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How to duplicate every element of a linked list recursively in Java?

I have to write a duplicate function that duplicates every element of a linked list and returns a linked list such that if L = [2,3,4] then duplicate(L) = [2,2,3,3,4,4]. I have to do this recursively. I realize that below is not the correct solution, but I got confused. =(
public class MyList {
int value;
MyList next;
public static MyList duplicate(MyList L){
if(L.next == null){
L.next.value = L.value;
L.next.next = null;
} else {
MyList temp = L.next;
L.next.value = L.value;
L.next.next = temp;
duplicate(L.next);
}
return L;
}
}

First, check that L isn't an empty list (null). If it contains a value, return a new list that has that value repeated twice, followed by duplicating the rest of the list.
By giving MyList a constructor, this is more readable.
public class MyList {
int value;
MyList next;
public MyList(int value, MyList next) {
this.value = value;
this.next = next;
}
public static MyList duplicate(MyList list) {
if (list == null) {
return null;
} else {
return new MyList(list.value,
new MyList(list.value,
duplicate(list.next)));
}
}
}

You currently add an item and then call the recursion on it, ending at endlessly adding items.
You either need to at elements behind your recursion when processing in a forward-direction, or after the recursion when processing backwards.
Let's create a backwards-direction version. We first recursively walk to the end of the list and then resolve the recursion backwards, adding items after our current element each time.
public <E> void duplicateEntries(MyLinkedList<E> list) {
// Do nothing if list is empty
if (list.size() != 0) {
// Call the recursive method on the head node
duplicateEntriesHelper(list.head);
}
}
public <E> void duplicateEntriesHelper(Node<E> node) {
// Walk to the end of the list
if (node.next != null) {
duplicateEntriesHelper(node.next);
}
// Resolve recursion, duplicate current
// entry by inserting it after the current element
Node<E> duplicatedEntry = new Node<>();
duplicatedEntry.data = node.data;
// Insert element after current node
duplicatedEntry.next = node.next;
node.next = duplicatedEntry;
}
The classes I used should look similar to:
public class MyLinkedList<E> {
public Node<E> head = null;
#Override
public String toString() {
// Build something like "MyLinkedList[2, 3, 4]"
StringBuilder sb = new StringBuilder();
sb.append("MyLinkedList[");
StringJoiner sj = new StringJoiner(",");
Node<E> node = head;
while (node != null) {
sj.add(node);
node = node.next;
}
sb.append(sj);
sb.append("]");
return sb.toString();
}
}
public class Node<E> {
public Node next = null;
public E data = null;
#Override
public String toString() {
return E;
}
}
And here is the demo:
public static void main(String[] args) {
// Setup the list
MyLinkedList<Integer> list = new MyLinkedList<>();
Node<Integer> first = new Node<>();
first.data = 2;
Node<Integer> second = new Node<>();
second.data = 3;
Node<Integer> third = new Node<>();
third.data = 4;
list.head = data;
first.next = second;
second.next = third;
// Demonstrate the method
System.out.println("Before: " + list);
duplicateEntries(list);
System.out.println("After: " + list);
}
Of course you can add additional methods and functionality to them. For example using some constructors or getter/setter methods.

Add Doubly Linked List Node in Singly Linked List

Given singly Linked List: 1 -> 2 -> 3 -> 4 -> 5 -> null
Modify middle element as doubly Linked List Node
here middle element is 3
3 -> next should point to 4
3 -> prev should point to 1
Can any one suggest how can it be done ? interviewer gave me hint use interface. but I couldn't figure it out how.
I have to iterate over this linked list and print all the node and when it reaches to the middle, print where next and prev is pointing to, then print till the end of the list.
Expected output : 1, 2, Middle: 3, Prev: 1, Next: 4, 5
I'm facing problem in adding the middle node.

So, this "works", but if this is expected to be answered on an interview, it is way too much work.
LinkedList
public class LinkedList {
public interface Linkable<V, L extends Linkable> {
V getValue();
L getNext();
void setNext(L link);
}
public static class Node implements Linkable<Integer, Linkable> {
int value;
Linkable next;
Node(int value) {
this.value = value;
}
#Override
public Integer getValue() {
return value;
}
#Override
public Linkable getNext() {
return next;
}
#Override
public void setNext(Linkable link) {
this.next = link;
}
}
private Linkable head;
public boolean isEmpty() {
return this.head == null;
}
public Linkable getHead() {
return head;
}
public void add(int v) {
Node next = new Node(v);
if (isEmpty()) {
this.head = next;
} else {
Linkable tmp = this.head;
while (tmp.getNext() != null) {
tmp = tmp.getNext();
}
tmp.setNext(next);
}
}
}
Interface
interface DoublyLinkable<V, L extends LinkedList.Linkable> extends LinkedList.Linkable<V,L> {
LinkedList.Linkable getPrev();
void setPrev(LinkedList.Linkable prev);
}
DoubleNode
public class DoubleNode extends LinkedList.Node implements DoublyLinkable<Integer, LinkedList.Linkable> {
LinkedList.Linkable prev;
public DoubleNode(int value) {
super(value);
}
#Override
public LinkedList.Linkable getPrev() {
return prev;
}
#Override
public void setPrev(LinkedList.Linkable prev) {
this.prev = prev;
}
}
Driver
Outputs
1, 2, Middle: 3, Prev: 1, Next: 4, 5
public class Driver {
public static LinkedList getList() {
LinkedList list = new LinkedList();
for (int i = 1; i <= 5; i++) {
list.add(i);
}
return list;
}
public static void main(String[] args) {
LinkedList list = getList();
LinkedList.Linkable head = list.getHead();
LinkedList.Linkable beforeMiddle = null;
LinkedList.Linkable middle = list.getHead();
LinkedList.Linkable end = list.getHead();
if (head != null) {
// find the middle of the list
while (true) {
if (end.getNext() == null || end.getNext().getNext() == null) break;
beforeMiddle = middle;
middle = middle.getNext();
end = end.getNext().getNext();
}
// Replace middle by reassigning the pointer to it
if (beforeMiddle != null) {
DoubleNode n = new DoubleNode((int) middle.getValue()); // same value
n.setPrev(list.getHead()); // point back to the front
n.setNext(middle.getNext()); // point forward to original value
beforeMiddle.setNext((DoublyLinkable) n);
middle = beforeMiddle.getNext();
}
// Build the "expected" output
StringBuilder sb = new StringBuilder();
final String DELIMITER = ", ";
head = list.getHead();
boolean atMiddle = false;
if (head != null) {
do {
if (head instanceof DoublyLinkable) {
atMiddle = true;
String out = String.format("Middle: %d, Prev: %d, ", (int) head.getValue(), (int) ((DoublyLinkable) head).getPrev().getValue());
sb.append(out);
} else {
if (atMiddle) {
sb.append("Next: ");
atMiddle = false;
}
sb.append(head.getValue()).append(DELIMITER);
}
head = head.getNext();
} while (head != null);
}
sb.setLength(sb.length() - DELIMITER.length());
System.out.println(sb.toString());
}
}
}

By definition, a single-linked list consists of single-linked nodes only, and a double-linked consists of double-linked nodes only. Otherwise. it is neither.
By definition the field prev of a double-linked list must point to the previous element.
Whatever you are supposed to build. It's something not well specified. So if you really were asked this in an interview (and did not misunderstand the question - maybe he wanted you to point out that ghis violates the interface?) this is a case for the code horror stories of http://thedailywtf.com/ - section "incompetent interviewers".

If you haven't, you'd better define a lenght() function so given one linked list you can know how many nodes does it have.
Thanks to the response of Cereal_Killer to the previous version of this answer, I noticed that the list is firstly a singly linked list, and you just have to make the middle node be linked both to the next node and to some previous node.
Now I guess that you have defined two structures (Struct, Class or whatever depending on the language you're using). So lets say you have Node_s defined as a node with only a next pointer, and Node_d with both a next and a prev pointer. (Node_d may inherite from Node_s so you just have to add the prev attribute in the child class). Knowing this, the code above should be doing what you need:
function do_it(my_LinkedList linkedList){
int i_middle;
int length = linkedList.length();
if ( (length ÷ 2 ) != 0 ) i_middle = length / 2;
else return -1;
Node_s aux = linkedList.first();
int index = 0;
Node_d middle= null;
while (aux != null) {
if (index == i_middle - 1){ //now aux is the middle's previous node
middle.data = aux.next.data; //aux.next is the middle singly node, we assignate the data to the new double linked node
middle.prev = aux; //as we said, aux is the prev to the middle
midle.next = aux.next.next; //so aux.next.next is the next to the middle
print(what you need to print);
}else {
print("Node " + index " next: "+ aux.next);
}//end if
index++;
aux = aux.next;
} //end while
}//end function
This previous code should be doing what you need. I wrote the answer in some kind of pseudo-java code so if you're not familiar with Java or don't understand what my pseudo-code does, please let me know. Anyway, the idea of my code may present some troubles depending on the language you're working with, so you'll have to adapt it.
Note that at the end of the execution of this program, your data structure won't be a singly linked list, and neither a double one, since you'll have linkedList.length() - 1 nodes linked in a signly way but the middle one will have two links.
Hope this helps.

Java Using Nodes with LinkedList

I've been working through some standard coding interview questions from a book I recently bought, and I came across the following question and answer:
Implement an algorithm to find the nth to last element in a linked list.
Here's the provided answer:
public static LinkedListNode findNtoLast(LinkedListNode head, int n) { //changing LinkedListNode to ListNode<String>
if(head == null || n < 1) {
return null;
}
LinkedListNode p1 = head;
LinkedListNode p2 = head;
for(int j = 0; j < n-1; ++j) {
if(p2 == null) {
return null;
}
p2 = p2.next;
}
if(p2 == null) {
return null;
}
while(p2.next != null) {
p1 = p1.next;
p2 = p2.next;
}
return p1;
}
I understand the algorithm, how it works, and why the book lists this as its answer, but I'm confused about how to access the LinkedListNodes to send as an argument to the method. I know that I'd have to create a LinkedListNode class (since Java doesn't already have one), but I can't seem to figure out how to do that. It's frustrating because I feel like I should know how to do this. Here's something that I've been working on. I'd greatly appreciate any clarification. You can expand/comment on my code or offer your own alternatives. Thanks.
class ListNode<E> {
ListNode<E> next;
E data;
public ListNode(E value) {
data = value;
next = null;
}
public ListNode(E value, ListNode<E> n) {
data = value;
next = n;
}
public void setNext(ListNode<E> n) {
next = n;
}
}
public class MyLinkedList<E> extends LinkedList {
LinkedList<ListNode<E>> list;
ListNode<E> head;
ListNode<E> tail;
ListNode<E> current;
ListNode<E> prev;
public MyLinkedList() {
list = null;
head = null;
tail = null;
current = null;
prev = null;
}
public MyLinkedList(LinkedList<E> paramList) {
list = (LinkedList<ListNode<E>>) paramList; //or maybe create a loop assigning each ListNode a value and next ptr
head = list.getFirst();
tail = list.getLast(); //will need to update tail every time add new node
current = null;
prev = null;
}
public void addNode(E value) {
super.add(value);
//ListNode<E> temp = tail;
current = new ListNode<E>(value);
tail.setNext(current);
tail = current;
}
public LinkedList<ListNode<E>> getList() {
return list;
}
public ListNode<E> getHead() {
return head;
}
public ListNode<E> getTail() {
return tail;
}
public ListNode<E> getCurrent() {
return current;
}
public ListNode<E> getPrev() {
return prev;
}
}
How can the LinkedListNode head from a LinkedList?
Update: I think part of my confusion comes from what to put in the main method. Do I need to create a LinkedList of ListNode? If I do that, how would I connect the ListNodes to each other? How would I connect them without using a LinkedList collection object? If someone could show me how they would code the main method, I think that would put things into enough perspective for me to solve my issues. Here's my latest attempt at the main method:
public static void main(String args[]) {
LinkedList<ListNode<String>> list = new LinkedList<ListNode<String>>();
//MyLinkedList<ListNode<String>> list = new MyLinkedList(linkedList);
list.add(new ListNode<String>("Jeff"));
list.add(new ListNode<String>("Brian"));
list.add(new ListNode<String>("Negin"));
list.add(new ListNode<String>("Alex"));
list.add(new ListNode<String>("Alaina"));
int n = 3;
//ListIterator<String> itr1 = list.listIterator();
//ListIterator<String> itr2 = list.listIterator();
LinkedListNode<String> head = new LinkedListNode(list.getFirst(), null);
//String result = findNtoLast(itr1, itr2, n);
//System.out.println("The " + n + "th to the last value: " + result);
//LinkedListNode<String> nth = findNtoLast(list.getFirst(), n);
ListNode<String> nth = findNtoLast(list.getFirst(), n);
System.out.println("The " + n + "th to the last value: " + nth);
}
In an attempt to connect the nodes without using a custom linked list class, I have edited my ListNode class to the following:
class ListNode<E> {
ListNode<E> next;
ListNode<E> prev; //only used for linking nodes in singly linked list
ListNode<E> current; //also only used for linking nodes in singly linked list
E data;
private static int size = 0;
public ListNode() {
data = null;
next = null;
current = null;
if(size > 0) { //changed from prev != null because no code to make prev not null
prev.setNext(this);
}
size++;
}
public ListNode(E value) {
data = value;
next = null;
current = this;
System.out.println("current is " + current);
if(size > 0) {
prev.setNext(current);//this line causing npe
}
else
{
prev = current;
System.out.println("prev now set to " + prev);
}
size++;
System.out.println("after constructor, size is " + size);
}
public ListNode(E value, ListNode<E> n) {
data = value;
next = n;
current = this;
if(size > 0) {
prev.setNext(this);
}
size++;
}
public void setNext(ListNode<E> n) {
next = n;
}
}
As is right now, the program will run until it reaches prev.setNext(current); in the single argument constructor for ListNode. Neither current nor prev are null at the time this line is reached. Any advice would be greatly appreciated. Thanks.

You don't actually need a separate LinkedList class; the ListNode class is a linked list. Or, to state it differently, a reference to the head of the list is a reference to the list.
The use of head, tail, current, prev in the sample code you posted has come from a double-linked list which is a data type that has links in both directions. This is more efficient for certain types of applications (such as finding the nth last item).
So I would recommend renaming your ListNode class to LinkedList and renaming next to tail.
To add a new item to the list you need a method that creates a new list with the new item at it's head. Here is an example:
class LinkedList<E> {
...
private LinkedList(E value, LinkedList<E> tail) {
this.data = value;
this.tail = tail;
}
public LinkedList<E> prependItem(E item) {
return new LinkedList(item, this);
}
}
Then to add a new item i to list you use list = list.prependItem(i);
If for some reason you need to always add the items to the end, then:
private LinkedList(E value) {
this.data = value;
this.tail = null;
}
public void appendItem(E item) {
LinkedList<E> list = this;
while (list.tail != null)
list = list.tail;
list.tail = new LinkedList<>(item);
}
However this is obviously pretty inefficient for long lists. If you need to do this then either use a different data structure or just reverse the list when you have finished adding to it.
Incidentally, an interesting side effect of this is that a reference to any item in the list is a reference to a linked list. This makes recursion very easy. For example, here's a recursive solution for finding the length of a list:
public int getLength(LinkedList list) {
if (list == null) {
return 0;
} else {
return 1 + getLength(list.getTail());
}
}
And using this a simple (but very inefficient!) solution to the problem you provided - I've renamed the method to make its function more obvious:
public LinkedList getTailOfListOfLengthN(LinkedList list, int n) {
int length = getLength(list);
if (length < n) {
return null;
} else if (length == n) {
return list;
} else {
return getTailOfLengthN(list.getTail(), n);
}
}
And to reverse the list:
public LinkedList<E> reverse() {
if (tail == null) {
return this;
} else {
LinkedList<E> list = reverse(tail);
tail.tail = this;
tail = null;
return list;
}
}
As I hope you can see this makes the methods a lot more elegant than separating the node list classes.

Actually you have created a linked list with you class ListNode.
A linked list is made of a node and a reference to another linked list (see the recursion?).

How to implement a Least Frequently Used (LFU) cache?

Least Frequently Used (LFU) is a type of cache algorithm used to manage memory within a computer. The standard characteristics of this method involve the system keeping track of the number of times a block is referenced in memory. When the cache is full and requires more room the system will purge the item with the lowest reference frequency.
What would be the best way to implement a most-recently-used cache of objects, say in Java?
I've already implemented one using LinkedHashMap(by maintaining the no. of times objects are accessed) But I'm curious if any of the new concurrent collections would be better candidates.
Consider this case : Suppose cache is full and we need to make space for another one. Say two objects are noted in cache which are accessed for one time only. Which one to remove if we come to know that other(which is not in cache)object is being accessed for more than once ?
Thanks!

You might benefit from the LFU implementation of ActiveMQ: LFUCache
They have provided some good functionality.

I think, the LFU data structure must combine priority queue (for maintaining fast access to lfu item) and hash map (for providing fast access to any item by its key); I would suggest the following node definition for each object stored in cache:
class Node<T> {
// access key
private int key;
// counter of accesses
private int numAccesses;
// current position in pq
private int currentPos;
// item itself
private T item;
//getters, setters, constructors go here
}
You need key for referring to an item.
You need numAccesses as a key for priority queue.
You need currentPos to be able to quickly find a pq position of item by key.
Now you organize hash map (key(Integer) -> node(Node<T>)) to quickly access items and min heap-based priority queue using number of accesses as priority. Now you can very quickly perform all operations (access, add new item, update number of acceses, remove lfu). You need to write each operation carefully, so that it maintains all the nodes consistent (their number of accesses, their position in pq and there existence in hash map). All operations will work with constant average time complexity which is what you expect from cache.

According to me, the best way to implement a most-recently-used cache of objects would be to include a new variable as 'latestTS' for each object. TS stands for timestamp.
// A static method that returns the current date and time as milliseconds since January 1st 1970
long latestTS = System.currentTimeMillis();
ConcurrentLinkedHashMap is not yet implemented in Concurrent Java Collections.
(Ref: Java Concurrent Collection API). However, you can try and use ConcurrentHashMap and DoublyLinkedList
About the case to be considered: in such case, as I have said that you can declare latestTS variable, based upon the value of latestTS variable, you can remove an entry and add the new object. (Don't forget to update frequency and latestTS of the new object added)
As you have mentioned, you can use LinkedHashMap as it gives element access in O(1) and also, you get the order traversal.
Please, find the below code for LFU Cache:
(PS: The below code is the answer for the question in the title i.e. "How to implement LFU cache")
import java.util.LinkedHashMap;
import java.util.Map;
public class LFUCache {
class CacheEntry
{
private String data;
private int frequency;
// default constructor
private CacheEntry()
{}
public String getData() {
return data;
}
public void setData(String data) {
this.data = data;
}
public int getFrequency() {
return frequency;
}
public void setFrequency(int frequency) {
this.frequency = frequency;
}
}
private static int initialCapacity = 10;
private static LinkedHashMap<Integer, CacheEntry> cacheMap = new LinkedHashMap<Integer, CacheEntry>();
/* LinkedHashMap is used because it has features of both HashMap and LinkedList.
* Thus, we can get an entry in O(1) and also, we can iterate over it easily.
* */
public LFUCache(int initialCapacity)
{
this.initialCapacity = initialCapacity;
}
public void addCacheEntry(int key, String data)
{
if(!isFull())
{
CacheEntry temp = new CacheEntry();
temp.setData(data);
temp.setFrequency(0);
cacheMap.put(key, temp);
}
else
{
int entryKeyToBeRemoved = getLFUKey();
cacheMap.remove(entryKeyToBeRemoved);
CacheEntry temp = new CacheEntry();
temp.setData(data);
temp.setFrequency(0);
cacheMap.put(key, temp);
}
}
public int getLFUKey()
{
int key = 0;
int minFreq = Integer.MAX_VALUE;
for(Map.Entry<Integer, CacheEntry> entry : cacheMap.entrySet())
{
if(minFreq > entry.getValue().frequency)
{
key = entry.getKey();
minFreq = entry.getValue().frequency;
}
}
return key;
}
public String getCacheEntry(int key)
{
if(cacheMap.containsKey(key)) // cache hit
{
CacheEntry temp = cacheMap.get(key);
temp.frequency++;
cacheMap.put(key, temp);
return temp.data;
}
return null; // cache miss
}
public static boolean isFull()
{
if(cacheMap.size() == initialCapacity)
return true;
return false;
}
}

Here's the o(1) implementation for LFU - http://dhruvbird.com/lfu.pdf

I have tried to implement this below LFU cache implementation. Took reference from this -
LFU paper. My implementation is working nicely.
If anyone wants to provide any further suggestion to improve it again, please let me know.
import java.util.HashMap;
import java.util.Map;
import java.util.Objects;
import java.util.TreeMap;
public class LFUCacheImplementation {
private Map<Integer, Node> cache = new HashMap<>();
private Map<Integer, Integer> counts = new HashMap<>();
private TreeMap<Integer, DoublyLinkedList> frequencies = new TreeMap<>();
private final int CAPACITY;
public LFUCache(int capacity) {
this.CAPACITY = capacity;
}
public int get(int key) {
if (!cache.containsKey(key)) {
return -1;
}
Node node = cache.get(key);
int frequency = counts.get(key);
frequencies.get(frequency).remove(new Node(node.key(), node.value()));
removeFreq(frequency);
frequencies.computeIfAbsent(frequency + 1, k -> new DoublyLinkedList()).add(new Node(node.key(), node.value()));
counts.put(key, frequency + 1);
return cache.get(key).value();
}
public void set(int key, int value) {
if (!cache.containsKey(key)) {
Node node = new Node(key, value);
if (cache.size() == CAPACITY) {
int l_count = frequencies.firstKey();
Node deleteThisNode = frequencies.get(l_count).head();
frequencies.get(l_count).remove(deleteThisNode);
int deleteThisKey = deleteThisNode.key();
removeFreq(l_count);
cache.remove(deleteThisKey);
counts.remove(deleteThisKey);
}
cache.put(key, node);
counts.put(key, 1);
frequencies.computeIfAbsent(1, k -> new DoublyLinkedList()).add(node);
}
}
private void removeFreq(int frequency) {
if (frequencies.get(frequency).size() == 0) {
frequencies.remove(frequency);
}
}
public Map<Integer, Node> getCache() {
return cache;
}
public Map<Integer, Integer> getCounts() {
return counts;
}
public TreeMap<Integer, DoublyLinkedList> getFrequencies() {
return frequencies;
}
}
class Node {
private int key;
private int value;
private Node next;
private Node prev;
public Node(int key, int value) {
this.key = key;
this.value = value;
}
public Node getNext() {
return next;
}
public void setNext(Node next) {
this.next = next;
}
public Node getPrev() {
return prev;
}
public void setPrev(Node prev) {
this.prev = prev;
}
public int key() {
return key;
}
public int value() {
return value;
}
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof Node)) return false;
Node node = (Node) o;
return key == node.key &&
value == node.value;
}
#Override
public int hashCode() {
return Objects.hash(key, value);
}
#Override
public String toString() {
return "Node{" +
"key=" + key +
", value=" + value +
'}';
}
}
class DoublyLinkedList {
private int size;
private Node head;
private Node tail;
public void add(Node node) {
if (null == head) {
head = node;
} else {
tail.setNext(node);
node.setPrev(tail);
}
tail = node;
size++;
}
public void remove(Node node) {
if(null == head || null == node) {
return;
}
if(this.size() == 1 && head.equals(node)) {
head = null;
tail = null;
} else if (head.equals(node)) {
head = node.getNext();
head.setPrev(null);
} else if (tail.equals(node)) {
Node prevToTail = tail.getPrev();
prevToTail.setNext(null);
tail = prevToTail;
} else {
Node current = head.getNext();
while(!current.equals(tail)) {
if(current.equals(node)) {
Node prevToCurrent = current.getPrev();
Node nextToCurrent = current.getNext();
prevToCurrent.setNext(nextToCurrent);
nextToCurrent.setPrev(prevToCurrent);
break;
}
current = current.getNext();
}
}
size--;
}
public Node head() {
return head;
}
public int size() {
return size;
}
}
Client code to use the above cache implementation -
import java.util.Map;
public class Client {
public static void main(String[] args) {
Client client = new Client();
LFUCache cache = new LFUCache(4);
cache.set(11, function(11));
cache.set(12, function(12));
cache.set(13, function(13));
cache.set(14, function(14));
cache.set(15, function(15));
client.print(cache.getFrequencies());
cache.get(13);
cache.get(13);
cache.get(13);
cache.get(14);
cache.get(14);
cache.get(14);
cache.get(14);
client.print(cache.getCache());
client.print(cache.getCounts());
client.print(cache.getFrequencies());
}
public void print(Map<Integer, ? extends Object> map) {
for(Map.Entry<Integer, ? extends Object> entry : map.entrySet()) {
if(entry.getValue() instanceof Node) {
System.out.println("Cache Key => "+entry.getKey()+", Cache Value => "+((Node) entry.getValue()).toString());
} else if (entry.getValue() instanceof DoublyLinkedList) {
System.out.println("Frequency Key => "+entry.getKey()+" Frequency Values => [");
Node head = ((DoublyLinkedList) entry.getValue()).head();
while(null != head) {
System.out.println(head.toString());
head = head.getNext();
}
System.out.println(" ]");
} else {
System.out.println("Count Key => "+entry.getKey()+", Count Value => "+entry.getValue());
}
}
}
public static int function(int key) {
int prime = 31;
return key*prime;
}
}

How about a priority queue? You can keep elements sorted there with keys representing the frequency. Just update the object position in the queue after visiting it. You can update just from time to time for optimizing the performance (but reducing precision).

Many implementations I have seen have runtime complexity O(log(n)). This means, when the cache size is n, the time needed to insert/remove an element into/from chache is logarithmic. Such implementations use usually a min heap to maintain usage frequencies of elements. The root of the heap contains the element with lowest frequency, and can be accessed in O(1) time. But to maintain the heap property we have to move an element, every time it is used (and frequency is incremented) inside of the heap, to place it into proper position, or when we have to insert new element into the cache (and so put it into the heap).
But the runtime complexity can be reduced to O(1), when we maintain a hashmap (Java) or unordered_map (C++) with the element as key. Additinally we need two sorts of lists, frequency list and elements lists. The elements lists contain elements that have same frequency, and the frequency list contain the element lists.
frequency list
1 3 6 7
a k y x
c l z
m n
Here in the example we see the frequency list that has 4 elements (4 elements lists). The element list 1 contains elements (a,c,m), the elements list 3 contains elements (k, l, n) etc.
Now, when we use say element y, we have to increment its frequency and put it in the next list. Because the elements list with frequency 6 becomes empty, we delete it. The result is:
frequency list
1 3 7
a k y
c l x
m n z
We place the element y in the begin of the elements list 7. When we have to remove elements from the list later, we will start from the end (first z, then x and then y).
Now, when we use element n, we have to increment its frequency and put it into the new list, with frequencies 4:
frequency list
1 3 4 7
a k n y
c l x
m z
I hope the idea is clear. I provide now my C++ implementation of the LFU cache, and will add later a Java implementation.
The class has just 2 public methods, void set(key k, value v)
and bool get(key k, value &v). In the get method the value to retrieve will be set per reference when the element is found, in this case the method returns true. When the element is not found, the method returns false.
#include<unordered_map>
#include<list>
using namespace std;
typedef unsigned uint;
template<typename K, typename V = K>
struct Entry
{
K key;
V value;
};
template<typename K, typename V = K>
class LFUCache
{
typedef typename list<typename Entry<K, V>> ElementList;
typedef typename list <pair <uint, ElementList>> FrequencyList;
private:
unordered_map <K, pair<typename FrequencyList::iterator, typename ElementList::iterator>> cacheMap;
FrequencyList elements;
uint maxSize;
uint curSize;
void incrementFrequency(pair<typename FrequencyList::iterator, typename ElementList::iterator> p) {
if (p.first == prev(elements.end())) {
//frequency list contains single list with some frequency, create new list with incremented frequency (p.first->first + 1)
elements.push_back({ p.first->first + 1, { {p.second->key, p.second->value} } });
// erase and insert the key with new iterator pair
cacheMap[p.second->key] = { prev(elements.end()), prev(elements.end())->second.begin() };
}
else {
// there exist element(s) with higher frequency
auto pos = next(p.first);
if (p.first->first + 1 == pos->first)
// same frequency in the next list, add the element in the begin
pos->second.push_front({ p.second->key, p.second->value });
else
// insert new list before next list
pos = elements.insert(pos, { p.first->first + 1 , {{p.second->key, p.second->value}} });
// update cachMap iterators
cacheMap[p.second->key] = { pos, pos->second.begin() };
}
// if element list with old frequency contained this singe element, erase the list from frequency list
if (p.first->second.size() == 1)
elements.erase(p.first);
else
// erase only the element with updated frequency from the old list
p.first->second.erase(p.second);
}
void eraseOldElement() {
if (elements.size() > 0) {
auto key = prev(elements.begin()->second.end())->key;
if (elements.begin()->second.size() < 2)
elements.erase(elements.begin());
else
elements.begin()->second.erase(prev(elements.begin()->second.end()));
cacheMap.erase(key);
curSize--;
}
}
public:
LFUCache(uint size) {
if (size > 0)
maxSize = size;
else
maxSize = 10;
curSize = 0;
}
void set(K key, V value) {
auto entry = cacheMap.find(key);
if (entry == cacheMap.end()) {
if (curSize == maxSize)
eraseOldElement();
if (elements.begin() == elements.end()) {
elements.push_front({ 1, { {key, value} } });
}
else if (elements.begin()->first == 1) {
elements.begin()->second.push_front({ key,value });
}
else {
elements.push_front({ 1, { {key, value} } });
}
cacheMap.insert({ key, {elements.begin(), elements.begin()->second.begin()} });
curSize++;
}
else {
entry->second.second->value = value;
incrementFrequency(entry->second);
}
}
bool get(K key, V &value) {
auto entry = cacheMap.find(key);
if (entry == cacheMap.end())
return false;
value = entry->second.second->value;
incrementFrequency(entry->second);
return true;
}
};
Here are examples of usage:
int main()
{
LFUCache<int>cache(3); // cache of size 3
cache.set(1, 1);
cache.set(2, 2);
cache.set(3, 3);
cache.set(2, 4);
rc = cache.get(1, r);
assert(rc);
assert(r == 1);
// evict old element, in this case 3
cache.set(4, 5);
rc = cache.get(3, r);
assert(!rc);
rc = cache.get(4, r);
assert(rc);
assert(r == 5);
LFUCache<int, string>cache2(2);
cache2.set(1, "one");
cache2.set(2, "two");
string val;
rc = cache2.get(1, val);
if (rc)
assert(val == "one");
else
assert(false);
cache2.set(3, "three"); // evict 2
rc = cache2.get(2, val);
assert(rc == false);
rc = cache2.get(3, val);
assert(rc);
assert(val == "three");
}

Here is a simple implementation of LFU cache in Go/Golang based on here.
import "container/list"
type LFU struct {
cache map[int]*list.Element
freqQueue map[int]*list.List
cap int
maxFreq int
lowestFreq int
}
type entry struct {
key, val int
freq int
}
func NewLFU(capacity int) *LFU {
return &LFU{
cache: make(map[int]*list.Element),
freqQueue: make(map[int]*list.List),
cap: capacity,
maxFreq: capacity - 1,
lowestFreq: 0,
}
}
// O(1)
func (c *LFU) Get(key int) int {
if e, ok := c.cache[key]; ok {
val := e.Value.(*entry).val
c.updateEntry(e, val)
return val
}
return -1
}
// O(1)
func (c *LFU) Put(key int, value int) {
if e, ok := c.cache[key]; ok {
c.updateEntry(e, value)
} else {
if len(c.cache) == c.cap {
c.evict()
}
if c.freqQueue[0] == nil {
c.freqQueue[0] = list.New()
}
e := c.freqQueue[0].PushFront(&entry{key, value, 0})
c.cache[key] = e
c.lowestFreq = 0
}
}
func (c *LFU) updateEntry(e *list.Element, val int) {
key := e.Value.(*entry).key
curFreq := e.Value.(*entry).freq
c.freqQueue[curFreq].Remove(e)
delete(c.cache, key)
nextFreq := curFreq + 1
if nextFreq > c.maxFreq {
nextFreq = c.maxFreq
}
if c.lowestFreq == curFreq && c.freqQueue[curFreq].Len() == 0 {
c.lowestFreq = nextFreq
}
if c.freqQueue[nextFreq] == nil {
c.freqQueue[nextFreq] = list.New()
}
newE := c.freqQueue[nextFreq].PushFront(&entry{key, val, nextFreq})
c.cache[key] = newE
}
func (c *LFU) evict() {
back := c.freqQueue[c.lowestFreq].Back()
delete(c.cache, back.Value.(*entry).key)
c.freqQueue[c.lowestFreq].Remove(back)
}

Merge Sort with Strings

I have a linked list in a class named WordNode, which holds the following attributes:
String _word; WordNode _next;
I have another class which is the "actual list" which only holds a reference to the head of the list, the class is called TextList and it receives a String and supposed to put every word of the String SORTED IN THE LIST. For example, for the sentence:
coding in Java is cool.
The linked list would look like:
coding >>> cool >>> Java >>> in >>> is.
The arrows are like pointers to the next node in the list.
I want to first take all the words and put them in a linked list (TextList class) and then make a MERGE SORT to sort the words in the linked list.
What I though of doing is taking a split method to split the list into two lists: "odd" and "evens", which is these methods:
private TextList splitOdds(){
boolean flag=true;
TextList odds=new TextList();
WordNode o=null;
WordNode ptr=_head;
while (ptr.getNext()!=null){
if(flag)
o=new WordNode(ptr.getWord(),o);
ptr=ptr.getNext();
flag=!flag;
}
odds._head=o;;
return odds;
}
private TextList splitEvens(){
boolean flag=true;
TextList evens=new TextList();
WordNode e=null;
WordNode ptr=this._head.getNext();
while (ptr!=null){
if(flag)
e=new WordNode(ptr.getWord(),e);
ptr=ptr.getNext();
flag=!flag;
}
evens._head=e;
return evens;
}
The split does work.
But I don't know where to continue from here. I want to call the split method, recursively and split the list until it is a list of one or two nodes, but I can't figure out how to do that.
Edit: can not use a third class, forhibited by the excercise. Also holding the length of TextList. Only holding number of times each word appears by an attribute on WordNode class.

This is of course assuming that you keep the length of the list at each insertion or deletion. All you have to do is split the list in half like this while keeping the head/root of the original list. You will not even need any intermediary list while implementing your merge sort.
LinkedList LinkedList::split(){
_Node_* p=head, *q=p;
for(int i=0; i<len/2; i++){
q=p;
p=p->next;
}
LinkedList ret;
ret.head=p;
ret.len=len-len/2;
len=len/2;
q->next=NULL;
return ret;
}

IMHO, the concept is wrong. You dont need to use merge-sort here. Try searching for PriorityQueue or actually BinaryHeap in order to solve this task. Secondly, merge sort on linked list is not a good idea since it will not be efficient at all. I think you should totally rework your solution.
NB. Just implement the operation YourLinkedList.getByIndex() for convinience, add the size atribute to hold the number of items in the linked list, then create one more linkedList and perform the bottom-up merge-sort like you would do with a simple array.
Structures:
public class Item {
private String word;
private Item next;
public Item(String word) {
this.word = word;
}
public Item getNext() {
return next;
}
public void setNext(Item next) {
this.next = next;
}
public String getWord() {
return word;
}
public void setWord(String word) {
this.word = word;
}
}
Linked List:
public class LinkedList {
private int size = 0;
private Item first = null;
public void swapFragment(LinkedList list, int from, int to) {
if (from >= 0 && to < size) {
list.get(to-from).setNext(this.get(to+1));
if (from > 0) {
this.get(from-1).setNext(list.get(0));
} else {
this.first = list.get(0);
}
}
}
public void addItem(String word) {
if (first == null) {
first = new Item(word);
} else {
Item item = first;
while (item.getNext() != null) {
item = item.getNext();
}
item.setNext(new Item(word));
}
this.size++;
}
public Item get(int index) {
if (index >= size) {
return null;
} else {
Item item = first;
for(int i = 1; i <= index; i++) {
item = item.getNext();
}
return item;
}
}
public int getSize() {
return size;
}
public void setSize(int size) {
this.size = size;
}
public String toString() {
Item item = first;
String message = null;
if (item != null) {
message = item.getWord() + " ";
} else {
return null;
}
while (item.getNext() != null) {
item = item.getNext();
message = message + item.getWord() + " ";
}
return message;
}
}
Merge Sort:
public class ListMergeSort {
public void sort(LinkedList list, int lo, int hi) {
if (hi <= lo) {
return;
}
int mid = lo + (hi-lo)/2;
sort(list, lo, mid);
sort(list, mid+1, hi);
merge(list, lo, hi, mid);
}
private void merge(LinkedList list, int lo, int hi, int mid) {
int i = lo;
int j = mid+1;
LinkedList newList = new LinkedList();
for (int k = lo; k <= hi; k++) {
if (i > mid) {
newList.addItem(list.get(j).getWord());
j++;
} else if (j > hi) {
newList.addItem(list.get(i).getWord());
i++;
} else if (list.get(i).getWord().compareTo(list.get(j).getWord()) < 0) {
newList.addItem(list.get(i).getWord());
i++;
} else {
newList.addItem(list.get(j).getWord());
j++;
}
}
list.swapFragment(newList, lo, hi);
}
}
Test Class for Strings:
import org.junit.*;
public class MergeTest {
#Test
public void testWords() {
LinkedList list = new LinkedList();
list.addItem("word");
list.addItem("pipe");
list.addItem("trainer");
list.addItem("stark");
list.addItem("33");
list.addItem("dmitry");
ListMergeSort lms = new ListMergeSort();
lms.sort(list, 0, list.getSize()-1);
}
}
Now you just need to create a class which receives a string as an argument, splits it with String.split() and adds the resulting words into the internal LinkedList datastructure. Then you sort them inside with the merge sort and you get the result.