Develop Reference

Guava's Streams::findLast implementation

I am looking into the implementation of Streams::findLast from Guava and while trying to understand it, there were a couple of things that simply I could not grasp. Here is it's implementation:
public static <T> java.util.Optional<T> findLast(Stream<T> stream) {
class OptionalState {
boolean set = false;
T value = null;
void set(#Nullable T value) {
set = true;
this.value = value;
}
T get() {
checkState(set);
return value;
}
}
OptionalState state = new OptionalState();
Deque<Spliterator<T>> splits = new ArrayDeque<>();
splits.addLast(stream.spliterator());
while (!splits.isEmpty()) {
Spliterator<T> spliterator = splits.removeLast();
if (spliterator.getExactSizeIfKnown() == 0) {
continue; // drop this split
}
// Many spliterators will have trySplits that are SUBSIZED even if they are not themselves
// SUBSIZED.
if (spliterator.hasCharacteristics(Spliterator.SUBSIZED)) {
// we can drill down to exactly the smallest nonempty spliterator
while (true) {
Spliterator<T> prefix = spliterator.trySplit();
if (prefix == null || prefix.getExactSizeIfKnown() == 0) {
break;
} else if (spliterator.getExactSizeIfKnown() == 0) {
spliterator = prefix;
break;
}
}
// spliterator is known to be nonempty now
spliterator.forEachRemaining(state::set);
return java.util.Optional.of(state.get());
}
Spliterator<T> prefix = spliterator.trySplit();
if (prefix == null || prefix.getExactSizeIfKnown() == 0) {
// we can't split this any further
spliterator.forEachRemaining(state::set);
if (state.set) {
return java.util.Optional.of(state.get());
}
// fall back to the last split
continue;
}
splits.addLast(prefix);
splits.addLast(spliterator);
}
return java.util.Optional.empty();
}
In essence the implementation is not that complicated to be honest, but here are the things that I find a bit weird (and I'll take the blame here if this question gets closed as "opinion-based", I understand it might happen).
First of all is the creation of OptionalState class, this could have been replaced with an array of a single element:
T[] state = (T[]) new Object[1];
and used as simple as:
spliterator.forEachRemaining(x -> state[0] = x);
Then the entire method could be split into 3 pieces:
when a certain Spliterator is known to be empty:
if (spliterator.getExactSizeIfKnown() == 0)
In this case it's easy - just drop it.
then if the Spliterator is known to be SUBSIZED. This is the "happy-path" scenario; as in this case we can split this until we get to the last element. Basically the implementation says: split until the prefix is either null or it's empty (in which case consume the "right" spliterator) or if after a split the "right" spliterator is known to be empty, consume the prefix one. This is done via:
// spliterator is known to be nonempty now
spliterator.forEachRemaining(state::set);
return java.util.Optional.of(state.get());
Second question I have is actually about this comment:
// Many spliterators will have trySplits that are SUBSIZED
// even if they are not themselves SUBSIZED.
This is very interesting, but I could not find such an example, would appreciate if someone would introduce me to one. As a matter of fact, because this comment exists, the code in the next (3-rd part of the method can not be done with a while(true) like the second), because it assumes that after a trySplit we could obtain a Spliterator that is SUBSIZED, even if our initial one was not, so it has to go to the very beginning of findLast.
this part of the method is when a Spliterator is known not to be SUBSIZED and in this case it does not have a known size; thus it relies on how the Spliterator from the source is implemented and in this case actually a findLast makes little sense... for example a Spliterator from a HashSet will return whatever the last entry is in the last bucket...

When you iterate a Spliterator of an unknown size, you have to track whether an element has been encountered. This can be done by calling tryAdvance and using the return value or by using forEachRemaining with a Consumer which records whether an element has been encountered. When you go the latter route, a dedicated class is simpler than an array. And once you have a dedicated class, why not use it for the SIZED spliterator as well.
What’s strange to me, is that this local class, which only exists to be used as a Consumer, doesn’t implement Consumer but requires the binding via state::set.
Consider
Stream.concat(
Stream.of("foo").filter(s -> !s.isEmpty()),
Stream.of("bar", "baz"))
The Spliterator representing the entire stream can’t have the SIZED characteristic. But when splitting off the first substream with the unknown size, the remaining stream has a known size.
Test code:
Spliterator<String> sp = Stream.concat(
Stream.of("foo").filter(s -> !s.isEmpty()),
Stream.of("bar", "baz"))
.spliterator();
do {
System.out.println(
"SIZED: "+sp.hasCharacteristics(Spliterator.SIZED)
+ ", SUBSIZED: "+sp.hasCharacteristics(Spliterator.SUBSIZED)
+ ", exact size if known: "+sp.getExactSizeIfKnown());
} while(sp.trySplit() != null);
Result:
SIZED: false, SUBSIZED: false, exact size if known: -1
SIZED: true, SUBSIZED: true, exact size if known: 2
SIZED: true, SUBSIZED: true, exact size if known: 1
But to me, it looks weird when someone tells in a comment to know that splitting can change the characteristics and then doing a pre-test with SUBSIZED, instead of just doing the split and check whether the result has a known size. After all, the code is doing the split anyway, in the alternative branch, when the characteristic is not present. In my old answer, I did the pretest to avoid allocating data structures, but here, the ArrayDeque is always created and used. But I think, even my old answer could be simplified.
I’m not sure what you are aiming at. When a Spliterator has the ORDERED characteristic, the order of traversal and splitting is well-defined. Since HashSet is not ordered, the term “last” is meaningless. If you are radical, you could optimize the operation to just return the first element for unordered streams; that’s valid and much faster.
What is strange, is this condition:
if (prefix == null || prefix.getExactSizeIfKnown() == 0) {
// we can't split this any further
(and a similar loop termination in the SUBSIZED path)
Just because one prefix happened to have a known zero size, it does not imply that the suffix can’t split further. Nothing in the specification says that.
As a consequence of this condition, Stream.concat(Stream.of("foo"), Stream.of("bar","baz")) can be handled optimally, whereas for Stream.concat(Stream.of(), Stream.of("bar", "baz")), it will fall back to a traversal, because the first prefix has a known size of zero.

Java Recursion Parameter Value

From a binary tree, you want to create a 2-d arraylist where each arraylist within the main arraylist contains all the nodes of a binary tree on a single level, for all of the levels. I understand how to do this recursively with DFS but I am very confused about why it possible to get the correct answer by passing in the arraylist I want to populate as a parameter in the recursive function.
The header of my recursive function looks like this
createLevelLinkedList(TreeNode current, ArrayList<LinkedList<TreeNode>> lists, int level)
With basecase:
if (current == null) return;
As the tree is traversed, the current node will be appended to its corresponding arraylist within "lists". Within the recursive function there there two recursive calls to move the function to the child nodes of current:
createLevelLinkedList(current.left, lists, level + 1);
createLevelLinkedList(current.right, lists, level + 1);
Suppose there is a tree that looks like this
5
3 8
2 4 9
With depth first tree traversal, after the first return, we will have TreeNode "2" is the top object in the stack, then "3". When TreeNode "3" is current and the function calls
createLevelLinkedList(current.right, lists, level + 1);
To push TreeNode "4" onto the stack, how can lists contains TreeNode "2" if it actually does and what is really going in memory?
The code I am referencing can be found on Github: https://github.com/gaylemcd/ctci/blob/master/java/Chapter%204/Question4_4/QuestionDFS.java
public static void createLevelLinkedList(TreeNode root, ArrayList<LinkedList<TreeNode>> lists, int level)
{
if (root == null) return;
LinkedList<TreeNode> list = null;
if (lists.size() == level) { // Level not contained in list
list = new LinkedList<TreeNode>();
/* Levels are always traversed in order. So, if this is the first time we've visited level i,
* we must have seen levels 0 through i - 1. We can therefore safely add the level at the end. */
lists.add(list);
} else {
list = lists.get(level);
}
list.add(root);
createLevelLinkedList(root.left, lists, level + 1);
createLevelLinkedList(root.right, lists, level + 1);
}

You should understand that there's only one ArrayList instance passed to the createLevelLinkedList method. Each recursive call to createLevelLinkedList receives a reference to the same instance of the ArrayList.
Therefore, once the TreeNode 2 is added to the ArrayList (or, to be exact, to one of the LinkedLists contained within the ArrayList), it stays there throughout the execution of the recursive method. It doesn't disappear when the invocation of createLevelLinkedList that added it to the list returns.

"Simple" Trie Implementation

I need to implement a Trie (in Java) for a college project. The Trie should be able to add and remove Strings (for phase 1).
I have spent several hours each day (for the last few days) trying to figure out how to do this and FAILED miserably each time.
I require some help, the examples on the internet and my textbook (Data Structures and Algorithms in Java By Adam Drozdek) are not helping.
Information
Node classes I am working with:
class Node {
public boolean isLeaf;
}
class internalNode extends Node {
public String letters; //letter[0] = '$' always.
//See image -> if letter[1] = 'A' then children[1] refers to child node "AMMO"
//See image -> if letter[2] = 'B' then children[2] refers to internal node "#EU"
public TrieNode[] children = new TrieNode[2];
public TrieInternalNode(char ch)
{
letters = "#" + String.valueOf(ch);//letter[0] = '$' always.
isLeaf = false;
}
}
class leafNode extends Node
{
public String word;
public TrieLeafNode(String word)
{
this.word = new String(word);
isLeaf = true;
}
}
And here is the pseudo code for insert that I need to follow: (warning it is very vague)
trieInsert(String K)
{
i = 0;
p = the root;
while (not inserted)
{
if the end of word k is reached
set the end-of-word marker in p to true;
else if (p.ptrs[K[i]] == 0)
create a leaf containing K and put its address in p.ptrs[K[i]];
else if reference p.ptrs[K[i]] refers to a leaf
{
K_L = key in leaf p.ptrs[K[i]]
do
{
create a nonleaf and put its address in p.ptrs[K[i]];
p = the new nonleaf;
} while (K[i] == K_L[i++]);
}
create a leaf containing K and put its address in p.ptrs[K[--i]];
if the end of word k is reached
set the end-of-word marker in p to true;
else
create a leaf containing K_L and put its address in p.ptrs[K_L[i]];
else
p = p.ptrs[K[i++]];
}
}
I need to implement the following methods.
public boolean add(String word){...}//adds word to trie structure should return true if successful and false otherwise
public boolean remove(String word){...}//removes word from trie structure should return true if successful and false otherwise
I cant find pseudo code for remove, but if insert does not work delete wont help me.
Here is a image of how the Trie that I need to implement should look like.
I am aware that the Trie will still be inefficient if implemented like this, but at the moment I need not worry about this.
The book provides an implementation that is similar to what I need to do but doesn't use the end of word char ('$') and only stores the words without their prefixes in the child nodes http://mathcs.duq.edu/drozdek/DSinJava/SpellCheck.java
Constraints
I need to implement the trie in JAVA.
I may not import or use any of Java's built-in data structures. (ie. no Map, HashMap, ArrayList etc)
I may use Arrays, Java primitive Types and Java Strings.
The Trie must use a $ (dollar) symbol to indicate a end-of-word. (see the image below )
I may asume that now word containing the $symbol will be inserted.
I need to implement the Trie it in the same style as the book does.
Case of words doesn't matter ie. all words will be considered to be lowercase
The Trie should only store the end-of-word character and the characters applicable to a word and not the entire alphabet(like some implementations).
I do not expect anyone to do the implementation for me(unless they have one lying around :P) I just really need help.

First of all, I don't think you should make leaf nodes and internal nodes separate classes. I recommend making a universal node class with an isLeaf() method. This method would return true if a node has no children.
Here is some higher-level pseudocode for the functions you need to implement. For simplicity, I assume the existence of a method called getIndex() which returns the index corresponding to a character.
Insert(String str)
Node current = null
for each character in str
int index = getIndex(character)
if current.children[index] has not been initialized
initialize current.children[index] to be a new Node
current = current.children[index]
You can easily augment this pseudocode to fit your needs. For example, if you want to return false whenever insertion isn't successful:
Return false if the input string is null
Return false if the input string contains invalid characters
Now, here is some higher-level pseudocode for remove.
Remove(String str)
Node current = null
for each character in str
int index = getIndex(character)
current = current.children[index]
// At this point, we found the node we want to remove. However, we want to
// delete as many ancestor nodes as possible. We can delete an ancestor node
// if it is not need it any more. That is, we can delete an ancestor node
// if it has exactly one child.
Node ancestor = current
while ancestor is not null
if ancestor has 2 or more children
break out of loop
else if ancestor has less than 2 children
Node grandAncestor = ancestor.parent
if grandAncestor is not null
reinitialize grandAncestor.children // this has the effect of removing ancestor
ancestor = ancestor.parent
At a very high level, we follow the input string to the node we want to remove. After this, we traverse up the tree following parent pointers and delete every node with 1 child (since it is no longer needed). Once we reach a node with 2 children, we stop.
Like Insert, we can easily augment this pseudocode to return false whenever deletion isn't successful:
Return false if the input string is null
Return false if the input string contains invalid characters
Return false if the input string leads to a Node which doesn't exist
It is easiest to implement delete if your Node class has a parent field. However, it is possible to implement the method without parent points, but it is more difficult. You can see an example of the trickier implementation here.

Java Stack Comparison

I'm wondering how to accomplish this:
Compare two Stack objects
Do this recursively
After the method that
does this is complete, the Stacks remain as they were to begin with
(i.e. same order, same items).
Only the push, pop and isEmpty methods for Stack is available.
I'm looking more for theoretical help than coding help, but any insight would be appreciated.

Two stacks are identical if their top level elements are identical, and the remaining stacks are identical (namely, the recursive condition).
Now, think what to do just before returning from the method call, in order to leave the stacks the same way they where given at the invocation time.
---EDIT---
The working Java code (derived from Markus A. solution, but with an interesting use of "finally" and with generics):
static <T> boolean compareStacks(Stack<T> a, Stack<T> b) {
if (a.isEmpty() != b.isEmpty()) return false;
if (a.isEmpty() && b.isEmpty()) return true;
T element_a = a.pop();
T element_b = b.pop();
try {
if (((element_a==null) && (element_b!=null)) || (!element_a.equals(element_b)))
return false;
return compareStacks(a, b);
} finally { // restore elements
a.push(element_a);
b.push(element_b);
}
}

In pseudo-code, you could do something like this:
boolean compareStacks(a, b) {
if (a.isEmpty() != b.isEmpty()) return false; // check if one is empty
if (a.isEmpty() && b.isEmpty()) return true; // check if both are empty
element_a = a.pop(); // grab elements and compare them
element_b = b.pop();
if (((element_a==null) && (element_b!=null)) || !element_a.equals(element_b)) {
a.push(element_a); // if they are not equal, restore them and return false
b.push(element_b);
return false;
}
result = compareStacks(a, b); // compare shortened stacks recursively
a.push(element_a); // restore elements
b.push(element_b);
return result; // return result from recursive call
}

With recursion it always helps to think of it as 2 parts, a "Setup" and a recursive function. Your setup would create the proper situation (create the two stacks, pass them in, etc) and then calls the recursive method and when the recursive method is done, report the results.
In your case you probably want this signature for the "recursive" method:
public boolean compareStacks(Stack one, Stack two)
If that method pops & compares the top tow elements of the stack, it can return false right then (saying they don't compare). If they do, you now have two stacks, each one shorter than before. You already know how to compare those two stacks, right (You just wrote the method to do so!).
at the end you can "Push" your one element back onto each stack to restore it's previous state before returning.
There will be a little trickiness in restoring the stack in the case where they don't compare, and ensuring that if the compareStack you call fails it properly passes that up to the previous state, even if the "current" compareStack succeeds, but those are implementation details--just thought I'd mention those so you don't overlook them.
There is a cute solution with Try/finally (no catch, return from within the try and push back onto the stack in the finally) that would make the code pretty slick, but it's easy enough without it.

Threaded Binary tree implementation from binary tree

I am working on assignment for school. It manly consists of a method that takes as input a binary tree and returns a double threaded tree. Eg(if left child = null then left child will be connected with preceding inorder parent and if right child = null the it will link to its inorder succesor. Now I have an idea for the implementation...
I iterate recursively trough the original BINARY tree and store into an array the inorder traversal. Now, because my teachers implementation requires that threaded trees be a different class from binary. I must traverse again trough the binary tree and convert each node from binaryNode to threadedNode thus having at the end a "duplicate" of the initial BinaryTree but as Threadedtree type. After I do this I traverse again trough this threadedTree and whenever i see a null left or right child I refer to the inorder arraylist and find the threads.
Now as you might have noticed this is extremely inefficient, i am essentially traversing the tree 3 times. My professor has stated that this could be done recursively with only one traversal, essentially converting to threadedNode and finding the threads all at once. I have tried multiple ways but i can not find one that works. Does anyone have any kind of tip or some way i can implement it? Thanks
This is the method as specified by the instructor
public static <T> ThreadedNode<T> thread(BinaryNode<T> root)
{
//threads a binary tree
}

The instructor is correct. One traversal is sufficient.
Traverse the original binary tree, creating new ThreadedNodes as you walk this tree.
public static <T> ThreadedNode<T> thread(BinaryNode<T> root) {
// We'll be keeping track of the "previous" node as we go, so use
// a recursive helper method. At first, there is no previous.
return threadHelper(root, null);
}
private static <T> ThreadedNode<T> threadHelper(BinaryNode<T> n, ThreadedNode<T> previous) {
// Create a new threaded node from the current root. Note that the threaded nodes
// are actually created in "preorder". Assume the ThreadedNode constructor sets
// the left, right, threadLeft, and threadRight fields to null.
ThreadedNode<T> t = new ThreadedNode<T>(n.getData());
// First go down the left side, if necessary.
if (n.getLeft() != null) {
// If there is a left child we have to descend. Note that as we go down the
// left side the previous doesn't change, until we start "backing up".
t.left = threadHelper(n.getLeft(), previous);
previous = t.left;
} else {
// If there is no left child, connect our left thread to the previous.
t.threadLeft = previous;
}
// Now before we go down the right side, see if the previous
// node (it will be in the left subtree) needs to point here.
if (previous != null && previous.right == null) {
previous.threadRight = t;
}
if (n.getRight() != null) {
// If there is a right child we can descend the right. As we go down we
// update previous to the current node. We do this just by passing the current
// node as the second parameter.
t.right = threadHelper(n.getRight(), t);
} else {
// No right child, no worries. We'll hook up our thread-right pointer
// later.
}
return t;
}
Consider the tree (A (B (D) ()) C). The first node you hit in an inorder traversal is D. There is no previous node. So save D as previous. Then the next node you hit is B. The previous node was D, which had no right child, so add a threaded right pointer from D to B. Then set previous to B and continue. Next you hit A. B had no right child, so add a threaded right link from B to A. A has a right child so continue, setting previous to A. The next node is C. C has no left child, so add a threaded left link from C to the current value of previous, which is A.

You could skip the second trip of traversal that you mention in your method. You could convert the nodes from BinaryNode to ThreadedNode on the fly. You'd still need to traverse twice, I think, for the inorder traversal, and for finding the threads and converting it to aThreadedTree.
For conversion on the fly, you could use the method that your instructor has given.
HTH!