Array object and maximum - java

I know how to find minimum and maximum in an array. If a method lets say was called fMax():
public static double fMax(Object[] stuff)
The parameter is an array object how would I go about finding the max of this array? I cannot just do. Okay so how would I do this if I want the method to return a double and if the memory hasnt been allocated for the parameter named stuff then it will return the value NEGATIVE_INFINITY in the Double class, otherwise the return value will be the maximum value from the elements in the stuff array
Object max = stuff[0];
for (int i = 0; i < stuff.length; i++) {
if (data[i] > max) {
max = stuff[i];
}
}

To find the maximum of something, either
a) that something needs to implement the Comparable interface
b) you need to have some sort of explicit criteria for determining what maximum is, so you can put that in an instance of Comparator
Object itself isn't going to have anything useful for sorting. If you subclass object, you could sort based on the components of that object.
public class Example implements Comparable
{
int sortableValue = 0;
public Example (int value)
{
this.sortableValue = value;
}
public int compareTo(Example other)
{
return Integer.compare(this.sortableValue, other.sortableValue);
}
}
That's an object definition that has a natural sorting order. Java can look at that with any of the built in sorting algorithms and know the order they belong in.
If you don't provide java with a means of determining how an object has greater or lesser relative value compared to another object of the same type, it won't figure it out on its own.

Object is not comparable, you need a definite type if you want to compare values, sort or find something.
Streams are the most powerful, versatile tools for the job, this here will solve your problem if your want to find min/max of an array of Double :
Double[] arr = {1d, 2d, 3d, 4d};
Double min = Arrays.asList(arr).stream().parallel().min(Double::compare).get();
Double max = Arrays.asList(arr).stream().parallel().max(Double::compare).get();

String[] stringArray = Arrays.copyOf(objectArray, objectArray.length, String[].class);
Now, just compare the new primitive array that we made from the object. If you don't need the object after this, and you aren't planning on returning an array object, then make your original array null, to take up less memory.
Check this:
How to compare two object arrays in Java?

Related

Need help using java.lang.reflect.Array to sort arrays

An interview question was to write this method to remove duplicate element in an array.
public static Array removeDuplicates(Array a) {
...
return type is java.lang.reflect.Array and parameter is also java.lang.reflect.Array type.
How would this method be called for any array?
Also not sure about my implementation:
public static Array removeDuplicates(Array a)
{
int end=Array.getLength(a)-1;
for(int i=0;i<=end-1;i++)
{
for(int j=i+1;j<=end;j++)
{
if(Array.get(a, i)==Array.get(a, j))
{
Array.set(a, j, Array.get(a, end));
end--;
j--;
}
}
}
Array b=(Array) Array.newInstance(a.getClass(), end+1);
for(int i=0;i<=end;i++)
Array.set(a, i, Array.get(a, i));
return b;
}
You may want to consider using a different data structure such as a hashmap to detect the duplicate (O(1)) instead of looping with nested for loops (O(n^2)). It should give you much better time complexity.
There are various problem with this code. Starting here:
if(Array.get(a, i)==Array.get(a, j))
Keep in mind that those get() calls return Object. So, when you pass in an array of strings, comparing with == simply will most likely result in wrong results (because many objects that are in fact equal still have different references --- so your check returns false all the time!)
So, the first thing to change: use equals() instead of == !
The other problem is:
end--;
Seriously: you never ever change the variable that controls your for loop.
Instead: have another counter, like
int numberOfOutgoingItems = end;
and then decrease that counter!
For your final question - check the javadoc; for example for get(). That reads get(Object array, int index)
So you should be able to do something like:
int a[] = ...;
Object oneValue = Array.get(a, 0);
for example.
Disclaimer. I have to admit: I don't know if the Array implementation is smart enough to automatically turn the elements of an int[] into an Integer object.
It could well be that you have to write code first to detect the exact type of array (if it is an array of int for example); to instead call getInt() instead of getObject().
Beyond that, some further reading how to use reflection/Array can be found here

How can I create copy of a libgdx array with custom objects?

I have an array which contains values pawnArray. I need to find the highest value in pawnArray so using a custom class method getPawn() I retrieve the highest value but I do
public static Pawn getPawn(Array<Pawn> strollpawns) {
Array<Pawn> pawns = strollpawns;
pawns.sort();
Pawn best = pawns.get(0);
return best;
}
I hence need to copy the array since this method doesn't work. How can I make a copy of this array?
If your problem is with Java arrays (the syntax is Pawn[]) then you have methods in class java.util.Arrays for many different operations on them. What you are asking for could be accomplished with:
Pawn[] newArr = Arrays.copyOf(oldArr, oldArr.length);
Or, since array classes implement Cloneable, also with:
Pawn[] newArr = (Pawn[]) oldArr.clone(); // I don't remember if the cast is necessary
Note that both of these provide shallow copies, that is, the arrays are independent of each other (you can sort one and the indexes in the other are unaffected) but their contents are not.
EDIT: it has been kindly pointed out to me that your Array<T> is actually a class in libgdx. Looking at the documentation, then, you could simply use the constructor taking another instance of Array to create your shallow copy, since the doc says that the new instance will have the same type of backing array (not the same instance). For example:
Array<T> newArr = new Array<>(oldArr); // oldArr can be either Array<? extends T> or T[]
I'm adding a separate answer to this, since you want to copy your array and sort it in order to retrieve the highest value. My other answer deals with copying the array, while tjago's answer deals with sorting with a custom Comparator in order to customize what the "max value" is. However, it seems that the libgdx Array<T> class has a method to do just what you want, without having to make a sorted copy of the array.
This solution saves you code, memory and time if you only need one value from the sorted array: the minimum, maximum, whatever. If you need more than one, it is likely that sorting the array will be faster.
The method I'm talking about is Array.selectRanked, which returns the nth element according to the provided Comparator. There is another method selectRankedIndex which returns the index of that element instead of the object itself. You could use it like this:
// If Pawn implements Comparable<Pawn>:
Pawn minVal = arr.selectRanked(Comparator.naturalOrder(), 1);
Pawn maxVal = arr.selectRanked(Comparator.naturalOrder(), arr.size);
// If it does not implement Comparable, you need to provide a Comparator<Pawn>:
// Assuming Pawn has an "int getValue()" method that we want to compare:
Pawn minVal = arr.selectRanked(Comparator.comparingInt(Pawn::getValue), 1);
// You could also write your own implementation directly:
Comparator<Pawn> comp = (a,b) -> /* your int-returning logic here */;
Pawn minVal = arr.selectRanked(comp, 1);
It seems you have a java related problem. To help you with sorting In java object programming there exist concept of method overriding and interfaces.
Special interface for sorting is Comparator, you can either put him inline in method like this.
Collections.sort(pawns ,new Comparator<Student>(){
public int compare(Pawn1 p1,Pawn2 p2){
// Write your logic here.
//ie.:
return p1.score - p2.score;
//or for different order
return p2.score - p1.score;
}});
if this comparator return value == 0 means the value are equal;
if value < 0 means p1 is bigger than p2, therefore swap them.
Or put him inside your Object class like:
Class Pawn implements Comparator {
private String name;
private Position[][] posXY;
private int value;
....
Pawn() { ... }
...
public int compare(Pawn1 p1,Pawn2 p2){
return p1.value- p2.value;
}
}
then in your code you can call as you originally intended:
pawns.sort();
Pawn best = pawns.get(0);
and as expected you should get an maximum value Pawn from ArrayList.
The above code is just sample and requires tunning. But You should get an good overview now that Java has no idea how to sort Objects defined by a programmer unless he implements the Comparator logic for Collection sorting.
for external reference I suggest running a simple example on tutorialpoint
Answer to your question: How can I create copy of a libgdx array
Array<Pawn> pawns = new Array<Pawn>(strollpawns);
or if the pawns Array object already exists
pawns.clear();
pawns.addAll(strollpawns);
The first solution will create a new Array object that will be deleted on completion of the function, meaning time lost by garbage collector!
But I agree with Tenfour04: Duplicating an array and sorting it is a very expensive way to select the biggest value.

Making a variable length array of ints

I want to have a array of integers where the length is variable. The obvious choice is to use ArrayList but can I do this for primitive types such as
ArrayList<int> myArray=new ArrayList<int>();
I dont want to use
ArrayList<Integer>
because the Integer class is clumsy in terms of coding.
EDIT: From the answers below I think the solution is to write my own Integer class.
To answer the question below about "clumsy" let me give a specific, and I would of thought common use for integers namely using the last member of the array in any place you would want the integer. If I just call the array "name" then to get the actual integer that can be operated on I need
name.get(name.size()-1).intValue();
To me this seems like an awfully unwieldy expression for a simple integer - particularly if it appears in an expression twice. It also seems that (most of the) methods available for the Integer class are absolutely redundant. Take two examples
static int compare(int a, int b)
Quite unbelievably, according to the documentation, this method returns a-b!!
static Integer valueOf(int a)
returns an Integer instance of the integer a. Can someone give me a single example where
new Integer(a)
does not achieve exactly the same result?
Method 1: (not recommended)
You can do something like this, but this doubles the code and is not efficient:
int[] a;
//get size (from command line maybe ow whatever method you want)
You can set size 0 initially, and for ex. you are transferring values from arraylist so you will have to write:
while(itr.hasNext()){
size++;} //itr is an object of Iterator
int i=0;
a=new int[size];
// then loop again to store values
while(itr.hasNext()){
a[i]=itr.next();
i++;}
Method 2:
Or you may use ArrayList without making it clumsy as follows:
ArrayList al=new ArrayList();
then you may declare Integer objects as volatile and perform operations on them just as you do with the primitive types.
Method 3: (not recommended)
Or simply write:
ArrayList al=new ArrayList();//ignore the warning about <E>
int x=2;
al.add(2);
Method 4: (recommended)
If I were you I would use ArrayList<Integer>.
UPDATE: Another thing that might work is that you may initially create an ArrayList<Integer> and store values there and later convert it to int[]
This SO answer tells about the conversion. Quoted the code form there:
public static int[] convertIntegers(List<Integer> integers)
{
int[] ret = new int[integers.size()];
for (int i=0; i < ret.length; i++)
{
ret[i] = integers.get(i).intValue();
}
return ret;
}
Hope this helps.
No it's not possible to use primitive types as generic type.
Well I would recommend you do use ArrayList and avoid primitive types in this case.
You can't change the size of an array once created. You have to allocate it bigger than you think you'll ever need
or
Accept the overhead of having to reallocate it to a new larger array and copy the data from the old to the new:
System.arraycopy(oldItems, 0, newItems, 0, 10);
But Much simpler to go with ArrayList.

Select subarray without copying into new buffer?

I have float[] array of length 100. Is there a way I can select (pseudocode):
x = array[10:19];
To get elements 10,11,12,...,19 without copying over into another buffer? I'm in a mobile application where I don't want to waste space or time doing this. I'd rather just reference the pointers the system uses for array.
The most efficient way to do this would be to use System.arrayCopy(), which is much faster and more efficient than copying manually using a loop. It will require another array, but any approach you use (beyond just passing the original array around with a couple of ints representing the offset to use) will do this, and it's relatively cheap - the memory consuming bit is usually the objects that it's referencing rather than the array itself, and they are not copied.
No, there is no API to do that. The closest solution to this would be building your own class that wraps an existing array, and does the re-indexing:
class SubArray {
private final float[] data;
private final int offset;
private final int length;
public SubArray(float[] data, int offset, int length) {
this.data = data;
this.offset = offset;
this.length = length;
}
public float get(int index) {
if (index >= length) throw ...
return data[index + offset];
}
public void set(int index, float value) {
if (index >= length) throw ...
data[index + offset] = value;
}
}
If the result that you need is a new object that behaves like an array in all respects, including the indexing operator, you would need to make a copy.
(Update) Precondition: You should store the data in a Float[] instead of a float[], the performance-hit should be minimal.
You can use: Arrays.asList(array).subList(10, 20).
The Arrays.asList(array) does the following:
Returns a fixed-size list backed by the specified array. (Changes to the returned list "write through" to the array.) This method acts as bridge between array-based and collection-based APIs, in combination with Collection.toArray(). The returned list is serializable and implements RandomAccess.
Source
And then .subList(10, 20) returns you a List.
Then if you really want to work with arrays in the end, you could take the following lines:
List<Float> subList = Arrays.asList((Float[])array).subList(10, 20);
Float[] subArray = subList.toArray(new Float[subList.size()]);
(Update) Changed Arrays.asList(array) to Arrays.asList((Float[])array) such that it is correct now.
From documentation:
Returns an array containing all of the elements in this list in proper sequence (from first to last element); the runtime type of the returned array is that of the specified array. If the list fits in the specified array, it is returned therein. Otherwise, a new array is allocated with the runtime type of the specified array and the size of this list.
If the list fits in the specified array with room to spare (i.e., the array has more elements than the list), the element in the array immediately following the end of the list is set to null. (This is useful in determining the length of the list only if the caller knows that the list does not contain any null elements.)
Like the toArray() method, this method acts as bridge between array-based and collection-based APIs. Further, this method allows precise control over the runtime type of the output array, and may, under certain circumstances, be used to save allocation costs.
Suppose x is a list known to contain only strings. The following code can be used to dump the list into a newly allocated array of String:
Source
This should ensure that no data is wasted, the only thing to be careful about could be autoboxing.
UPDATE: Changed my answer such that it now is correct under a precondition.
What is the problem of using a simple for loop? Objects are in java called by reference.
So, executing copying the array does not copy the objects.
float[] subarray = new float[10];
for(int i = 10, j = 0; i < 19; i++, j++) {
subarray[j] = x[i];
}
The array[0] is a reference to the object of x[0].
edit: This only applies for objects, and i don't know if it also applies to a float

Not sure how to sort an ArrayList based on parts of Objects in that ArrayList (Java)

I have a Sorts class that sorts (based on insertion sort, which was the assignment's direction) any ArrayList of any type passed through it, and uses insertion sort to sort the items in the list lexicographically:
public class Sorts
{
public static void sort(ArrayList objects)
{
for (int i=1; i<objects.size(); i++)
{
Comparable key = (Comparable)objects.get(i);
int position = i;
while (position>0 && (((Comparable)objects.get(position)).compareTo(objects.get(position-1)) < 0))
{
objects.set(position, objects.get(position-1));
position--;
}
objects.set(position, key);
}
}
}
In one of my other files, I use a method (that is called in main later) that sorts objects of type Owner, and we have to sort them by last name (if they are the same, then first name):
Directions: "Sort the list of owners by last name from A to Z. If more than one owner have the same last name, compare their first names. This method calls the sort method defined in the Sorts class."
What I thought first was to get the last name of each owner in a for loop, add it to a temporary ArrayList of type string, call Sorts.sort(), and then re-add it back into the ArrayList ownerList:
public void sortOwners() {
ArrayList<String> temp = new ArrayList<String>();
for (int i=0; i<ownerList.size(); i++)
temp.add(((Owner)ownerList.get(i)).getLastName());
Sorts.sort(temp);
for (int i=0; i<temp.size(); i++)
ownerList.get(i).setLastName(temp.get(i));
}
I guess this was the wrong way to approach it, as it is not sorting when I compile.
What I now think I should do is create two ArrayLists (one is firstName, one is LastName) and say that, in a for loop, that if (lastName is the same) then compare firstName, but I'm not sure if I would need two ArrayLists for that, as it seems needlessly complicated.
So what do you think?
Edit: I am adding a version of compareTo(Object other):
public int compareTo(Object other)
{
int result = 0;
if (lastName.compareTo(((Owner)other).getLastName()) < 0)
result = -1;
else if (lastName.compareTo(((Owner)other).getLastName()) > 0)
result = 1;
else if (lastName.equals(((Owner)other).getLastName()))
{
if (firstName.compareTo(((Owner)other).getFirstName()) < 0)
result = -1;
else if (firstName.compareTo(((Owner)other).getFirstName()) > 0)
result = 1;
else if (firstName.equals(((Owner)other).getFirstName()))
result = 0;
}
return result;
}
I think the object should implement a compareTo method that follows the normal Comparable contract--search for sorting on multiple fields. You are correct that having two lists is unnecessary.
If you have control over the Owner code to begin with, then change the code so that it implements Comparable. Its compareTo() method performs the lastName / firstName test described in the assignment. Your sortOwners() method will pass a List<Owner> directly to Sorts.sort().
If you don't have control over Owner, then create a subclass of Owner that implements Comparable. Call it OwnerSortable or the like. It accepts a regular Owner object in its constructor and simply delegates all methods other than compareTo() to the wrapped object. Its compareTo() will function as above. Your sortOwners() method will create a new List<OwnerSortable> out of the Owner list. It can then pass this on to Sorts.sort().
Since you have an ArrayList of objects, ordinarily we would use the Collections.sort() method to accomplish this task. Note the method signature:
public static <T extends Comparable<? super T>> void sort(List<T> list)
What's important here is that all the objects being sorted must implement the Comparable interface, which allows objects to be compared to another in numerical fashion. To clarify, a Comparable object has a method called compareTo with the following signature:
int compareTo(T o)
Now we're getting to the good part. When an object is Comparable, it can be compared numerically to another object. Let's look at a sample call.
String a = "bananas";
String b = "zebras";
System.out.println(a.compareTo(b));
The result will be -24. Semantically, since zebras is farther in the back of the dictionary compared to bananas, we say that bananas is comparatively less than zebras (not as far in the dictionary).
So the solution should be clear now. Use compareTo to compare your objects in such a way that they are sorted alphabetically. Since I've shown you how to compare strings, you should hopefully have a general idea of what needs to be written.
Once you have numerical comparisons, you would use the Collections class to sort your list. But since you have your own sorting ability, not having access to it is no great loss. You can still compare numerically, which was the goal all along! So this should make the necessary steps clearer, now that I have laid them out.
Since this is homework, here's some hints:
Assuming that the aim is to implement a sort algorithm yourself, you will find that it is much easier (and more performant) to extract the list elements into an array, sort the array and then rebuild the list (or create a new one).
If that's not the aim, then look at the Collections class.
Implement a custom Comparator, or change the object class to implement Comparable.

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