)

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;
}

Get objects' index in a Java List

I have a list of strings in my (Android) Java program, and I need to get the index of an object in the list. The problem is, I can only find documentation on how to find the first and last index of an object. What if I have 3 or more of the same object in my list? How can I find every index?
Thanks!

You need to do a brute force search:
static <T> List<Integer> indexesOf(List<T> source, T target)
{
final List<Integer> indexes = new ArrayList<Integer>();
for (int i = 0; i < source.size(); i++) {
if (source.get(i).equals(target)) { indexes.add(i); }
}
return indexes;
}
Note that this is not necessarily the most efficient approach. Depending on the context and the types/sizes of lists, you might need to do some serious optimizations. The point is, if you need every index (and know nothing about the structure of the list contents), then you need to do a deathmarch through every item for at best O(n) cost.
Depending on the type of the underlying list, get(i) may be O(1) (ArrayList) or O(n) (LinkedList), so this COULD blow up to a O(n2) implementation. You could copy to an ArrayList, or you could walk the LinkedList incrementing an index counter manually.

If documentation is not helping me in my logic in this situation i would have gone for a raw approach for Traversing the list in a loop and saving the index where i found a match
ArrayList<String> obj = new ArrayList<String>();
obj.add("Test Data"): // fill the list with your data
String dataToFind = "Hello";
ArrayList<Integer> intArray = new ArrayList<Integer>();
for(int i = 0 ; i<obj.size() ; i++)
{
if(obj.get(i).equals(dataToFind)) intArray.add(i);
}
now intArray would have contained all the index of matched element in the list

An alternative brute force approach that will also find all null indexes:
static List<Integer> indexesOf(List<?> list, Object target) {
final List<Integer> indexes = new ArrayList<Integer>();
int offset = 0;
for (int i = list.indexOf(target); i != -1; i = list.indexOf(target)) {
indexes.add(i + offset);
list = list.subList(i + 1, list.size());
offset += i + 1;
}
return indexes;
}