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Sorting an ArrayList of arrays in java

I'd like to know if you know a way to sort an ArrayList of arrays in Java.
I have a function that gives a score between 0 and 1 to a specific array. And I'd like to sort the ArrayList so that arrays having the highest score come first.
public double evaluate(int[] toEvaluate) {
double result = 0.0;
for (int i = 0; i < toEvaluate.length; i++) {
result += table[i][casesMap.get(toEvaluate[i])];
}
return result / toEvaluate.length;
}
Any ideas ?

You should use Collections.sort() together with a custom Comparator:
List<Integer[]> arrays = new ArrayList<>();
arrays.add(new Integer[]{1, 2});
arrays.add(new Integer[]{3, 4});
Collections.sort(arrays, new Comparator<Integer[]>() {
public int compare(Integer[] a, Integer[] b) {
return 1; // FIX this according to your needs
}
});
compare() above is just a stub, you should implement it according to the documentation and it could be replaced with a lambda expression. In the code above that would be: Collections.sort(arrays, (a, b) -> 1).

You have to write a Comparator and compare method override where you can use your function to calculate comp value
#Override
public int compare(Integer[] o1, Integer[] o2) {
int o1Number=ratingFunction(o1) ;
int o2Number=ratingFunction(o2) ;
int cmp=o1Number.compareTo(o2Number);
return cmp;
}

You can either use comparator to sort list in descending order or you can use Collections sort method and then use reverse method to make it descending order,
something like this :
List<Integer> numberList =new ArrayList<Integer>();
numberList.add(3);
numberList.add(1);
numberList.add(2);
//before sort
for (Integer integer : numberList) {
System.out.println(integer);
}
//sorting
Collections.sort(numberList);
Collections.reverse(numberList);
//after sort
for (Integer integer : numberList) {
System.out.println(integer);
}

You might want to use stream api for that. So let's say we have the scoring function (I simplified it for the sake of example).
public static double evaluate(int[] arr){
return Arrays.stream(arr).sum() / arr.length;
}
Now we can use it with Comparator.comparing method:
List<int[]> list = Arrays.asList(new int[]{4, 5},
new int[]{2, 3}, new int[]{0, 1});
List<int[]> sorted = list.stream().
sorted(Comparator.comparing(Main::evaluate)).
collect(Collectors.toList());
sorted.forEach(x -> System.out.println(Arrays.toString(x)));
The idea behind the code is quite simple, you provide a comparator which defines how to sort int[] arrays. I hope this helps.

An efficient way to convert List<Integer> to int[] ( array ) without iteration

public static int[] convertListToArray(List<Integer> listResult) {
int[] result = new int[listResult.size()];
int i= 0;
for (int num : listResult) {
result[i++] = num;
}
return result;
}
Is there an efficient way to convert List to array without iterating List explicitly ?
Maybe it is possible by using methods like:
Arrays.copyOf(int [] origin , int newLength );
System.arraycopy(Object src, int srcPos,
Object dest, int destPos,
int length);
I know that there is a solution described here. However, I particularly interested in an efficient way of converting List<Integer> to int[]

Given the need to convert from Integer to int, I don't think you're going to find something more efficient than what you have, if I assume you're talking about runtime efficiency.
You might find converting to Integer[] first and then looping might be more efficient (below), but you might not, too. You'd have to test it in your specific scenario and see.
Here's that example:
int size = listResult.size();
int[] result = new int[size];
Integer[] temp = listResult.toArray(new Integer[size]);
for (int n = 0; n < size; ++n) {
result[n] = temp[n];
}

If efficiency is your primary concern, I think you can use your solution and make it more efficient by using an indexed for loop on the listResult if it is RandomAccess. However this makes the code much less readable, and you'd have to benchmark it for your use cases to see if it is more efficient.
public static int[] convertListToArray(List<Integer> listResult) {
int size = listResult.size();
int[] result = new int[size];
if (listResult instanceof RandomAccess)
{
for (int i = 0; i < size; i++)
{
result[i] = listResult.get(i);
}
}
else
{
int i = 0;
for (int num : listResult) {
result[i++] = num;
}
}
return result;
}
If you use Java 8 and would like to write less code you can use the Streams library.
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
int[] array = list.stream().mapToInt(i -> i).toArray();
If you are open to using a third party library, you can Eclipse Collections as follows.
MutableList<Integer> list = Lists.mutable.with(1, 2, 3, 4, 5);
int[] array = list.collectInt(i -> i).toArray();
The following is slightly more code, but is the most efficient solution I could come up with using Eclipse Collections.
MutableList<Integer> list = Lists.mutable.with(1, 2, 3, 4, 5);
int[] array = new int[list.size()];
list.forEachWithIndex((each, index) -> array[index] = each);
If you need to use the java.util.List interface, the ListIterate utility class can be used from Eclipse Collections.
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5);
int[] array = new int[list.size()];
ListIterate.forEachWithIndex(list, (each, index) -> array[index] = each);
The ListIterate utility will use different iteration code for RandomAccess lists and non-RandomAccess lists.
The most efficient thing to do would be to change the List<Integer> to a MutableIntList in Eclipse Collections or another library that has support for primitive collections.
Note: I am a committer for Eclipse Collections.

In Java 8:
int[] anArray = list.stream()
.filter(Objects::nonNull)
.mapToInt(Integer::intValue)
.toArray();

There is efficient way you could do this Java. However, this could open room for someone to create the generic function (depend on demand).
Just like this sample i wrote, I suggest you do the same to the specific knowledge of your program.
// Generic function to convert set to list
public static <T> ArrayList<T> convertSetToList(Set<T> set)
{
// create an empty list
ArrayList<T> list = new ArrayList<>();
// push each element in the set into the list
for (T t : set)
list.add(t);
// return the list
return list;
}

Java element-wise sum 2 arrays

Given that I have two arrays in Java, A and B I want to add the elements, element-wise, which results in a sum array. Doing this implicitly with a loop is easy but I was wondering if there was a more elegant solution, perhaps with the guava collections or build in java utils. Or perhaps a python-ish way which makes is easy with list comprehensions.
Example:
A = [2,6,1,4]
B = [2,1,4,4]
sum = [4,7,5,8]

You can do it like this:
private void sum() {
int a[] = {2, 6, 1, 4};
int b[] = {2, 1, 4, 4};
int result[] = new int[a.length];
Arrays.setAll(result, i -> a[i] + b[i]);
}
This will first create int result[] of the correct size.
Then with Java 8, released yesterday, the easy part comes:
You can do an Arrays.setAll(int[] array, IntUnaryOperator);
As IntUnaryOperator you can create a lambda mapping the index to the result, in here we choose to map i to a[i] + b[i], which exactly produces our sum.
For very big arrays we can even use Arrays.parallelSetAll

You can use java8 stream and operation on array like this:
//in this example a[] and b[] are same length
int[] a = ...
int[] b = ...
int[] result = new int[a.length];
IntStream.range(0, a.length)
.forEach(i -> result[i] = a[i] + b[i]);
The answer in java8

Java, find intersection of two arrays

I have already read a few other stack overflow threads on this:
to find the intersection of two multisets in java
How do I get the intersection between two arrays as a new array?
public static int[] intersection (int [] x, int numELementsInX, int [] y, int numElementsInY) {
I am trying to examine two arrays as well as their number of elements (numElementsInX and numElementsInY), and return a new array which contains the common values of array x and y. Their intersection.
Example,if x is{1,3,5,7,9}and y is{9,3,9,4} then
intersection(x, 5, y, 4} should return {3, 9} or {9, 3}
I've read I need to use the LCS algorithm. Can anyone give me an example as to how to do this? Both the array and values in array are initialized and generated in another method, then passed into intersection.
Any help/clarification is appreciated.
EDIT CODE
for (int i=0; i<numElementsInX; i++){
for (int j=0; j<numElementsInY; j++){
if (x[j]==x[i]) { //how to push to new array?;
}
else{
}
}
}

The simplest solution would be to use sets, as long as you don't care that the elements in the result will have a different order, and that duplicates will be removed. The input arrays array1 and array2 are the Integer[] subarrays of the given int[] arrays corresponding to the number of elements that you intend to process:
Set<Integer> s1 = new HashSet<Integer>(Arrays.asList(array1));
Set<Integer> s2 = new HashSet<Integer>(Arrays.asList(array2));
s1.retainAll(s2);
Integer[] result = s1.toArray(new Integer[s1.size()]);
The above will return an Integer[], if needed it's simple to copy and convert its contents into an int[].

If you are fine with java-8, then the simplest solution I can think of is using streams and filter. An implementation is as follows:
public static int[] intersection(int[] a, int[] b) {
return Arrays.stream(a)
.distinct()
.filter(x -> Arrays.stream(b).anyMatch(y -> y == x))
.toArray();
}

General test
The answers provide several solutions, so I decided to figure out which one is the most effective.
Solutions
HashSet based by Óscar López
Stream based by Bilesh Ganguly
Foreach based by Ruchira Gayan Ranaweera
HashMap based by ikarayel
What we have
Two String arrays that contain 50% of the common elements.
Every element in each array is unique, so there are no duplicates
Testing code
public static void startTest(String name, Runnable test){
long start = System.nanoTime();
test.run();
long end = System.nanoTime();
System.out.println(name + ": " + (end - start) / 1000000. + " ms");
}
With use:
startTest("HashMap", () -> intersectHashMap(arr1, arr2));
startTest("HashSet", () -> intersectHashSet(arr1, arr2));
startTest("Foreach", () -> intersectForeach(arr1, arr2));
startTest("Stream ", () -> intersectStream(arr1, arr2));
Solutions code:
HashSet
public static String[] intersectHashSet(String[] arr1, String[] arr2){
HashSet<String> set = new HashSet<>(Arrays.asList(arr1));
set.retainAll(Arrays.asList(arr2));
return set.toArray(new String[0]);
}
Stream
public static String[] intersectStream(String[] arr1, String[] arr2){
return Arrays.stream(arr1)
.distinct()
.filter(x -> Arrays.asList(arr2).contains(x))
.toArray(String[]::new);
}
Foreach
public static String[] intersectForeach(String[] arr1, String[] arr2){
ArrayList<String> result = new ArrayList<>();
for(int i = 0; i < arr1.length; i++){
for(int r = 0; r < arr2.length; r++){
if(arr1[i].equals(arr2[r]))
result.add(arr1[i]);
}
}
return result.toArray(new String[0]);
}
HashMap
public static String[] intersectHashMap(String[] arr1, String[] arr2){
HashMap<String, Integer> map = new HashMap<>();
for (int i = 0; i < arr1.length; i++)
map.put(arr1[i], 1);
ArrayList<String> result = new ArrayList<>();
for(int i = 0; i < arr2.length; i++)
if(map.containsKey(arr2[i]))
result.add(arr2[i]);
return result.toArray(new String[0]);
}
Testing process
Let's see what happens if we give the methods an array of 20 elements:
HashMap: 0.105 ms
HashSet: 0.2185 ms
Foreach: 0.041 ms
Stream : 7.3629 ms
As we can see, the Foreach method does the best job. But the Stream method is almost 180 times slower.
Let's continue the test with 500 elements:
HashMap: 0.7147 ms
HashSet: 4.882 ms
Foreach: 7.8314 ms
Stream : 10.6681 ms
In this case, the results have changed dramatically. Now the most efficient is the HashMap method.
Next test with 10 000 elements:
HashMap: 4.875 ms
HashSet: 316.2864 ms
Foreach: 505.6547 ms
Stream : 292.6572 ms
The fastest is still the HashMap method. And the Foreach method has become quite slow.
Results
If there are < 50 elements, then it is best to use the Foreach method. He strongly breaks away in speed in this category.
In this case, the top of the best will look like this:
Foreach
HashMap
HashSet
Stream - Better not to use in this case
But if you need to process big data, then the best option would be use the HashMap based method.
So the top of the best look like this:
HashMap
HashSet
Stream
Foreach

With duplicate elements in array finding intersection.
int [] arr1 = {1,2,2,2,2,2,2,3,6,6,6,6,6,6,};
int [] arr2 = {7,5,3,6,6,2,2,3,6,6,6,6,6,6,6,6,};
Arrays.sort(arr1);
Arrays.sort(arr2);
ArrayList result = new ArrayList<>();
int i =0 ;
int j =0;
while(i< arr1.length && j<arr2.length){
if (arr1[i]>arr2[j]){
j++;
}else if (arr1[i]<arr2[j]){
i++;
}else {
result.add(arr1[i]);
i++;
j++;
}
}
System.out.println(result);

If you don't want to use other data structures such as a Set, then the basic idea is that you want to iterate through the elements of one of the arrays and for each value see if it appears in the other. How do you see whether it appears in the other array? Walk through the elements in the other array and for each one, see if its value is equal to the value you are looking for. I suspect that you will be best served by trying to work through this problem on your own beyond this point if your goal in taking the class is to learn to write Java well, but it you get stuck you might consider updating your question with the code that you have written so you can get more detailed feedback and pointers in the right direction.

Try this:
public static void main(String[] args) {
int[] arr1 = new int[]{1, 2, 3, 4, 5};
int[] arr2 = new int[]{3, 2, 5, 9, 11};
getIntersection(arr1, arr2);
}
public static Object[] getIntersection(int[] arr1, int[] arr2) {
List<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < arr1.length; i++) {
for (int j = 0; j < arr2.length; j++) {
if (arr1[i] == arr2[j]) {
list.add(arr1[i]);
}
}
}
return list.toArray();
}

You can find the intersection of two arrays with:
T[] result = Arrays.stream(a1)
.filter(new HashSet<>(Arrays.asList(a2))::contains)
.toArray(T[]::new);
where T should be substitutable by a reference type e.g. String, Integer, etc.
although the above may seem like it's creating a new set for each element, it's not the case at all. instead only one set instance is created.
The above code is equivalent to:
List<T> list = new ArrayList<>();
HashSet<T> container = new HashSet<>(Arrays.asList(a2));
for (T s : a1) {
if (container.contains(s)) list.add(s);
}
T[] result = list.toArray(new T[0]);

finding intersection includes duplicate using the hash map.
Output: 1 2 2 15 9 7 12
public static void main(String[] args) {
int[] arr1 = {1, 2, 2, 1, 5, 9, 15, 9, 7, 7, 12};
int[] arr2 = {1, 2, 2, 3, 4, 15, 9, 7, 12, 14};
printIntersect(arr1, arr2);
}
private static void printIntersect(int[] arr1, int[] arr2) {
Map<Integer, Integer> map = new HashMap<>();
//put first array to map
for (int i = 0; i < arr1.length; i++) {
if (!map.containsKey(arr1[i])) {
map.put(arr1[i], 1);
} else {
map.put(arr1[i], map.get(arr1[i]) + 1);
}
}
//check all value in array two
for (int i = 0; i < arr2.length; i++) {
//if exist and value>1 then decrement value
//if value is 1 remove from map
if (map.containsKey(arr2[i])) {
System.out.print(arr2[i] + " ");
if (map.get(arr2[i]) > 1) {
map.put(arr2[i], map.get(arr2[i]) - 1);
} else {
map.remove(arr2[i]);
}
}
}
}

if the arrays are sorted
int a1[]=new int[] {1,2,3,5,7,8};
int a2[]=new int [] {1,5,6,7,8,9};
// get the length of both the array
int n1=a1.length;
int n2=a2.length;
//create a new array to store the intersection
int a3[]=new int[n1];
//run the loop and find the intersection
int i=0,j=0,k=0;
while(i<n1&& j<n2) {
if(a1[i]<a2[j]) {
// a1 element at i are smaller than a2 element at j so increment i
i++;
}else if(a1[i]>a2[j]) {
// a2 element at i are smaller than a2 element at j so increment j
j++;
}else {
// intersection element store the value and increment i, j, k to find the next element
a3[k]=a1[i];
i++;
j++;
k++;
}
}
for(int l=0;l<a3.length;l++) {
System.out.println(a3[l]);
}

How to Find the Intersection of 3 unsorted arrays in Java:-
I have used the Core Java approach using for loops & using Arrays.copyOf to achieve this.
public class Intersection {
public void intersection3Arrays(int ar1[], int ar2[], int ar3[]) {
Arrays. sort(ar1);
Arrays. sort(ar2);
Arrays. sort(ar3);
int ar1Len = ar1.length;
int ar2Len = ar2.length;
int ar3Len = ar3.length;
int larArray = ar3Len > (ar1Len > ar2Len ? ar1Len : ar2Len) ? ar3Len : ((ar1Len > ar2Len) ? ar1Len : ar2Len);
System.out.println("The largest array is " +larArray);
int[] inputArray1 = Arrays.copyOf(ar1, larArray);
int[] inputArray2 = Arrays.copyOf(ar2, larArray);
int[] inputArray3 = Arrays.copyOf(ar3, larArray);
Integer[] inputArray11 = new Integer[inputArray1.length];
Integer[] inputArray22 = new Integer[inputArray2.length];
Integer[] inputArray33 = new Integer[inputArray3.length];
for (int i = 0; i < inputArray11.length; i++) {
if (inputArray11[i] == null){
inputArray1[i] = 0;
}
}
for (int i = 0; i < inputArray22.length; i++) {
if (inputArray22[i] == null){
inputArray1[i] = 0;
}
}
for (int i = 0; i < inputArray33.length; i++) {
if (inputArray33[i] == null){
inputArray1[i] = 0;
}
}
for (int i = 0; i < inputArray11.length; i++)
for (int j = 0; j < inputArray22.length; j++)
for (int k = 0; k < inputArray33.length; j++)
if (inputArray11[i] == inputArray22[j] && inputArray11[i] == inputArray33[k]) {
System.out.print(inputArray11[i]+" ");
}
}
public static void main(String[] args) {
Intersection3Arrays arrays = new Intersection3Arrays();
int ar1[] = { 1, 2, 5, 10, 20, 40, 80 };
int ar2[] = { 80, 100, 6, 2, 7, 20 };
int ar3[] = {3, 4, 15, 20, 30, 70, 80, 120};
arrays.intersection3Arrays(ar1, ar2, ar3);
}
}

If you ever wanted to implement this in python, this is one way that you can find intersection.
#find intersection
def find_intersec(list_a, list_b):
return set(list_a).intersection(list_b)
#since lists are kind of like arrays in python we use two lists
list_a = [ 4, 9, 1, 17, 11, 26, 28, 10,28, 26, 66, 91]
list_b = [9, 9, 74, 21, 45, 11, 63,10]
print(find_intersec(list_a, list_b))

I hope this example will simple one.pass two arrays and you will definitely get INTERSECTION of array without duplicate items.
private static int[] findInterserctorOfTwoArray(int[] array1, int[] array2) {
Map<Integer,Integer> map=new HashMap<>();
for (int element : array1) {
for (int element2 : array2) {
if(element==element2) {
map.put(element, element);
}
}
}
int[] newArray=new int[map.size()];
int con=0;
for(Map.Entry<Integer, Integer> lst:map.entrySet()) {
newArray[con]=lst.getValue();
con++;
}
return newArray;
}

optimised for sorted arrays using only one loop.
int a1[]=new int[] {1,2,3,5,7,8};
int a2[]=new int [] {1,5,6,7,8,9};
// sort both the array
Arrays.sort(a1);
Arrays.sort(a2);
// get the length of both the array
int n1=a1.length;
int n2=a2.length;
//create a new array to store the intersection
int a3[]=new int[n1];
//run the loop and find the intersection
int i=0,j=0,k=0;
while(i<n1&& j<n2) {
if(a1[i]<a2[j]) {
// a1 element at i are smaller than a2 element at j so increment i
i++;
}else if(a1[i]>a2[j]) {
// a2 element at i are smaller than a2 element at j so increment j
j++;
}else {
// intersection element store the value and increment i, j, k to find the next element
a3[k]=a1[i];
i++;
j++;
k++;
}
}
for(int l=0;l<a3.length;l++) {
System.out.println(a3[l]);
}

Primitive Iterator: 6 Times Faster than HashSet
Tested on sorted arrays of 10,000,000 random elements, values between 0 and 200,000,000. Tested on 10 processor i9 with 4GB heap space. Sort time for two arrays was 1.9 seconds.
results:
primitive() - 1.1 seconds
public static int[] primitive(int[] a1, int[] a2) {
List<Integer> list = new LinkedList<>();
OfInt it1 = Arrays.stream(a1).iterator();
OfInt it2 = Arrays.stream(a2).iterator();
int i1 = it1.next();
int i2 = it2.next();
do {
if (i1==i2) {
list.add(i1);
i1 = it1.next();
}
if (i1 < i2) i1 = it1.next();
if (i2 < i1) i2 = it2.next();
} while(it1.hasNext() && it2.hasNext());
if (i1==i2) list.add(i1);
return list.stream().mapToInt(Integer::intValue).toArray();
}
boxed() - 6.8 seconds
public static int[] boxed(int[] a1, int[] a2) {
return Arrays.stream(a1)
.filter(new HashSet<>(Arrays.stream(a2).boxed()
.collect(Collectors.toList()))::contains)
.toArray();
}

How to find the permutation of a sort in Java

I want to sort an array and find the index of each element in the sorted order.
So for instance if I run this on the array:
[3,2,4]
I'd get:
[1,0,2]
Is there an easy way to do this in Java?

Let's assume your elements are stored in an array.
final int[] arr = // elements you want
List<Integer> indices = new ArrayList<Integer>(arr.length);
for (int i = 0; i < arr.length; i++) {
indices.add(i);
}
Comparator<Integer> comparator = new Comparator<Integer>() {
public int compare(Integer i, Integer j) {
return Integer.compare(arr[i], arr[j]);
}
}
Collections.sort(indices, comparator);
Now indices contains the indices of the array, in their sorted order. You can convert that back to an int[] with a straightforward enough for loop.

import java.util.*;
public class Testing{
public static void main(String[] args){
int[] arr = {3, 2, 4, 6, 5};
TreeMap map = new TreeMap();
for(int i = 0; i < arr.length; i++){
map.put(arr[i], i);
}
System.out.println(Arrays.toString(map.values().toArray()));
}
}

One way to achieve this is to make a list of pairs with the starting index as the second part of the pair. Sort the list of pairs lexicographically, then read off the starting positions from the sorted array.
Starting array:
[3,2,4]
Add pairs with starting indexes:
[(3,0), (2,1), (4,2)]
Sort it lexicographically
[(2,1), (3,0), (4,2)]
then read off the second part of each pair
[1,0,2]

import java.io.*;
public class Sample {
public static void main(String[] args) {
int[] data = {0, 3, 2, 4, 6, 5, 10};//case:range 0 - 10
int i, rangeHigh = 10;
int [] rank = new int[rangeHigh + 1];
//counting sort
for(i=0; i< data.length ;++i) ++rank[data[i]];
for(i=1; i< rank.length;++i) rank[i] += rank[i-1];
for(i=0;i<data.length;++i)
System.out.print((rank[data[i]]-1) + " ");//0 2 1 3 5 4 6
}
}

As an update, this is relatively easy to do in Java 8 using the streams API.
public static int[] sortedPermutation(final int[] items) {
return IntStream.range(0, items.length)
.mapToObj(value -> Integer.valueOf(value))
.sorted((i1, i2) -> Integer.compare(items[i1], items[i2]))
.mapToInt(value -> value.intValue())
.toArray();
}
It somewhat unfortunately requires a boxing and unboxing step for the indices, as there is no .sorted(IntComparator) method on IntStream, or even an IntComparator functional interface for that matter.
To generalize to a List of Comparable objects is pretty straightforward:
public static <K extends Comparable <? super K>> int[] sortedPermutation(final List<K> items) {
return IntStream.range(0, items.size())
.mapToObj(value -> Integer.valueOf(value))
.sorted((i1, i2) -> items.get(i1).compareTo(items.get(i2)))
.mapToInt(value -> value.intValue())
.toArray();
}