First method:
public static List<List<Integer>> applicationPairs(int deviceCapacity, List<List<Integer>> foregroundAppList, List<List<Integer>> backgroundAppList) {
List<List<Integer>> result = new ArrayList<>();
int max = Integer.MIN_VALUE;
for (List<Integer> foregroundApp : foregroundAppList) {
for (List<Integer> backgroundApp : backgroundAppList) {
int memoryRequired = foregroundApp.get(1) + backgroundApp.get(1);
if (memoryRequired <= deviceCapacity) {
if (memoryRequired > max) {
result.clear();
max = memoryRequired;
result.add(Arrays.asList(foregroundApp.get(0), backgroundApp.get(0)));
} else if (memoryRequired == max) {
result.add(Arrays.asList(foregroundApp.get(0), backgroundApp.get(0)));
}
}
}
}
/*
return empty pair if no pair is found
*/
if (result.size() == 0) {
return new ArrayList<>(Collections.emptyList());
}
return result;
}
Second method:
public static List<String> sortOrders(List<String> orderList) {
// Write your code here
var result = orderList.stream()
.filter(e -> e.split(" ")[1].matches("[a-z]+"))
.sorted((s, t1) -> {
int r = s.substring(s.indexOf(" ")).compareTo(t1.substring(t1.indexOf(" ")));
if (r == 0) {
return s.substring(0, s.indexOf(" ")).compareTo(t1.substring(0, t1.indexOf(" ")));
}
return r;
})//.thenComparing(s -> s.substring(0, s.indexOf(" "))))
.collect(toList());
return Stream.concat(result.stream(), orderList.stream().filter(e -> !result.contains(e)))
.collect(toList());
}
I was asked to write these solutions as amazon's online assessment thing. Question also said to write the time complexity of the solutions, which I am not good at calculating.
Anyone?
In the first method if you see you are iterating over two lists, foregroundAppList, and backgroundAppList. As you are iterating each element backgroundAppList per element if foregroundAppList, it's clearly a time complexity of O(n*m) Where n is the size of backgroundAppList and m is the size of foregroundAppList.
In the second method, you have implemented a custom sort method and as you might be aware the Java sorted method uses Quicksort for list of primitives and merge sort for list of objects, where in both the cases the average time complexity fall under O(n*log(n)) where n is the size of the list.
public class MyArrayList<T> implements MyList<T>{
int num; //number of things in the list
T[] vals; //to store the contents
#SuppressWarnings("unchecked")
public MyArrayList() {
num = 0;
vals = (T[]) new Object[3];
}
public T getUnique(){
T distinct = null;
int count = 0;
for (int i=0; i<vals.length; i++){
distinct = vals[i];
for (int j = 0; j<vals.length; j++){
if (vals[j] == vals[i]){
count++;
}
if (count == 1){
return distinct;
}
}
}
if (distinct == null){
throw new IllegalArgumentException();
}
return distinct;
}
I am trying to work on a get Unique Method. A method getUnique that takes no arguments and returns the first value in the list that appears only once. (For example, calling the method on the list [1,2,3,1,2,4] would return 3 since 1 and
2 both appear more than once.) If the list is empty or all its values appear more than once, the method throws a NoSuchElementException
I have added some FIXME's to your code:
public T getUnique(){
T distinct = null;
int count = 0; // FIXME: move this initialization inside the i loop
for (int i=0; i<vals.length; i++){
distinct = vals[i];
for (int j = 0; j<vals.length; j++){
if (vals[j] == vals[i]){ // FIXME: use .equals() not ==
count++;
}
if (count == 1){ // FIXME: move this check outside the j loop
return distinct;
}
}
}
if (distinct == null){ //FIXME: no check needed, just throw it
throw new IllegalArgumentException();
}
return distinct; //FIXME: no valid return can reach this point
}
Patrick Parker's advice will fix your code, but I wanted to provide a cleaner and faster solution to the problem of finding a unique element in a list. This algorithm runs in time O(n) instead of O(n^2).
public static <T> Optional<T> getUnique(List<T> ls) {
// Create a map whose keys are elements of the list and whose values are
// lists of their occurences. E.g. [1,2,3,1,2,4] becomes {1->[1, 1],
// 2->[2, 2], 3->[3], 4->[4]}. Then elements.get(x).size() tells us how
// many times x occured in ls.
Map<T, List<T>> elements = ls.stream()
.collect(Collectors.groupingBy(x -> x));
// Find the first element that occurs exactly one time in ls.
return ls.stream().filter(x -> elements.get(x).size() == 1)
.findFirst();
}
You might call it like this:
Integer[] vals = {1,2,3,1,2,4};
System.out.println(getUnique(Arrays.asList(vals))
.orElseThrow(NoSuchElementException::new));
This code uses Java 8 streams and Optional. Below is another implementation of the same algorithm that doesn't use Java 8 language features; if you've never encountered streams, you may find it more understandable.
private static <T> T getUnique(List<T> arr) {
Map<T, Integer> numOccurrences = new HashMap<>();
for (T item : arr) {
numOccurrences.put(item, 1 + numOccurrences.getOrDefault(item, 0));
}
for (T item : arr) {
if (numOccurrences.get(item) == 1) {
return item;
}
}
throw new NoSuchElementException();
}
How do I pick a random element from a set?
I'm particularly interested in picking a random element from a
HashSet or a LinkedHashSet, in Java.
Solutions for other languages are also welcome.
int size = myHashSet.size();
int item = new Random().nextInt(size); // In real life, the Random object should be rather more shared than this
int i = 0;
for(Object obj : myhashSet)
{
if (i == item)
return obj;
i++;
}
A somewhat related Did You Know:
There are useful methods in java.util.Collections for shuffling whole collections: Collections.shuffle(List<?>) and Collections.shuffle(List<?> list, Random rnd).
In Java 8:
static <E> E getRandomSetElement(Set<E> set) {
return set.stream().skip(new Random().nextInt(set.size())).findFirst().orElse(null);
}
Fast solution for Java using an ArrayList and a HashMap: [element -> index].
Motivation: I needed a set of items with RandomAccess properties, especially to pick a random item from the set (see pollRandom method). Random navigation in a binary tree is not accurate: trees are not perfectly balanced, which would not lead to a uniform distribution.
public class RandomSet<E> extends AbstractSet<E> {
List<E> dta = new ArrayList<E>();
Map<E, Integer> idx = new HashMap<E, Integer>();
public RandomSet() {
}
public RandomSet(Collection<E> items) {
for (E item : items) {
idx.put(item, dta.size());
dta.add(item);
}
}
#Override
public boolean add(E item) {
if (idx.containsKey(item)) {
return false;
}
idx.put(item, dta.size());
dta.add(item);
return true;
}
/**
* Override element at position <code>id</code> with last element.
* #param id
*/
public E removeAt(int id) {
if (id >= dta.size()) {
return null;
}
E res = dta.get(id);
idx.remove(res);
E last = dta.remove(dta.size() - 1);
// skip filling the hole if last is removed
if (id < dta.size()) {
idx.put(last, id);
dta.set(id, last);
}
return res;
}
#Override
public boolean remove(Object item) {
#SuppressWarnings(value = "element-type-mismatch")
Integer id = idx.get(item);
if (id == null) {
return false;
}
removeAt(id);
return true;
}
public E get(int i) {
return dta.get(i);
}
public E pollRandom(Random rnd) {
if (dta.isEmpty()) {
return null;
}
int id = rnd.nextInt(dta.size());
return removeAt(id);
}
#Override
public int size() {
return dta.size();
}
#Override
public Iterator<E> iterator() {
return dta.iterator();
}
}
This is faster than the for-each loop in the accepted answer:
int index = rand.nextInt(set.size());
Iterator<Object> iter = set.iterator();
for (int i = 0; i < index; i++) {
iter.next();
}
return iter.next();
The for-each construct calls Iterator.hasNext() on every loop, but since index < set.size(), that check is unnecessary overhead. I saw a 10-20% boost in speed, but YMMV. (Also, this compiles without having to add an extra return statement.)
Note that this code (and most other answers) can be applied to any Collection, not just Set. In generic method form:
public static <E> E choice(Collection<? extends E> coll, Random rand) {
if (coll.size() == 0) {
return null; // or throw IAE, if you prefer
}
int index = rand.nextInt(coll.size());
if (coll instanceof List) { // optimization
return ((List<? extends E>) coll).get(index);
} else {
Iterator<? extends E> iter = coll.iterator();
for (int i = 0; i < index; i++) {
iter.next();
}
return iter.next();
}
}
If you want to do it in Java, you should consider copying the elements into some kind of random-access collection (such as an ArrayList). Because, unless your set is small, accessing the selected element will be expensive (O(n) instead of O(1)). [ed: list copy is also O(n)]
Alternatively, you could look for another Set implementation that more closely matches your requirements. The ListOrderedSet from Commons Collections looks promising.
In Java:
Set<Integer> set = new LinkedHashSet<Integer>(3);
set.add(1);
set.add(2);
set.add(3);
Random rand = new Random(System.currentTimeMillis());
int[] setArray = (int[]) set.toArray();
for (int i = 0; i < 10; ++i) {
System.out.println(setArray[rand.nextInt(set.size())]);
}
List asList = new ArrayList(mySet);
Collections.shuffle(asList);
return asList.get(0);
This is identical to accepted answer (Khoth), but with the unnecessary size and i variables removed.
int random = new Random().nextInt(myhashSet.size());
for(Object obj : myhashSet) {
if (random-- == 0) {
return obj;
}
}
Though doing away with the two aforementioned variables, the above solution still remains random because we are relying upon random (starting at a randomly selected index) to decrement itself toward 0 over each iteration.
Clojure solution:
(defn pick-random [set] (let [sq (seq set)] (nth sq (rand-int (count sq)))))
Java 8+ Stream:
static <E> Optional<E> getRandomElement(Collection<E> collection) {
return collection
.stream()
.skip(ThreadLocalRandom.current()
.nextInt(collection.size()))
.findAny();
}
Based on the answer of Joshua Bone but with slight changes:
Ignores the Streams element order for a slight performance increase in parallel operations
Uses the current thread's ThreadLocalRandom
Accepts any Collection type as input
Returns the provided Optional instead of null
Perl 5
#hash_keys = (keys %hash);
$rand = int(rand(#hash_keys));
print $hash{$hash_keys[$rand]};
Here is one way to do it.
C++. This should be reasonably quick, as it doesn't require iterating over the whole set, or sorting it. This should work out of the box with most modern compilers, assuming they support tr1. If not, you may need to use Boost.
The Boost docs are helpful here to explain this, even if you don't use Boost.
The trick is to make use of the fact that the data has been divided into buckets, and to quickly identify a randomly chosen bucket (with the appropriate probability).
//#include <boost/unordered_set.hpp>
//using namespace boost;
#include <tr1/unordered_set>
using namespace std::tr1;
#include <iostream>
#include <stdlib.h>
#include <assert.h>
using namespace std;
int main() {
unordered_set<int> u;
u.max_load_factor(40);
for (int i=0; i<40; i++) {
u.insert(i);
cout << ' ' << i;
}
cout << endl;
cout << "Number of buckets: " << u.bucket_count() << endl;
for(size_t b=0; b<u.bucket_count(); b++)
cout << "Bucket " << b << " has " << u.bucket_size(b) << " elements. " << endl;
for(size_t i=0; i<20; i++) {
size_t x = rand() % u.size();
cout << "we'll quickly get the " << x << "th item in the unordered set. ";
size_t b;
for(b=0; b<u.bucket_count(); b++) {
if(x < u.bucket_size(b)) {
break;
} else
x -= u.bucket_size(b);
}
cout << "it'll be in the " << b << "th bucket at offset " << x << ". ";
unordered_set<int>::const_local_iterator l = u.begin(b);
while(x>0) {
l++;
assert(l!=u.end(b));
x--;
}
cout << "random item is " << *l << ". ";
cout << endl;
}
}
Solution above speak in terms of latency but doesn't guarantee equal probability of each index being selected.
If that needs to be considered, try reservoir sampling. http://en.wikipedia.org/wiki/Reservoir_sampling. Collections.shuffle() (as suggested by few) uses one such algorithm.
Since you said "Solutions for other languages are also welcome", here's the version for Python:
>>> import random
>>> random.choice([1,2,3,4,5,6])
3
>>> random.choice([1,2,3,4,5,6])
4
Can't you just get the size/length of the set/array, generate a random number between 0 and the size/length, then call the element whose index matches that number? HashSet has a .size() method, I'm pretty sure.
In psuedocode -
function randFromSet(target){
var targetLength:uint = target.length()
var randomIndex:uint = random(0,targetLength);
return target[randomIndex];
}
PHP, assuming "set" is an array:
$foo = array("alpha", "bravo", "charlie");
$index = array_rand($foo);
$val = $foo[$index];
The Mersenne Twister functions are better but there's no MT equivalent of array_rand in PHP.
Icon has a set type and a random-element operator, unary "?", so the expression
? set( [1, 2, 3, 4, 5] )
will produce a random number between 1 and 5.
The random seed is initialized to 0 when a program is run, so to produce different results on each run use randomize()
In C#
Random random = new Random((int)DateTime.Now.Ticks);
OrderedDictionary od = new OrderedDictionary();
od.Add("abc", 1);
od.Add("def", 2);
od.Add("ghi", 3);
od.Add("jkl", 4);
int randomIndex = random.Next(od.Count);
Console.WriteLine(od[randomIndex]);
// Can access via index or key value:
Console.WriteLine(od[1]);
Console.WriteLine(od["def"]);
Javascript solution ;)
function choose (set) {
return set[Math.floor(Math.random() * set.length)];
}
var set = [1, 2, 3, 4], rand = choose (set);
Or alternatively:
Array.prototype.choose = function () {
return this[Math.floor(Math.random() * this.length)];
};
[1, 2, 3, 4].choose();
In lisp
(defun pick-random (set)
(nth (random (length set)) set))
How about just
public static <A> A getRandomElement(Collection<A> c, Random r) {
return new ArrayList<A>(c).get(r.nextInt(c.size()));
}
For fun I wrote a RandomHashSet based on rejection sampling. It's a bit hacky, since HashMap doesn't let us access it's table directly, but it should work just fine.
It doesn't use any extra memory, and lookup time is O(1) amortized. (Because java HashTable is dense).
class RandomHashSet<V> extends AbstractSet<V> {
private Map<Object,V> map = new HashMap<>();
public boolean add(V v) {
return map.put(new WrapKey<V>(v),v) == null;
}
#Override
public Iterator<V> iterator() {
return new Iterator<V>() {
RandKey key = new RandKey();
#Override public boolean hasNext() {
return true;
}
#Override public V next() {
while (true) {
key.next();
V v = map.get(key);
if (v != null)
return v;
}
}
#Override public void remove() {
throw new NotImplementedException();
}
};
}
#Override
public int size() {
return map.size();
}
static class WrapKey<V> {
private V v;
WrapKey(V v) {
this.v = v;
}
#Override public int hashCode() {
return v.hashCode();
}
#Override public boolean equals(Object o) {
if (o instanceof RandKey)
return true;
return v.equals(o);
}
}
static class RandKey {
private Random rand = new Random();
int key = rand.nextInt();
public void next() {
key = rand.nextInt();
}
#Override public int hashCode() {
return key;
}
#Override public boolean equals(Object o) {
return true;
}
}
}
The easiest with Java 8 is:
outbound.stream().skip(n % outbound.size()).findFirst().get()
where n is a random integer. Of course it is of less performance than that with the for(elem: Col)
With Guava we can do a little better than Khoth's answer:
public static E random(Set<E> set) {
int index = random.nextInt(set.size();
if (set instanceof ImmutableSet) {
// ImmutableSet.asList() is O(1), as is .get() on the returned list
return set.asList().get(index);
}
return Iterables.get(set, index);
}
In Mathematica:
a = {1, 2, 3, 4, 5}
a[[ ⌈ Length[a] Random[] ⌉ ]]
Or, in recent versions, simply:
RandomChoice[a]
Random[] generates a pseudorandom float between 0 and 1. This is multiplied by the length of the list and then the ceiling function is used to round up to the next integer. This index is then extracted from a.
Since hash table functionality is frequently done with rules in Mathematica, and rules are stored in lists, one might use:
a = {"Badger" -> 5, "Bird" -> 1, "Fox" -> 3, "Frog" -> 2, "Wolf" -> 4};
PHP, using MT:
$items_array = array("alpha", "bravo", "charlie");
$last_pos = count($items_array) - 1;
$random_pos = mt_rand(0, $last_pos);
$random_item = $items_array[$random_pos];
you can also transfer the set to array use array
it will probably work on small scale i see the for loop in the most voted answer is O(n) anyway
Object[] arr = set.toArray();
int v = (int) arr[rnd.nextInt(arr.length)];
If you really just want to pick "any" object from the Set, without any guarantees on the randomness, the easiest is taking the first returned by the iterator.
Set<Integer> s = ...
Iterator<Integer> it = s.iterator();
if(it.hasNext()){
Integer i = it.next();
// i is a "random" object from set
}
A generic solution using Khoth's answer as a starting point.
/**
* #param set a Set in which to look for a random element
* #param <T> generic type of the Set elements
* #return a random element in the Set or null if the set is empty
*/
public <T> T randomElement(Set<T> set) {
int size = set.size();
int item = random.nextInt(size);
int i = 0;
for (T obj : set) {
if (i == item) {
return obj;
}
i++;
}
return null;
}
I'm doing a program where user input five numbers and in the end the numbers are printed out which is working fine. What I can't get to work is a boolean function to check for duplicates. It should check for duplicates as the user write them in, so e.g. if number one is 5 and the second numbers is also 5, you should get an error until you write in a different number. Meaning if the user input a duplicate it should NOT be saved in the array. This is obviously an assignment, so I'm just asking for a hint or two.
This program is written based on pseudo-code given to me, and therefore I have to use a boolean to check for duplicates with the public boolean duplicate( int number ) class.
I've tried getting my head around it and tried something by myself, but obviously I'm doing a stupid mistake. E.g.:
if(int i != myNumbers[i])
checkDuplicates = false
else
checkDuplicates = true;
return checkDuplicates;
DuplicatesTest class:
public class DuplicatesTest {
public final static int AMOUNT = 5;
public static void main(String[] args) {
Duplicates d = new Duplicates(AMOUNT);
d.inputNumber();
d.duplicate(AMOUNT);
d.printInputNumbers();
}
}
Duplicates class:
public class Duplicates {
private int amount;
private int[] myNumbers;
private boolean checkDuplicates;
public Duplicates(int a) {
amount = a;
myNumbers = new int[amount];
}
public void inputNumber() {
for(int i = 0; i < amount; i++ ) {
int input = Integer.parseInt(JOptionPane.showInputDialog("Input 5 numbers"));
myNumbers[i] = input;
}
}
public boolean duplicate( int number ) {
<BOOLEAN TO CHECK FOR DUPLICATES, RETURN FALSE OR TRUE>
}
public void printInputNumbers() {
JTextArea output = new JTextArea();
output.setText("Your numbers are:" + "\n");
for(int i = 0; i < myNumbers.length; i++) {
if (i % 5 == 0) {
output.append("\n");
}
output.append(myNumbers[i] + "\t");
}
JOptionPane.showMessageDialog(null, output, "Numbers", JOptionPane.PLAIN_MESSAGE);
}
}
Sorry if the code tag is messy, I had some trouble with white fields in between and such. I'm new here.
Don't store the numbers in an array. Use a Set<Integer> instead. And then do a Set#contains() operation. It's O(1) operation which is actually far better than iterating over the array to search for duplicates.
Ok, if it's a compulsion to use an array, then you should modify your current approach, to return true as soon as you find a duplicate, instead of iterating over the array again. In your current approach, since you are setting the boolean variable to false in the else block, your method will return false if the last element of the array is not the same as what you are checking. So, just modify your approach to:
// loop over the array
if (number == myNumbers[i])
return true;
// outside the loop, if you reach, return false
return false;
Note that your current if statement will not compile. You are declaring an int variable there, which you can't do.
if (int i == myNumbers[i]) // this is not a valid Java code.
int nums[] = new int[5];
int count = 0;
public boolean duplicate(int number)
{
boolean isDup = false;
for (int i = 0; i <= count; i++)
{
if (number == nums[i])
{
isDup = true;
break;
}
}
if (!isDup)
{
count++;
nums[count] = number;
}
return isDup;
}
I am trying to have a method (duplicates) return true if a given array called x (entered by user in another method), contains duplicate values. Otherwise it would return false. Rather then checking the entire array, which is initialized to 100, it will check only the amount of values entered, which is kept track of with a global counter: numElementsInX.
What is the best way to accomplish this?
public static boolean duplicates (int [] x)
I am prompting for user data like so:
public static void readData (int [] x, int i){
Scanner input = new Scanner(System.in);
System.out.println("Please enter integers, enter -999 to stop");
while (i <= 99) {
int temp = input.nextInt();
if(temp == -999){
break;
}
else {
x[i++]=temp;
}
// else
}//end while
printArray(x,i);
}//end readData
public static void printArray(int [] x, int numElementsInX){
int n = numElementsInX;
for (int i = 0; i < n; i++){
System.out.print(x[i] + " ");
}//end for
System.out.println();
}//end printArray
I am sure there is a better way to do this, but this is how I have been taught so far.
Here is a solution that:
Compiles and executes without throwing.
Uses numElementsInX as you requested.
Returns as soon as it finds a duplicate.
This approach tests whether each member of the array has been seen before. If it has, the method can return immediately. If it hasn't, then the member is added to the set seen before.
public static boolean duplicates (int [] x, int numElementsInX ) {
Set<Integer> set = new HashSet<Integer>();
for ( int i = 0; i < numElementsInX; ++i ) {
if ( set.contains( x[i])) {
return true;
}
else {
set.add(x[i]);
}
}
return false;
}
Here's a sample program containing the above code.
this should do it.
public boolean containsDuplicates(Integer[] x) {
return new HashSet<Integer>(Arrays.asList(x)).size() != x.length
}
You dont need numElementsInX as this is the same as x.length
edit after comment from Louis. Arrays.asList does not work with int arrays.
To convert int[] to Integer try this question How to convert int[] to Integer[] in Java?
or do soemthing like this (not tested but from memory)
Integer[] newArray = new Integer[a.length];
System.arraycopy(a, 0, newArray, 0, a.length);
This certainly isn't the most efficient way, but since you don't know about Sets yet, you can use two loops:
public static boolean duplicates (int [] x){
for (int i=0; i<numElementsInX; i++){
for (int j=i+1; j<numElementsInX; j++){
if (x[j]==x[i]) return true;
}
}
return false;
}
"set.add()" returns true if the element is not already present in the set and false otherwise. We could make use of that and get rid of "set.contains()" as in the above solution.
public static boolean duplicates (int[] x, int numElementsInX) {
Set<Integer> myset = new HashSet<>();
for (int i = 0; i < numElementsInX; i++) {
if (!myset.add(x[i])) {
return true;
}
}
return false;
}
For java, return true if the array contains a duplicate value,
boolean containsDuplicates(int[] a) {
HashSet<Integer> hs = new HashSet<>();
for(int i = 0; i<a.length; i++) {
if(!hs.add(a[i])){
return true;
}
}
return false;
}