Develop Reference

How to iterate through a bunch of streams and read one element at a time?

I have a stream class which consists of sorted in ascending order.
class Stream {
boolean hasNext();
int peek();
int next();
}
I need to write a program to merge multiple streams and print out the elements that exist more than a number of times in different streams without repeating.
This is how I design it with pseudo code:
// print out numbers with occurance greater than (k).
void mergeStreams(List<Stream> streams, int k) {
int minCount = 0;
int min = Integer.MAX_VALUE;
List<Stream> minStreams;
loop through each stream and read in elements on head {
if (streams.size() < k) return;
if (!stream.hasNext()) {
streams.remove(stream); continue;
}
if (stream.peek() <= min) {
if (stream.peek() == min) {
minCount++;
}
else {
minStreams = new ArrayList<Stream>();
min = stream.peek();
minCount = 1;
}
minStreams.add(stream);
}
if last stream in streams {
if (minCount >= k) {
System.out.println(min);
for (Stream s : minStreams) {
if (s.hasNext()) s.next();
else streams.remove(s);
}
}
}
Problem is I don't know how exactly to iterate through streams and read on element at a time and hence the pseudo code part. Plus, I am not sure if streams.remove(s) at the bottom of the code really deletes stream s from original list of Streams since it has been added to new list minStreams. I can really use some advice here.

To remove elements from a list while iterating over it, you should use Iterator.remove(). You'll also need a while loop around the whole thing to check more than one value.
void mergeStreams(List<Stream> streams, int k) {
while (streams.size() >= k) { // each loop checks one value (the current minimum)
int minCount = 0;
int min = Integer.MAX_VALUE;
List<Stream> minStreams = new ArrayList<>();
Iterator<Stream> streamIter = streams.iterator(); // supports remove during iteration
while (streamIter.hasNext()) {
Stream stream = streamIter.next();
if (!stream.hasNext()) {
streamIter.remove(); // remove stream from original list
continue;
}
if (stream.peek() <= min) {
if (stream.peek() == min) {
minCount++;
} else {
minStreams = new ArrayList<>();
min = stream.peek();
minCount = 1;
}
minStreams.add(stream);
}
}
if (minCount >= k) { // runs after iterating over streams
System.out.println(min);
}
for (Stream s : minStreams) { // must advance past the current min value
if (s.hasNext()) {
s.next();
} else {
streams.remove(s);
}
}
}
}
I also fixed a bug where the streams in minStreams didn't advance past the minimum value when minCount < k, so it would get stuck on that value.

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();

How to start back at beginning of arraylist in a while loop?

I'm trying to solve a puzzle that goes like this: 100 people stand in a circle. The first person kills the person next to him and hands the gun to the next person. Which person is left at the end?
This is what I have so far, but when I run it, it shows an out of bounds exception. I realized that when I write people.remove(i+1), the program runs to the end of the arraylist and has no way to start back at the beginning to continue the pattern. How do I do this?
Thanks for any help!
private void btnEnterActionPerformed(java.awt.event.ActionEvent evt) {
int input = Integer.parseInt(txtInput.getText());
ArrayList <Integer> people = new ArrayList <> ();
for (int i = 0; i < input; i++) {
people.add(i);
}
while (people.size() != 0) {
int i = 1;
people.remove(i+1);
i++;
}
for (int i = 0; i < people.size(); i++) {
lblOutput.setText(" " + people.get(i));
}

The reason you get an out of bound exception is that you check the size to be non-zero, but the call of remove(i+1) with i set to 1 means removing from the third spot in the list, which may not be there. Only the initial element at index zero is guaranteed to be there.
Also note that i++ has no effect, because i is reset back to 1 at the top of the loop's body.
With the condition of people.size() != 0 the only guaranteed thing is that you would be able to remove at index zero. However, this is rather inefficient, because all elements past that index need to be copied. This makes removal an O(n2), which could be slow when the list is really long.
Generally, though, the idiomatic way of managing removals from a list is using ListIterator<T> for removal of zero to a few items, or copying into a separate list and replacing the original list with the new one when you need to remove a significant portion of the list.

As I understand the problem, you need to remove every second person from the list until only one person remains.
The basic problem with your current implementation, is, first, you don't do any range checking (how do you know an element actually exists at i+1) and secondly, you loop until the list is empty, which isn't what you really want.
The basic requirement could use compounding loops, the outer loop checks the size of the list and keeps looping while the size of the List is greater then 1, the second loop processes the list, removing every other person from the list. Note, I don't reset the hasGun flag in the outer loop, this means that on each iteration of the inner loop, the gun continues to pass to the next survivor.
ArrayList<Integer> people = new ArrayList<>();
for (int i = 0; i < 10; i++) {
people.add(i);
}
boolean hasGun = true;
while (people.size() > 1) {
Iterator<Integer> iterator = people.iterator();
while (iterator.hasNext()) {
System.out.print("> " + iterator.next());
if (!hasGun) {
// You get shot...
iterator.remove();
System.out.println(" got shot");
} else {
System.out.println(" shoots");
}
hasGun = !hasGun;
}
}
for (Integer person : people) {
System.out.println(person);
}
This example also makes uses the List's Iterator, this over comes, in part, the issue of the array out of bounds, but you could also use a for-next loop and the hasGun flag as well.

To circulate through your array with indexing, use the remainder operator:
int actual = 0;
while (people.size() != 1) {
people.remove( (actual+1) % people.size() );
actual = (actual+1) % people.size();
}

I just think an ArrayList is not the best data structure for this problem. I find a LinkedList would be more fit. Actually, I found a very easy recursive solution using one. Have a look at this code:
public class Main {
public static int kill(LinkedList<Integer> people) {
assert people.size() > 0;
System.out.println("people: " + people);
if (people.size() < 3)
return people.getFirst();
else {
System.out.println("kill: " + people.remove(1));
people.addLast(people.removeFirst());
return kill(people);
}
}
public static void main(String[] args) {
LinkedList<Integer> people = new LinkedList<>();
for (int i = 0; i <=100; i++) {
people.add(i);
}
int survivor = kill(people);
System.out.println("Last survivor: " + survivor);
}
}
I just remove (kill?) the second member on the list and send the first one back to the end of the list. This process can be repeated until there are 2 people left, in which case you can guess the last survivor will be the first one in the list cause he will kill the second person.

If I had to resolve this problem, I would create my own Person class with a next property pointing to the next person.
Person class:
public class Person {
private int id;
private Person next;
public Person(int id) {
this.id = id;
}
public int getId() {
return this.id;
}
public Person getNext() {
return this.next;
}
public void setNext(Person next) {
this.next = next;
}
public void killNext() {
this.next = this.next.next;
}
}
Once that is in place, it's trivial to setup a circular set of linked persons. The algorithm then simply becomes looping each person by following the next property, killing the next person on each iteration. And the loop exits when the next property points to himself, indicating that there is no one left.
Algorithm:
public static void main(String[] args) {
// Setup 100 persons in a linked circle.
Person startingPerson = new Person(1);
Person currentPerson = startingPerson;
for (int i = 2; i <= 100; i++) {
currentPerson.setNext(new Person(i));
currentPerson = currentPerson.getNext();
}
currentPerson.setNext(startingPerson);
// Loop around until a single person is left.
currentPerson = startingPerson;
while (currentPerson != currentPerson.getNext()) {
currentPerson.killNext();
currentPerson = currentPerson.getNext();
}
System.out.println("Surviving person: " + currentPerson.getId());
}
Output:
Surviving person: 73

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.

How to leftshift an ArrayList

I'm using an ArrayList to hold a history of objects. Each new object I add using the .add method, like:
if(event.getAction() == MotionEvent.ACTION_UP)
{
if(currentWord != null)
{
wordHist.add(currentWord);
}
if(wordHist.size() > WORDHIST_MAX_COUNT)
{
wordHist.remove(0);
}
}
However I don't want this to grow indefinitely, but to be limited to a certain value. If it reaches this maximum value, I want the oldest object (index 0) to be removed, and the rest to be left shifted, so previous index 1 is now index 0, etc.
How can this be done?
Thanks

ArrayList is not really a good choice in this case, but it can by done by calling remove(0) method. But if you want to do that efficiently, a linked list is better
(edited to make it clear that LinkedList is not generally better than ArrayList, but only in this case)

If it reaches this maximum value, I want the oldest object (index 0) to be removed
Then do wordHist.remove(0). That will remove the element at index 0.
To be precise:
wordHist.add(new Word("hello"));
if (wordHist.size() > MAX_SIZE)
wordHist.remove(0);
As user658991 states however, you should be aware of that this is a linear operation, i.e., takes time proportional to the number of elements in the list.
You could do this in constant time using LinkedList methods add and removeFirst.
Another option would be to wrap an array, or ArrayList in a class called something like CircularArrayList. In circular list structures you'll override the oldest element when adding a new one.
Edit:
Your code works fine:
import java.util.*;
class Test {
static int WORDHIST_MAX_COUNT = 3;
static List<String> wordHist = new ArrayList<String>();
public static void add(String currentWord) {
// VERBATIM COPY OF YOUR CODE
if (true/*event.getAction() == MotionEvent.ACTION_UP*/)
{
if(currentWord != null)
{
wordHist.add(currentWord);
}
if(wordHist.size() > WORDHIST_MAX_COUNT)
{
wordHist.remove(0);
}
}
}
public static void main(String[] args) {
add("a");
add("b");
add("c");
for (int i = 0; i < wordHist.size(); i++)
System.out.printf("i: %d, word: %s%n", i, wordHist.get(i));
System.out.println();
add("d");
for (int i = 0; i < wordHist.size(); i++)
System.out.printf("i: %d, word: %s%n", i, wordHist.get(i));
}
}
Prints:
i: 0, word: a
i: 1, word: b
i: 2, word: c
i: 0, word: b <-- b is now at index 0.
i: 1, word: c
i: 2, word: d

Use the remove( ) method.
Using remove(0) will remove the element from the 0th index.

U can use list.remove(index)// here index being '0', this internally shifts rest of the array up. An alternative solution wud be to use a queue or dequeue.

One simple implementation of what Op De Cirkel suggested
import java.util.ArrayList;
import java.util.List;
public class SimpleCircularHistory {
private int sizeLimit, start = 0, end = 0;
boolean empty = false;
private List<String> history;
public SimpleCircularHistory(int sizeLimit) {
this.sizeLimit = sizeLimit;
history = new ArrayList<String>(sizeLimit);
}
public void add(String state){
empty = false;
end = (end + 1) % sizeLimit;
if(history.size() < sizeLimit){
history.add(state);
}else {
history.set(end, state);
start = (end + 1) % sizeLimit;
}
}
public String rollBack(){
if(empty){ // Empty
return null;
}else {
String state = history.get(end);
if(start == end){
empty = true;
}else {
end = (end + sizeLimit - 1) % sizeLimit;
}
return state;
}
}
public void print(){
if(empty){
System.out.println("Empty");
}else {
for(int i = start;; i = (i + 1) % sizeLimit){
System.out.println(history.get(i));
if(i == end) break;
}
System.out.println();
}
}
public static void main(String[] args) {
SimpleCircularHistory h = new SimpleCircularHistory(3);
h.add("a");
h.add("b");
h.add("c");
h.add("d");
h.add("e");
h.add("f");
h.print();
h.add("X");
h.add("Y");
h.rollBack();
h.rollBack();
h.print();
h.add("t");
h.add("v");
h.add("w");
h.print();
h.rollBack();
h.rollBack();
h.rollBack();
h.print();
h.rollBack();
h.print();
}
}
This would print out :
d
e
f
f
t
v
w
Empty
Empty

Yeah, I've noticed this behaviour in adroid's lists too. It's REALLY irritating.
Anyway, there is a way to get around it if I don't mind object creation/destruction and the resulting garbage collection (NEVER do this in a onDraw of a surfaceview or something).
What I do is basically have two tracking int's; one to place the new object, and one to remove it:
int trackInt = 0;
int removeInt = 0;
//and then, in the method/class you use this:
Object newobject = new Object();
//add to list
objectList.add(trackInt, newobject);
trackInt++;
if (bugList.size() > 20) //20 is the max number of object you want, ie the maximum size of the list
{
objectList.remove(removeInt);
trackInt = removeInt;
removeInt++;
if (removeInt > 19) //remember, the list is zero indexed!
{
removeInt = 0;
}
}

Commons-collections has exactly what you're looking for:
http://commons.apache.org/collections/apidocs/org/apache/commons/collections/buffer/CircularFifoBuffer.html