I'm doing the exercise on leetcode using Scala. The problem I'm working on is "Maximum Depth of Binary Tree", which means find the maximum depth of a binary tree.
I've passed my code with IntelliJ, but I keep having compile error(type mismatch) when submitting my solution in Leetcode. Here is my code, is there any problem or any other solution please?
object Solution {
abstract class BinTree
case object EmptyTree extends BinTree
case class TreeNode(mid: Int, left: BinTree, right: BinTree) extends BinTree
def maxDepth(root: BinTree): Int = {
root match {
case EmptyTree => 0
case TreeNode(_, l, r) => Math.max(maxDepth(l), maxDepth(r)) + 1
}
}
}
The error is here : Line 17: error: type mismatch; Line 24: error: type mismatch; I know it is quite strange because I just have 13 lines of codes, but I didn't made mistakes, trust me ;)
This looks like an error specific of the leetcode problem.
I assume you're referring to https://leetcode.com/problems/maximum-depth-of-binary-tree/description/
Perhaps you're not supposed to re-implement the data structure but just to provide the implementation for maxDepth, i.e. TreeNode is already given. Try this:
object Solution {
def maxDepth(root: TreeNode): Int = {
if (root == null) {
0
} else {
Math.max(maxDepth(root.left), maxDepth(root.right)) + 1
}
}
}
This assumes that the TreeNode data structure is the one given in the comment:
/**
* Definition for a binary tree node.
* class TreeNode(var _value: Int) {
* var value: Int = _value
* var left: TreeNode = null
* var right: TreeNode = null
* }
*/
Related
I am working on a compiler and one aspect currently is how to wait for interpolated variable names to be resolved. So I am wondering how to take a nested interpolated variable string and build some sort of simple data model/schema for unwrapping the evaluated string so to speak. Let me demonstrate.
Say we have a string like this:
foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}
That has 1, 2, and 3 levels of nested interpolations in it. So essentially it should resolve something like this:
wait for x, y, one, two, and c to resolve.
when both x and y resolve, then resolve a{x}-{y} immediately.
when both one and two resolve, resolve baz{one}-{two}.
when a{x}-{y}, baz{one}-{two}, and c all resolve, then finally resolve the whole expression.
I am shaky on my understanding of the logic flow for handling something like this, wondering if you could help solidify/clarify the general algorithm (high level pseudocode or something like that). Mainly just looking for how I would structure the data model and algorithm so as to progressively evaluate when the pieces are ready.
I'm starting out trying and it's not clear what to do next:
{
dependencies: [
{
path: [x]
},
{
path: [y]
}
],
parent: {
dependency: a{x}-{y} // interpolated term
parent: {
dependencies: [
{
}
]
}
}
}
Some sort of tree is probably necessary, but I am having trouble figuring out what it might look like, wondering if you could shed some light on that with some pseudocode (or JavaScript even).
watch the leaf nodes at first
then, when the children of a node are completed, propagate upward to resolving the next parent node. This would mean once x and y are done, it could resolve a{x}-{y}, but then wait until the other nodes are ready before doing the final top-level evaluation.
You can just simulate it by sending "events" to the system theoretically, like:
ready('y')
ready('c')
ready('x')
ready('a{x}-{y}')
function ready(variable) {
if ()
}
...actually that may not work, not sure how to handle the interpolated nodes in a hacky way like that. But even a high level description of how to solve this would be helpful.
export type SiteDependencyObserverParentType = {
observer: SiteDependencyObserverType
remaining: number
}
export type SiteDependencyObserverType = {
children: Array<SiteDependencyObserverType>
node: LinkNodeType
parent?: SiteDependencyObserverParentType
path: Array<string>
}
(What I'm currently thinking, some TypeScript)
Here is an approach in JavaScript:
Parse the input string to create a Node instance for each {} term, and create parent-child dependencies between the nodes.
Collect the leaf Nodes of this tree as the tree is being constructed: group these leaf nodes by their identifier. Note that the same identifier could occur multiple times in the input string, leading to multiple Nodes. If a variable x is resolved, then all Nodes with that name (the group) will be resolved.
Each node has a resolve method to set its final value
Each node has a notify method that any of its child nodes can call in order to notify it that the child has been resolved with a value. This may (or may not yet) lead to a cascading call of resolve.
In a demo, a timer is set up that at every tick will resolve a randomly picked variable to some number
I think that in your example, foo, and a might be functions that need to be called, but I didn't elaborate on that, and just considered them as literal text that does not need further treatment. It should not be difficult to extend the algorithm with such function-calling features.
class Node {
constructor(parent) {
this.source = ""; // The slice of the input string that maps to this node
this.texts = []; // Literal text that's not part of interpolation
this.children = []; // Node instances corresponding to interpolation
this.parent = parent; // Link to parent that should get notified when this node resolves
this.value = undefined; // Not yet resolved
}
isResolved() {
return this.value !== undefined;
}
resolve(value) {
if (this.isResolved()) return; // A node is not allowed to resolve twice: ignore
console.log(`Resolving "${this.source}" to "${value}"`);
this.value = value;
if (this.parent) this.parent.notify();
}
notify() {
// Check if all dependencies have been resolved
let value = "";
for (let i = 0; i < this.children.length; i++) {
const child = this.children[i];
if (!child.isResolved()) { // Not ready yet
console.log(`"${this.source}" is getting notified, but not all dependecies are ready yet`);
return;
}
value += this.texts[i] + child.value;
}
console.log(`"${this.source}" is getting notified, and all dependecies are ready:`);
this.resolve(value + this.texts.at(-1));
}
}
function makeTree(s) {
const leaves = {}; // nodes keyed by atomic names (like "x" "y" in the example)
const tokens = s.split(/([{}])/);
let i = 0; // Index in s
function dfs(parent=null) {
const node = new Node(parent);
const start = i;
while (tokens.length) {
const token = tokens.shift();
i += token.length;
if (token == "}") break;
if (token == "{") {
node.children.push(dfs(node));
} else {
node.texts.push(token);
}
}
node.source = s.slice(start, i - (tokens.length ? 1 : 0));
if (node.children.length == 0) { // It's a leaf
const label = node.texts[0];
leaves[label] ??= []; // Define as empty array if not yet defined
leaves[label].push(node);
}
return node;
}
dfs();
return leaves;
}
// ------------------- DEMO --------------------
let s = "foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}";
const leaves = makeTree(s);
// Create a random order in which to resolve the atomic variables:
function shuffle(array) {
for (var i = array.length - 1; i > 0; i--) {
var j = Math.floor(Math.random() * (i + 1));
[array[j], array[i]] = [array[i], array[j]];
}
return array;
}
const names = shuffle(Object.keys(leaves));
// Use a timer to resolve the variables one by one in the given random order
let index = 0;
function resolveRandomVariable() {
if (index >= names.length) return; // all done
console.log("\n---------------- timer tick --------------");
const name = names[index++];
console.log(`Variable ${name} gets a value: "${index}". Calling resolve() on the connected node instance(s):`);
for (const node of leaves[name]) node.resolve(index);
setTimeout(resolveRandomVariable, 1000);
}
setTimeout(resolveRandomVariable, 1000);
your idea of building a dependency tree it's really likeable.
Anyway I tryed to find a solution as simplest possible.
Even if it already works, there are many optimizations possible, take this just as proof of concept.
The background idea it's produce a List of Strings which you can read in order where each element it's what you need to solve progressively. Each element might be mandatory to solve something that come next in the List, hence for the overall expression. Once you solved all the chunks you have all pieces to solve your original expression.
It's written in Java, I hope it's understandable.
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Objects;
public class StackOverflow {
public static void main(String[] args) {
String exp = "foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}";
List<String> chunks = expToChunks(exp);
//it just reverse the order of the list
Collections.reverse(chunks);
System.out.println(chunks);
//output -> [c, two, one, baz{one}-{two}, y, x, a{x}-{y}]
}
public static List<String> expToChunks(String exp) {
List<String> chunks = new ArrayList<>();
//this first piece just find the first inner open parenthesys and its relative close parenthesys
int begin = exp.indexOf("{") + 1;
int numberOfParenthesys = 1;
int end = -1;
for(int i = begin; i < exp.length(); i++) {
char c = exp.charAt(i);
if (c == '{') numberOfParenthesys ++;
if (c == '}') numberOfParenthesys --;
if (numberOfParenthesys == 0) {
end = i;
break;
}
}
//this if put an end to recursive calls
if(begin > 0 && begin < exp.length() && end > 0) {
//add the chunk to the final list
String substring = exp.substring(begin, end);
chunks.add(substring);
//remove from the starting expression the already considered chunk
String newExp = exp.replace("{" + substring + "}", "");
//recursive call for inner element on the chunk found
chunks.addAll(Objects.requireNonNull(expToChunks(substring)));
//calculate other chunks on the remained expression
chunks.addAll(Objects.requireNonNull(expToChunks(newExp)));
}
return chunks;
}
}
Some details on the code:
The following piece find the begin and the end index of the first outer chunk of expression. The background idea is: in a valid expression the number of open parenthesys must be equal to the number of closing parenthesys. The count of open(+1) and close(-1) parenthesys can't ever be negative.
So using that simple loop once I find the count of parenthesys to be 0, I also found the first chunk of the expression.
int begin = exp.indexOf("{") + 1;
int numberOfParenthesys = 1;
int end = -1;
for(int i = begin; i < exp.length(); i++) {
char c = exp.charAt(i);
if (c == '{') numberOfParenthesys ++;
if (c == '}') numberOfParenthesys --;
if (numberOfParenthesys == 0) {
end = i;
break;
}
}
The if condition provide validation on the begin and end indexes and stop the recursive call when no more chunks can be found on the remained expression.
if(begin > 0 && begin < exp.length() && end > 0) {
...
}
I am unable to get the std::map find to locate the correct row in the std::map. The key is a class pointer and I have created a struct (tdEcApplDataMapEq) to compare the class's binary arrays for a match.'
The problem is it doesn't work. I call FoEcApplData::operator== when the find starts. It says the first entry does not a match and then the find returns out pointing to the first item on the std::map list. There is no attempt by find to search the other map entries. Also the one match test failed (false), so why is find saying its a match?
This probably has something to do with the std::map declaration. std::map says the third argument is for std::less, but I am doing a == vs. <.
If I change it to do < the same this happens. It enters FoEcApplData::operator< which return a true on the first check and find search stops with the search pointing to the 1st entry in the list.
How do I get find() to use the custom struct for the search?
My example adds 10 rows to FdTEcApplDataMap. It copies the CDH_DISABLE_XACT182 class into hold for the search later. I then do the find() test using hold as the search key.
Inside entry1
Inside entry2
Inside entry3<== this is the one I am searching for
Inside entry4
Inside entry5
Inside entry6
Inside entry7
Inside entry8
Inside entry9
Inside entry10
Inside entry1
This is the find:
auto hazard = ExcludedCmdDict.find(&hold);
if(hazard != ExcludedCmdDict.end())
{
std::cout << "found it " << hazard->second << std::endl;
}
This is the compare function being used:
bool FoEcApplData::operator==( const FoEcApplData& FoEcApplDataObject) const {
if(myNumOfBytes <= FoEcApplDataObject.NumOfBytes())
{
const EcTOctet* temp;
temp = FoEcApplDataObject.Data() ;
for(EcTInt i = 0; i < myNumOfBytes ; i++)
{
if(myData[i] != temp[i])
{
return false ;
}
}
return true;
}
else // myNumOfBytes > FoEcApplDataObject.NumOfBytes()
{
const EcTOctet* temp;
temp = FoEcApplDataObject.Data() ;
for(EcTInt i = 0; i < FoEcApplDataObject.NumOfBytes(); i++)
{
if(myData[i] != temp[i])
{
return false ;
}
}
return true;
}
}
This is the declaration for the std::map.
/*
Custom less for find on the FdTEcApplDataMap.
Needed since we are using pointers.
Returns - true - match, false - no match
node - pointer to the item you are looking for
node2 - pointer to an item on the list
*/
struct tdEcApplDataMapEq {
bool operator()(FoEcApplData *const& node, FoEcApplData *const& node2) const
{
return *node == *node2;
}
};
typedef std::map< FoEcApplData *, std::string, tdEcApplDataMapEq> FdTEcApplDataMap;
std::map expects the compare function to work like std::less. You need to use something along the lines of:
struct tdEcApplDataMapEq {
bool operator()(FoEcApplData *const& node, FoEcApplData *const& node2) const
{
return (*node < *node2); // Implement operator<() function for FoEcApplData
}
};
While at it, change the name of the struct to reflect the fact that it is trying to compute "less than".
struct tdEcApplDataMapLess {
bool operator()(FoEcApplData *const& node, FoEcApplData *const& node2) const
{
return (*node < *node2); // Implement operator<() function for FoEcApplData
}
};
I am in the process of learning D (I decided it would be a better beginner friendly language than C++) and I decided to give myself the excercise of implementing a general quicksort in D. My program runs fine when sorting integers but it doesn't compile and throws a strange error when sorting strings.
Here is my code:
import std.stdio, std.algorithm;
T[] quickSort(T)(T[] input) {
if (input.length <= 1) {return input;}
ulong i = input.length/2;
auto pivot = input[i];
input = input.remove(i);
T[] lesser = [];
T[] greater = [];
foreach (x; input) {
if (x<=pivot)
{
lesser ~= x;
}
else
{
greater ~=x;
}
}
return (quickSort(lesser) ~ cast(T)pivot ~ quickSort(greater));
}
void main() {
//Sort integers, this works fine
//writeln(quickSort([1,4,3,2,5]));
//Sort string, throws weird error
writeln(quickSort("oidfaosnuidafpsbufiadsb"));
}
When I run it on a string it throws this error:
/usr/share/dmd/src/phobos/std/algorithm.d(7397): Error: template std.algorithm.move does not match any function template declaration. Candidates are:
/usr/share/dmd/src/phobos/std/algorithm.d(1537): std.algorithm.move(T)(ref T source, ref T target)
/usr/share/dmd/src/phobos/std/algorithm.d(1630): std.algorithm.move(T)(ref T source)
/usr/share/dmd/src/phobos/std/algorithm.d(1537): Error: template std.algorithm.move cannot deduce template function from argument types !()(dchar, dchar)
/usr/share/dmd/src/phobos/std/algorithm.d(7405): Error: template std.algorithm.moveAll does not match any function template declaration. Candidates are:
/usr/share/dmd/src/phobos/std/algorithm.d(1786): std.algorithm.moveAll(Range1, Range2)(Range1 src, Range2 tgt) if (isInputRange!(Range1) && isInputRange!(Range2) && is(typeof(move(src.front, tgt.front))))
/usr/share/dmd/src/phobos/std/algorithm.d(7405): Error: template std.algorithm.moveAll(Range1, Range2)(Range1 src, Range2 tgt) if (isInputRange!(Range1) && isInputRange!(Range2) && is(typeof(move(src.front, tgt.front)))) cannot deduce template function from argument types !()(string, string)
helloworld.d(9): Error: template instance std.algorithm.remove!(cast(SwapStrategy)2, string, ulong) error instantiating
helloworld.d(31): instantiated from here: quickSort!(immutable(char))
helloworld.d(31): Error: template instance helloworld.quickSort!(immutable(char)) error instantiating
the problem is that strings are immutable so remove won't work (as it manipulates the string)
you can fix that by not removing and not inserting the pivot in the concat:
auto pivot = input[i];
//input = input.remove(i); //<- remove this line
T[] lesser = [];
//...
return (quickSort(lesser) ~ quickSort(greater)); //<- remove cast(T)pivot ~
or by passing in a dup:
writeln(quickSort("oidfaosnuidafpsbufiadsb".dup));
You have to put a "d" behind the string to make it utf-32, otherwise remove won't accept it.
writeln(quickSort("oidfaosnuidafpsbufiadsb"d.dup));
I want to create a tree (using a Node or ADT) in which every node has an annotation pointing back to its parent. Below is an example with a simple linked list data structure:
import util::Math;
import IO;
import Node;
anno LinkedList LinkedList#parent;
anno int LinkedList#something;
data LinkedList = item(int val, LinkedList next)
| last(int val)
;
public LinkedList linkedList = item(5,
item(4,
item(3,
item(2,
last(1)[#something=99]
)[#something=99]
)[#something=99]
)[#something=99]
)[#something=99];
public LinkedList addParentAnnotations(LinkedList n) {
return top-down visit (n) {
case LinkedList x: {
/* go through all children whose type is LinkedList */
for (LinkedList cx <- getChildren(x), LinkedList ll := cx) {
/* setting the annotation like this doesn't seem to work */
cx#parent = x;
// DOESN'T WORK EITHER: setAnnotation(cx, getAnnotations(cx) + ("parent": x));
}
}
}
}
Executing addParentAnnotations(linkedList) yields the following result:
rascal>addParentAnnotations(linkedList);
LinkedList: item(
5,
item(
4,
item(
3,
item(
2,
last(1)[
#something=99
])[
#something=99
])[
#something=99
])[
#something=99
])[
#something=99
]
The thing is that Rascal data is immutable, so you can not update anything using an assignment. The assignment will simply give you a new binding for cx with the annotation set, but will not change the original tree.
To change the original tree, you can either use the => operator for case statements, or the insert statement as follows:
case LinkedList x => x[#parent=...] // replace x by a new x that is annotated
or:
case LinkedList x : {
...
x#parent= ...;
insert x; // replace original x by new x in tree
}
Some other tip, in the Traversal.rsc library you can find a function called getTraversalContext() which produces a list of parents of the currently visited node if called from the body of a case:
import Traversal;
visit (...) {
case somePattern: {
parents = getTraversalContext();
parent = parents[1];
}
}
I am working on a binary search tree and i have been given an insertnode function that looks like
void insertNode(Node **t, Node *n)
{
if(!(*t))
*t=n;
else if((*t)->key<n->key)insertNode(&(*t)->right,n);
else if((*t)->key>n->key) insertNode(&(*t)->left,n);
}
I am trying to write a function that removes nodes recursively so far i have come up with:
void remove(int huntKey,Node **t)
{
bool keyFound=false;
if(!(*t))
cout<<"There are no nodes"<<endl;
while(keyFound==false)
{
if((*t)->key==huntKey)
{
keyFound=true;
(*t)->key=0;
}
else if((*t)->key < huntKey)remove(huntKey,&(*t)->right);
else if((*t)->key> huntKey) remove(huntKey,&(*t)->left);
}
}
Both of these functions are getting called from a switch in my main function which looks like:
int main()
{
int key=0,countCatch=0;char q;
Node *t, *n;
t=0;
while((q=menu()) !=0)
{
switch(q)
{
case'?': menu(); break;
case'i': inOrderPrint(t); break;
case'a': preOrderPrint(t); break;
case'b': postOrderPrint(t); break;
case'c': {cout<<"enter key: ";cin>>key;
n=createNode(key);insertNode(&t,n);break;}
case'r':{cout<<"enter the key you want removed: ";
cin>>key;
remove(key,&t);
break;}
case'n': {countCatch=countNodes(t);cout<<countCatch<<"\n"; };break;
}
}
return 0;
}
my remove node function is not working properly....any advice would help....
When you remove the node, you are only setting its key to '0', not actually removing it.
Example:
'4' has child '2,' which has children '1' and '3.'
In your code, removing '2' gives you this tree: 4 has child 0, which has children 1 and 3.
To remove an internal node (a node with children), you must handle its parent pointer and its children. You must set the parent's child-pointer to one of the removed node's children. Check this article for more:
http://en.wikipedia.org/wiki/Binary_tree#Deletion
Look at the code, it is not recursive though
http://code.google.com/p/cstl/source/browse/src/c_rb.c