Develop Reference

Using Queue Data Structure to solve problem

My code involves me to calculate the amount of time required for a person in a certain position in a queue to buy burgers with the index of the burgers array showing an integer of the amount of burgers a person ordered. Each time a person at the head buys a burger they are taken to the tail and have one burger removed and 1 time value added. I have to find the amount of time it takes for the person at what position I inputted to find the time to buy all of their burgers.
Here is my code:
import java.util.Queue;
public class A4Q2 {
public int calcTimeReqToBuyBurgers(int[] burgers, int p)
{
Queue<Integer> q = new LinkedList<Integer>();
q.addAll(burgers);
int time = 0;
int i = 0;
numBurg = burgers[p];
while(numBurg > 0) {
int x = q.remove();
x--;
if(x != 0) {
q.add(x);
}
time +=1;
}
return time;
}
public static void main(String[] args) {
A4Q2 ex = new A4Q2();
}
}
I do not know where to go from here or if I am even doing this right. I know my question might look a little messy but I am still new to this website. Can someone help me?

You need not use queue for this, and also in your solution it seems that you are not reducing numBurg(which makes the queue run infinitely) and also you need to keep track of index of numBurg. Then I guess it would work. Below is my solution
int time = 0;
int p_value = burgers[p];
for(int i=0;i<burgers.length;i++){
if(burgers[p] > p_value) time+=p_value;
else time+=burgers[p];
if(i == p) p_value-=1;
}
return time;
Explanation: you need to reduce burgers[p] value from all the elements in array until you reach p but after that you can stop so the prior elements will have only p-1 reduced in them as we can stop after we reach p. and also if someone needs less burgers they will only participate until they reach 0. This would be O(n) with no extra space solution.

public int calcTimeReqToBuyBurgers(int[] burgers, int p)
{
Queue<Integer> q = new LinkedList<Integer>();
for (int burger : burgers){
q.add(burger);}
int time = 0;
int i = 0;
int numBurg = burgers[p];
while(numBurg > 0) {
int curr = q.remove();
if(curr == 1){time ++;}
if(curr > 1){
q.add(curr);
time ++;
}
if(curr == numBurg){
q.add(curr -- );
numBurg --;
time ++;
}
}
return time;
}
public static void main(String[] args) {
A4Q2 ex = new A4Q2 ();
int [] burgers= new int [] {3,2,4,1};
int time=ex.calcTimeReqToBuyBurgers(burgers,1);
System.out.println("time "+time);
}

Print Tree with 4 nodes (simple forest) for checking a benchmark

I implemented an experimental OOP language and now benchmark garbage collection using a Storage benchmark. Now I want to check/print the following benchmark for small depths (n=2, 3, 4,..).
The tree (forest with 4 subnode) is generated by the buildTreeDepth method. The code is as follows:
import java.util.Arrays;
public final class StorageSimple {
private int count;
private int seed = 74755;
public int randomNext() {
seed = ((seed * 1309) + 13849) & 65535;
return seed;
}
private Object buildTreeDepth(final int depth) {
count++;
if (depth == 1) {
return new Object[randomNext() % 10 + 1];
} else {
Object[] arr = new Object[4];
Arrays.setAll(arr, v -> buildTreeDepth(depth - 1));
return arr;
}
}
public Object benchmark() {
count = 0;
buildTreeDepth(7);
return count;
}
public boolean verifyResult(final Object result) {
return 5461 == (int) result;
}
public static void main(String[] args) {
StorageSimple store = new StorageSimple();
System.out.println("Result: " + store.verifyResult(store.benchmark()));
}
}
Is there a somewhat simple/straight forward way to print the tree generated by buildTreeDepth? Just the short trees of n=3, 4, 5.

As other has already suggested, you may choose some lib to do so. But if you just want a simple algo to test in command line, you may do the following, which I always use when printing tree in command line (write by handle, may have some bug. Believe you can get what this BFS algo works):
queue.add(root);
queue.add(empty);
int count = 1;
while (queue.size() != 1) {
Node poll = queue.poll();
if (poll == empty) {
count = 1;
queue.add(empty);
}
for (Node n : poll.getChildNodes()) {
n.setNodeName(poll.getNodeName(), count++);
queue.add(n);
}
System.out.println(poll.getNodeName());
}
Sample output:
1
1-1 1-2 1-3 1-4
1-1-1 1-1-2 1-1-3 1-2-1 1-2-2 1-3-1 1-3-2 1-4-1
...
And in your case you use array, which seems even easier to print.

Instead of using object arrays, use a List implementation like ArrayList. For an improved better result subclass ArrayList to also hold a 'level' value and add indentation to the toString() method.

String list get an item starting without loop

I have a array list contains thousands of data.
For Example:
List<String> custNames = new ArrayList<String>();
custNames.add("John");
custNames.add("Tom");
custNames.add("Bart");
custNames.add("Tim");
custNames.add("Broad");
Now I want to get count of names only starting with 'T'. I used looping mechanism for my solution.
List<String> filterNames = new ArrayList<String>();
String nameStarts="T";
for(int i=0;i<custNames.size();i++)
{
if(custNames.get(i).toLowerCase().startsWith(nameStarts.toLowerCase()))
{
filterNames.add(custNames.get(i));
}
}
System.out.println(filterNames.size());
But I have very large collection of data in this custNames list.
Is there any different solution without using loop?
Thanks.

There is very good solution from Java 8 for your problem.
Try this,
long filterNameCount = custNames
.stream()
.parallel()
.filter((s) -> s.startsWith(nameStarts.toLowerCase()))
.count();
System.out.println(filterNameCount);

If you are open to using a third-party library, there are a few interesting options you could use with Eclipse Collections.
If you use the ArrayList as you have it above, you can use the LazyIterate utility as follows:
int count = LazyIterate.collect(custNames, String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
Assert.assertEquals(2, count);
If you use the Eclipse Collections replacement for ArrayList, you can use the rich functional protocols available directly on MutableList:
MutableList<String> custNames =
Lists.mutable.with("John", "Tom", "Bart", "Tim", "Broad");
String nameStarts= "T";
int count = custNames.asLazy()
.collect(String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
System.out.println(count);
Assert.assertEquals(2, count);
The serial API in Eclipse Collections is eager-by-default, which is why I called asLazy() first. The collect method would otherwise create another MutableList.
If you benchmark your code with your full set of data, the following parallel version of the code may be more performant:
MutableList<String> custNames =
Lists.mutable.with("John", "Tom", "Bart", "Tim", "Broad");
String nameStarts= "T";
int processors = Runtime.getRuntime().availableProcessors();
int batchSize = Math.max(1, custNames.size() / processors);
ExecutorService executor = Executors.newFixedThreadPool(processors);
int count = custNames.asParallel(executor, batchSize)
.collect(String::toLowerCase)
.countWith(String::startsWith, nameStarts.toLowerCase());
executor.shutdown();
Assert.assertEquals(2, count);
The asParallel() API in Eclipse Collections is lazy-by-default. The API forces you to pass in a an ExecutorService and an int batchSize. This gives you complete control over the parallelism.
You can also use the Stream API with all MutableCollections in Eclipse Collections because they extend java.util.Collection.
Note: I am a committer for Eclipse Collections.

You could also use a tree storage : it would very efficient for this kind of search. If you are stucked with a list the previous answered is a way to do.

remove all the items which dont start with "T" like this:
custNames.removeIf(p->!p.startsWith("T"));
you can make a copy out of your list and remove items not starting with "T".

First, you can shorten your initialization with Arrays.asList(T); Second, I would use a simple loop to build a table of counts once and then use that to determine the subsequent queries. Something like,
List<String> custNames = new ArrayList<String>(Arrays.asList("John", "Tom",
"Bart", "Tim", "Broad"));
int[] counts = new int[26];
for (String name : custNames) {
char ch = Character.toLowerCase(name.charAt(0));
counts[ch - 'a']++;
}
for (int i = 0; i < counts.length; i++) {
if (counts[i] > 0) {
System.out.printf("There are %d words that start with %c%n",
counts[i], (char) ('a' + i));
}
}
Which outputs
There are 2 words that start with b
There are 1 words that start with j
There are 2 words that start with t
Or, in the specific case - counts['t' - 'a'] is the count of words starting with t.

If you have more or less static list and perform search operation often you can sort your list or use TreeMap.
Also you don't need to create new list and get its size then. You can simply create a counter variable and increment it.

You can create your own sorting and finding implementation.
Consider the following:
public class ContainingArrayList<E> extends ArrayList<E> {
private Comparator<E> comparator;
public ContainingArrayList(Comparator<E> comparator) {
this.setComparator(comparator);
}
#Override
public boolean add(E e) {
// If the collection is empty or the new element is bigger than the last one, append it to the end of the collection
if(size() == 0 || comparator.compare(e, get(size()-1)) >= 0)
return super.add(e);
else {
for (int i = 0; i < size(); i++) {
int result = comparator.compare(e, get(i));
// If the new element is bigger than the current element, continue with the next element
if (result > 0) continue;
// If the new element is equal to the current element, no need to insert (you might insert of course)
if (result == 0) return false;
// Otherwise the new element is smaller than the current element, so insert it between the previous and the current element
super.add(i, e);
return true;
}
return super.add(e);
}
}
public E get(E containingElement) {
int start = 0;
int end = size()-1;
// If the element is the first one, return the first element
if(comparator.compare(containingElement, super.get(start)) == 0)
return super.get(start);
// If the element is the last one, return the last element
if(comparator.compare(containingElement, super.get(end)) == 0)
return super.get(end);
// Otherwise do a binary search
while(start != end) {
// Get the element between start and end positions
E mid = super.get(start + (end/2));
// Compare the two elements
int result = comparator.compare(containingElement, mid);
// If the middle element compared to the containing element is equal, return the middle element
if(result == 0) {
return mid;
}
// If the containing element is smaller than the middle, halve the end position
else if(result < 0) {
end = start + (end/2);
}
// If the containing element is bigger than the middle, set the start position to the middle position
else if(result > 0) {
start = start + (end/2);
}
}
return null;
}
public Comparator<E> getComparator() {
return comparator;
}
public void setComparator(Comparator<E> comparator) {
this.comparator = comparator;
}
}
The custom comparator is used to sort the elements and to find the element that starts with a specific character. This means that you can change the comparator implementation for your needs at any time or you can create a more dynamic finding solution.
Test:
public class SortFindTest {
public SortFindTest() {
ContainingArrayList<String> t = new ContainingArrayList<String>(new MyComparator());
t.add("John");
t.add("Tom");
t.add("Bart");
t.add("Tim");
t.add("Broad");
System.out.println(t.get("T"));
}
class MyComparator implements Comparator<String> {
#Override
public int compare(String o1, String o2) {
int o1c = o1.charAt(0);
int o2c = o2.charAt(0);
if(o1c == o2c)
return 0;
if(o1c > o2c)
return 1;
return -1;
}
}
public static void main(String[] args) {
new SortFindTest();
}
}
I'm not sure if this would be faster than Java 8 Stream API but it worth a try.

If the order in which the items are stored does not matter, you could store the names in a HashMap, where the first character of each name is the key, and an ArrayList of names with that first character are the values. And then all you need to do, assuming the HashMap is named customerList, is customerList.get("T").size().
Initializing HashList and Adding Customers
HashMap<Character, ArrayList<String>> customerList = new HashMap<Character, ArrayList<String>>();
int NUM_ALPHABETS = 26;
int ascii_char = 97;
for(int i = 0; i < NUM_ALPHABETS; i++){
char c = (char) ascii_char;
customerList.add(c, new ArrayList<String>());
ascii_char++;
}
customerList.get("t").add("Tony");
customerList.get("a").add("Alice");
customerList.get("b").add("Ben");
Getting Number of Customers Starting with "t"
int num_t = customerList.get("t").size();

Backtracking Recursion through a List of Strings

I'm trying to write a method that returns a sub lists and each containin the max value of strigns from a larger list.
I'm using iterator and backtrack recursion to achieve this, however - I'm not familiar with backtrack recursion. Any suggestions on how to make this work. Or if my code needs adjustments?
This is what I have so far:
private void printAnagrams(List<String> anagrams, int max, List<List<String>> listofLists) {
Iterator<String> iterate = anagrams.iterator();
String word = "";
while (iterate.hasNext()) {
for(int i = 0; i < anagrams.size(); i++) {
word = iterate.next();
listofLists.add(new ArrayList<>());
listofLists.get(i).add(word);
if (listofLists.size() == max) {
listofLists.add(new ArrayList<>());
// Continue new list
}
}
}
}
This is the method that calls the private method:
public void printAnagrams(String phrase, int max, List<List<String>> anagrams) {
if (anagrams == null || max < 0) {
throw new IllegalArgumentException();
} else if (max == 0) {
getWords(phrase);
} else
printAnagrams(phrase, max, anagrams);
}// End of printAnagram method
Sample output:
List:
[core, course, cs, cure, for, force, forces, four, of, off, offer, offers, or, our, ours, re, score, so, source, suffer, sure, us, use, user]
SubLists with max = 3:
[core, off, us]
[core, us, off]
[course, off]
[cure, off, so]
[cure, so, off]
[force, of, us]
[force, us, of]

I have to return a value from a function that is called multiple times and I am unable to do that

I am stuck.
The following function is supposed to return currVm, an integer. But if I make a return I will break the loop and next time when this function is called,the same process will begin again.
What shall I do, so that I continue from where I left off ? I tried making static variables but I that didn't help me.
#Override
public int getNextAvailableVm() {
Set<String> dataCenters = confMap.keySet();
for (String dataCenter : dataCenters) {
LinkedList<DepConfAttr> list = confMap.get(dataCenter);
Collections.sort(list, new MemoryComparator());
int size = list.size() - 1;
int count = 0;
while(size >= 0) {
DepConfAttr dca = (DepConfAttr)list.get(count);
int currVm = dca.getVmCount();
int c = 0;
while(c <= currVm) {
allocatedVm(currVm);
c++;
return currVm;
}
count++;
size--;
}
}
return 0;
}
The for-each loop assigns a new data center that acts as a key for the confMap.The list that I get as a value, is sorted.Then a loop is run till it escapes its size.Inside this while loop, another while loop is run from where a function named allocatedVm of the inherited class is called. A parameter named currVm is passed to it.
This is the variable that I need to return. What shall I do to return this variable ? I have to start from I left off. I mean the next call should appear to be the next step, whatever it was, while executing the loop.

Add List<Integer> object to your class, and change your method as follows:
private Iterator<Integer> availableVms = null;
#Override
public int getNextAvailableVm() {
if (availableVms != null) {
if (availableVms.hasNext()) {
return availableVms.next();
}
return 0;
}
List<Integer> tmp = new ArrayList<Integer>();
Set<String> dataCenters = confMap.keySet();
for (String dataCenter : dataCenters) {
LinkedList<DepConfAttr> list = confMap.get(dataCenter);
Collections.sort(list, new MemoryComparator());
int size = list.size() - 1;
int count = 0;
while(size >= 0) {
DepConfAttr dca = (DepConfAttr)list.get(count);
int currVm = dca.getVmCount();
int c = 0;
while(c <= currVm) {
allocatedVm(currVm);
c++;
tmp.add(currVm);
}
count++;
size--;
}
}
availableVms = tmp.iterator();
return availableVms.hasNext() ? availableVms.next() : 0;
}
The idea is to pre-generate the entire list, and store its iterator for future use. Before entering the method you check if the availableVms iterator has been prepared. If it has been prepared, grab the next item off of it if it's available; otherwise, return zero.
If the list has not been prepared yet, run your algorithm, and add the results to a temporary list tmp. Once the list is ready, grab its iterator, and use it for subsequent invocations.