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I have an array and I want to check if all the values from array are equal to what I want using edittext

Here is an example of my array
List<Integer> sampleList = new ArrayList<>();
all values inside this array are:
sampleList = [1,2,3,4,5,6]
Basically I'm using for to loop each values from this array
for (int i = 0; i < sampleList.size(); i++) {
int to = sampleList.get(i);
if (norooms == to) { //norooms = 1
//display a toast
}
}
I want to check if position 0-5 are all equal to 1 but I cant achieve it. Any help

You can try something like this.
int editTextValue = 3; // Your edit text value
boolean isEqual = true;//Flag to check if all the values are equal or not.
for (int i=0; i<sampleList.size(); i++){
if (sampleList.get(i) != editTextValue){
isEqual = false;
break;
}
}
Then check the condition of isEqual after the loop is over and handle accordingly.

public Check {
int value;
int index;
boolean match;
// constructor, setters and getters
}
For about all the information you want:
public List<Check> checkValues(List<Integer> originalList, Integer testValue) {
List<Check> result = new ArrayList<>();
for ( int i = 0; i< originalList.size(); i++ ) {
result.add(new Check(i, originalList.get(i), originalList.get(i) == testValue);
}
return result;
}
This is 'a way' to do it. You'll have information on how many matches there were, which other values there were and on which index in your List they were found.

I'd use Java Streams, check out more on: Streams
Basically turns the list into a stream of elements and then you can process the data.
//To get only the elements equal to 1
sampleList.stream()
.filter((num) -> num == 1);
//to get a list of booleans whose value depends on the element being equal to 1.
List<Boolean> listBooleans = sampleList.stream()
.map((num) -> num == 1)
.collect(Collectors.toList())
filter uses a predicate to determine if the element matches a certain condition, in this case number == 1.
map transforms the stream of Integers into a Stream of booleans where the result is number == 1.
collect terminates the stream, with Collectors.toList transforms it into a List.
Hope this helps.

Arraylist<String> matchedValues = new Arraylist<>();
String editTextValue = edittext.getText();
String arrayValue;
boolean isEqual ;
boolean isEveryElementPresent = false;
for (int i = 0 ; i < sampleList.size() ; i++){
arrayValue = sampleList.get(i);
if(arrayValue.equals(editTextValue)){
isEqual = true;
matchedValues.add(arrayValue);
}else{
isEqual =false;
}
}
if(matchedValues.size() == sampleList.size()){
isEveryElementPresent = true;
}

How to remove duplicate words containing in ArrayList<String> in java [duplicate]

I'm working on a program that uses an ArrayList to store Strings. The program prompts the user with a menu and allows the user to choose an operation to perform. Such operations are adding Strings to the List, printing the entries etc. What I want to be able to do is create a method called removeDuplicates(). This method will search the ArrayList and remove any duplicated values. I want to leave one instance of the duplicated value(s) within the list. I also want this method to return the total number of duplicates removed.
I've been trying to use nested loops to accomplish this but I've been running into trouble because when entries get deleted, the indexing of the ArrayList gets altered and things don't work as they should. I know conceptually what I need to do but I'm having trouble implementing this idea in code.
Here is some pseudo code:
start with first entry;
check each subsequent entry in the list and see if it matches the first entry;
remove each subsequent entry in the list that matches the first entry;
after all entries have been examined, move on to the second entry;
check each entry in the list and see if it matches the second entry;
remove each entry in the list that matches the second entry;
repeat for entry in the list
Here's the code I have so far:
public int removeDuplicates()
{
int duplicates = 0;
for ( int i = 0; i < strings.size(); i++ )
{
for ( int j = 0; j < strings.size(); j++ )
{
if ( i == j )
{
// i & j refer to same entry so do nothing
}
else if ( strings.get( j ).equals( strings.get( i ) ) )
{
strings.remove( j );
duplicates++;
}
}
}
return duplicates;
}
UPDATE: It appears that Will is looking for a homework solution that involves developing the algorithm to remove duplicates, rather than a pragmatic solution using Sets. See his comment:
Thx for the suggestions. This is part of an assignment and I believe the teacher had intended for the solution to not include sets. In other words, I am to come up with a solution that will search for and remove duplicates without implementing a HashSet. The teacher suggested using nested loops which is what I'm trying to do but I've been having some problems with the indexing of the ArrayList after certain entries are removed.

Why not use a collection such as Set (and an implementation like HashSet) which naturally prevents duplicates?

You can use nested loops without any problem:
public static int removeDuplicates(ArrayList<String> strings) {
int size = strings.size();
int duplicates = 0;
// not using a method in the check also speeds up the execution
// also i must be less that size-1 so that j doesn't
// throw IndexOutOfBoundsException
for (int i = 0; i < size - 1; i++) {
// start from the next item after strings[i]
// since the ones before are checked
for (int j = i + 1; j < size; j++) {
// no need for if ( i == j ) here
if (!strings.get(j).equals(strings.get(i)))
continue;
duplicates++;
strings.remove(j);
// decrease j because the array got re-indexed
j--;
// decrease the size of the array
size--;
} // for j
} // for i
return duplicates;
}

You could try this one liner to take a copy of the String preserving order.
List<String> list;
List<String> dedupped = new ArrayList<String>(new LinkedHashSet<String>(list));
This approach is also O(n) amortized instead of O(n^2)

Just to clarify my comment on matt b's answer, if you really want to count the number of duplicates removed, use this code:
List<String> list = new ArrayList<String>();
// list gets populated from user input...
Set<String> set = new HashSet<String>(list);
int numDuplicates = list.size() - set.size();

List<String> lst = new ArrayList<String>();
lst.add("one");
lst.add("one");
lst.add("two");
lst.add("three");
lst.add("three");
lst.add("three");
Set se =new HashSet(lst);
lst.clear();
lst = new ArrayList<String>(se);
for (Object ls : lst){
System.out.println("Resulting output---------" + ls);
}

I've been trying to use nested loops to accomplish this but I've been running into trouble because when entries get deleted, the indexing of the ArrayList gets altered and things don't work as they should
Why don't you just decrease the counter each time you delete an entry.
When you delete an entry the elements will move too:
ej:
String [] a = {"a","a","b","c" }
positions:
a[0] = "a";
a[1] = "a";
a[2] = "b";
a[3] = "c";
After you remove your first "a" the indexes are:
a[0] = "a";
a[1] = "b";
a[2] = "c";
So, you should take this into consideration and decrease the value of j ( j--) to avoid "jumping" over a value.
See this screenshot:

public Collection removeDuplicates(Collection c) {
// Returns a new collection with duplicates removed from passed collection.
Collection result = new ArrayList();
for(Object o : c) {
if (!result.contains(o)) {
result.add(o);
}
}
return result;
}
or
public void removeDuplicates(List l) {
// Removes duplicates in place from an existing list
Object last = null;
Collections.sort(l);
Iterator i = l.iterator();
while(i.hasNext()) {
Object o = i.next();
if (o.equals(last)) {
i.remove();
} else {
last = o;
}
}
}
Both untested.

Assuming you can't use a Set like you said, the easiest way of solving the problem is to use a temporary list, rather than attempting to remove the duplicates in place:
public class Duplicates {
public static void main(String[] args) {
List<String> list = new ArrayList<String>();
list.add("one");
list.add("one");
list.add("two");
list.add("three");
list.add("three");
list.add("three");
System.out.println("Prior to removal: " +list);
System.out.println("There were " + removeDuplicates(list) + " duplicates.");
System.out.println("After removal: " + list);
}
public static int removeDuplicates(List<String> list) {
int removed = 0;
List<String> temp = new ArrayList<String>();
for(String s : list) {
if(!temp.contains(s)) {
temp.add(s);
} else {
//if the string is already in the list, then ignore it and increment the removed counter
removed++;
}
}
//put the contents of temp back in the main list
list.clear();
list.addAll(temp);
return removed;
}
}

You could do something like this, must of what people answered above is one alternative, but here's another.
for (int i = 0; i < strings.size(); i++) {
for (int j = j + 1; j > strings.size(); j++) {
if(strings.get(i) == strings.get(j)) {
strings.remove(j);
j--;
}`
}
}
return strings;

Using a set is the best option to remove the duplicates:
If you have a list of of arrays you can remove the duplicates and still retain array list features:
List<String> strings = new ArrayList<String>();
//populate the array
...
List<String> dedupped = new ArrayList<String>(new HashSet<String>(strings));
int numdups = strings.size() - dedupped.size();
if you can't use a set, sort the array (Collections.sort()) and iterate over the list, checking if the current element is equal to the previous element, if it is, remove it.

Using a set is the best option (as others suggested).
If you want to compare all elements in a list with eachother you should slightly adapt your for loops:
for(int i = 0; i < max; i++)
for(int j = i+1; j < max; j++)
This way you don't compare each element only once instead of twice. This is because the second loop start at the next element compared to the first loop.
Also when removing from a list when iterating over them (even when you use a for loop instead of an iterator), keep in mind that you reduce the size of the list. A common solution is to keep another list of items you want to delete, and then after you finished deciding which to delete, you delete them from the original list.

public ArrayList removeDuplicates(ArrayList <String> inArray)
{
ArrayList <String> outArray = new ArrayList();
boolean doAdd = true;
for (int i = 0; i < inArray.size(); i++)
{
String testString = inArray.get(i);
for (int j = 0; j < inArray.size(); j++)
{
if (i == j)
{
break;
}
else if (inArray.get(j).equals(testString))
{
doAdd = false;
break;
}
}
if (doAdd)
{
outArray.add(testString);
}
else
{
doAdd = true;
}
}
return outArray;
}

You could replace the duplicate with an empty string*, thus keeping the indexing in tact. Then after you've completed you can strip out the empty strings.
*But only if an empty string isn't valid in your implementation.

The problem you are seeing in your code is that you remove an entry during iteration, thus invalidating the iteration location.
For example:
{"a", "b", "c", "b", "b", "d"}
i j
Now you are removing strings[j].
{"a", "b", "c", "b", "d"}
i j
The inner loop ends and j is incremented.
{"a", "b", "c", "b", "d"}
i j
Only one duplicate 'b' detected...oops.
best practice in these cases is to store the locations that have to be removed, and remove them after you have finished iterating through the arraylist. (One bonus, the strings.size() call can be optimized outside of the loops by you or the compiler)
Tip, you can start iterating with j at i+1, you've already checked the 0 - i!

The inner for loop is invalid. If you delete an element, you cannot increment j, since j is now pointing at the element after the one you deleted, and you will need to inspect it.
In other words, you should use a while loop instead of a for loop, and only increment j if the elements at i and j do not match. If they do match, remove the element at j. size() will decrease by 1 and j will now be pointing at the following element, so there is no need to increase j.
Also, there is no reason to inspect all elements in the inner loop, just the ones following i, since duplicates before i have already been removed by prior iterations.

public <Foo> Entry<Integer,List<Foo>> uniqueElementList(List<Foo> listWithPossibleDuplicates) {
List<Foo> result = new ArrayList<Foo>();//...might want to pre-size here, if you have reliable info about the number of dupes
Set<Foo> found = new HashSet<Foo>(); //...again with the pre-sizing
for (Foo f : listWithPossibleDuplicates) if (found.add(f)) result.add(f);
return entryFactory(listWithPossibleDuplicates.size()-found.size(), result);
}
and then some entryFactory(Integer key, List<Foo> value) method. If you want to mutate the original list (possibly not a good idea, but whatever) instead:
public <Foo> int removeDuplicates(List<Foo> listWithPossibleDuplicates) {
int original = listWithPossibleDuplicates.size();
Iterator<Foo> iter = listWithPossibleDuplicates.iterator();
Set<Foo> found = new HashSet<Foo>();
while (iter.hasNext()) if (!found.add(iter.next())) iter.remove();
return original - found.size();
}
for your particular case using strings, you may need to deal with some additional equality constraints (e.g., are upper and lower case versions the same or different?).
EDIT: ah, this is homework. Look up Iterator/Iterable in the Java Collections framework, as well as Set, and see if you don't come to the same conclusion I offered. The generics part is just gravy.

I am bit late to join this question, but I have come with a better solution regarding the same using GENERIC type. All the above provided solutions are just a solution. They are increasing a lead to the complexity of whole runtime thread.
RemoveDuplicacy.java
We can minimize it using a technique which should do the required , at the Load Time.
Example : For suppose when you are using a arraylist of the class type as :
ArrayList<User> usersList = new ArrayList<User>();
usersList.clear();
User user = new User();
user.setName("A");
user.setId("1"); // duplicate
usersList.add(user);
user = new User();
user.setName("A");
user.setId("1"); // duplicate
usersList.add(user);
user = new User();
user.setName("AB");
user.setId("2"); // duplicate
usersList.add(user);
user = new User();
user.setName("C");
user.setId("4");
usersList.add(user);
user = new User();
user.setName("A");
user.setId("1"); // duplicate
usersList.add(user);
user = new User();
user.setName("A");
user.setId("2"); // duplicate
usersList.add(user);
}
The Class for which is the base for the arraylist used above : User class
class User {
private String name;
private String id;
/**
* #param name
* the name to set
*/
public void setName(String name) {
this.name = name;
}
/**
* #return the name
*/
public String getName() {
return name;
}
/**
* #param id
* the id to set
*/
public void setId(String id) {
this.id = id;
}
/**
* #return the id
*/
public String getId() {
return id;
}
}
Now in java there are two Overrided methods present of Object (parent) Class, which can help here in the means to serve our purpose better.They are :
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((id == null) ? 0 : id.hashCode());
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
User other = (User) obj;
if (id == null) {
if (other.id != null)
return false;
} else if (!id.equals(other.id))
return false;
return true;
}
You have to override these methods in the User class
Here is the complete code :
https://gist.github.com/4584310
Let me know if you have any queries.

You can add the list into a HashSet and then again convert that hashset into list to remove the duplicates.
public static int removeDuplicates(List<String> duplicateList){
List<String> correctedList = new ArrayList<String>();
Set<String> a = new HashSet<String>();
a.addAll(duplicateList);
correctedList.addAll(a);
return (duplicateList.size()-correctedList.size());
}
here it will return the number of duplicates. You can also use the correctList with all unique values

Below is the code to remove duplicate elements from a list without changing the order of the list,without using temporary list and without using any set variables.This code saves the memory and boosts performance.
This is a generic method which works with any kind of list.
This was the question asked in one of the interviews.
Searched in many forums for the solution but could not find one,so thought this is the correct forum to post the code.
public List<?> removeDuplicate(List<?> listWithDuplicates) {
int[] intArray = new int[listWithDuplicates.size()];
int dupCount = 1;
int arrayIndex = 0;
int prevListIndex = 0; // to save previous listIndex value from intArray
int listIndex;
for (int i = 0; i < listWithDuplicates.size(); i++) {
for (int j = i + 1; j < listWithDuplicates.size(); j++) {
if (listWithDuplicates.get(j).equals(listWithDuplicates.get(i)))
dupCount++;
if (dupCount == 2) {
intArray[arrayIndex] = j; // Saving duplicate indexes to an array
arrayIndex++;
dupCount = 1;
}
}
}
Arrays.sort(intArray);
for (int k = intArray.length - 1; k >= 0; k--) {
listIndex = intArray[k];
if (listIndex != 0 && prevListIndex != listIndex){
listWithDuplicates.remove(listIndex);
prevListIndex = listIndex;
}
}
return listWithDuplicates;
}

Best way to remove null values from string array

Below are the 2 ways to remove null values, which one is the best approach?
public static String[] clean(final String[] v) {
List<String> list = new ArrayList<String>(Arrays.asList(v));
list.removeAll(Collections.singleton(null));
return list.toArray(new String[list.size()]);
}
public static String[] clean(final String[] v) {
List<String> list = new ArrayList<String>(v.length);
for (String aString : v)
{
if (aString != null)
{
list.add(aString);
}
}
return list.toArray(new String[list.size()]);
}

For removing null values from a single string, I would use a regular expression like this,
private static Pattern pattern = Pattern.compile("(?i)[(\\[{]?null[)\\]}]?");
public static String removeNullString(String value) {
if (StringUtils.isEmpty(value)) {
return StringUtils.EMPTY;
}
Matcher matcher = pattern.matcher(value);
return matcher.replaceAll(StringUtils.EMPTY);
}
It covers up all "null" and empty character from string.
For removing null value from a string array in Java 7,
String[] firstArray = {"test1", "", "test2", "test4", "", null};
List<String> list = new ArrayList<String>();
for(String s : firstArray) {
if(s != null && s.length() > 0) {
list.add(s);
}
}
firstArray = list.toArray(new String[list.size()]);
For removing null value from a string array in Java 8,
String[] firstArray = {"test1", "", "test2", "test4", "", null};
firstArray = Arrays.stream(firstArray)
.filter(s -> (s != null && s.length() > 0))
.toArray(String[]::new);

Performance wise, it is usually better to minimize calls outside of the scope of the current code block (ie method). Also, since memory allocation is relatively slow compared most other instructions, avoiding object creation is typically a goal. The best I can come up with in terms of performance (I chose to make it flexible enough to take any type of array):
public <T> T[] removeNulls(Class<T> type, final T[] original){
// first, shift all non-null instances to the head, all nulls to the end.
int nonNullCount=0;
T tempT = null;
for(int i=0; i < original.length; i++){
if(original[i] != null){
nonNullCount++;
}else if(i != original.length - 1){
// Swap the next non-null value with this null value
int j = i + 1;
// In case there are multiple null values in a row
// scan ahead until we find the next non-null value
while(j < original.length && (tempT = original[j]) == null){
j++;
}
original[nonNullCount] = tempT;
if(tempT != null){
nonNullCount++;
}
if(j < original.length){
original[j] = null;
}
i = j - 1;
}
}
// The case where there are no nulls in the array
if(nonNullCount == original.length){
return original;
}
final T[] noNulls = (T[]) Array.newInstance(type,nonNullCount);
System.arraycopy(original,0,noNulls,0,nonNullCount);
return noNulls;
}
But I'm not sure why you would want this complexity over the 3 or 4 lines to do the same thing when performance is not likely to be an issue. You would need to have HUGE arrays to see any benefit (if any) between my code and your clean example.

in Java 8 you should be able to do something like:
List<String> list = new ArrayList<String>(Arrays.asList(v));
list.removeIf(Objects::isNull);
return list.toArray(new String[list.size()]);

if you want to do it in same space i will suggest the follwing solution. But final array will also be having same size. I mean it will not shrink in size but all elements will get aggregated in same order.
public static void removeNullFromArray(String[] args) {
int location = 0;
for(int i=0; i<args.length; i++){
String arg = args[i];
if(arg!=null){
if(location<i){
args[location] = arg;
args[i] = null;
}
location++;
}
}
}

Java 8 code using streams and lambda. Filters non-nulls from an array and converts to a list.
Arrays.stream(arr).filter(Objects::nonNull).collect(Collectors.toList());

How to shift each elements in array using loops?

I want to shift each elements in array to left if there is a null. E.g
public static void main(String[] args) {
String asd[] = new String[5];
asd[0] = "zero";
asd[1] = "one";
asd[2] = null;
asd[3] = "three";
asd[4] = "four;
I want the output to be
zero, one, three, four.
The length should also be adjusted
How can i do this using loops? I tried using if statements to check if an element is not null copy that value to another array. But i dont know how to copy if there is a null.

Given the kind of question, I suppose you want a simple, loop only and array only based solution, to understand how it works.
You have to iterate on the array, keeping an index of the new insertion point. At the end, using that same index, you can "shrink" the array (actually copy to a new smaller array).
String[] arr = {"a","b",null,"c",null,"d"};
// This will move all elements "up" when nulls are found
int p = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[i] == null) continue;
arr[p] = arr[i];
p++;
}
// This will copy to a new smaller array
String[] newArr = new String[p];
System.arraycopy(arr,0,newArr,0,p);
Just tested this code.
EDIT :
Regarding the possibility of shrinking the array without using System.arraycopy, unfortunately in Java arrays size must be declared when they are instantiated, and can't be changed (nor made bigger nor smaller) after.
So if you have an array of length 6, and find 2 nulls, you have no way of shrinking it to a length of 4, if not creating a new empty array and then copying elements.
Lists can grow and shrink, and are more handy to use. For example, the same code with a list would be :
String[] arr = {"a","b",null,"c",null,"d"};
List<String> list = new ArrayList<>(Arrays.asList(arr));
Iterator<String> iter = list.iterator();
while (iter.hasNext()) if (iter.next() == null) iter.remove();
System.out.println(list);

Try:
int lengthNoNull = 0;
for(String a : asd) {
if(a != null) {
lengthNoNull++;
}
}
String[] newAsd = new String[lengthNoNull];
int i = 0;
for(String a : asd) {
if(a != null) {
newAsd[i++] = a;
}
}

Piece of code using only arrays.
String[] x = {"1","2","3",null,"4","5","6",null,"7","8","9"};
String[] a = new String[x.length];
int i = 0;
for(String s : x) {
if(s != null) a[i++] = s;
}
String[] arr = Arrays.copyOf(a, i);
Or this:
String[] xx = {"1","2","3",null,"4","5","6",null,"7","8","9"};
int pos = 0, i = 0;
String tmp;
for(String s : xx) {
if(s == null) {
tmp = xx[pos];
xx[pos] = s;
xx[i] = tmp;
pos++;
}
i++;
}
String[] arr = Arrays.copyOfRange(xx, pos, xx.length);

Detail formatter error: java.util.Arrays cannot be resolved to a type

I have developed a BlackBerry application where in I am reading in a HEX String values. The values returned are as follows:
String result = response.toString();
where result is:
["AC36C71DF3CB315A35BFE49A17F483B6","CF5B717ACC460E3C4545BE709E9BCB83","E1EE334738CA4FA14620639DD6750DC3","DD40E2822539C2184B652D1FC3D2B4E6","6AF4B1EAC8D8210D64A944BFD487B9F2"]
These are passed into the following split method to separate the values. The method is as follows:
private static String[] split(String original, String separator) {
Vector nodes = new Vector();
int index = original.indexOf(separator);
while (index >= 0) {
nodes.addElement(original.substring(0, index));
original = original.substring(index + separator.length());
index = original.indexOf(separator);
}
nodes.addElement(original);
String[] result = new String[nodes.size()];
if (nodes.size() > 0) {
for (int loop = 0; loop < nodes.size(); loop++) {
result[loop] = (String) nodes.elementAt(loop);
System.out.println(result[loop]);
}
}
return result;
}
The above array is passed is as the String original in the method. This part is working fine. However, when a single value is passed in as String original, i.e. ["6AF4B1EAC8D8210D64A944BFD487B9F2"], I get an error :
Detail formatter error:java.util.Arrays cannot be resolved to a type.
Please help !!! The values posted above are exact values as read including the parenthesis [] and quotations ""

The Blackberry libraries are based on Java ME and not Java SE. In Java ME some classes have been removed to reduce the runtime footprint such as the Arrays class.
Take a look at the Blackberry JDE java.util package, see there is no Arrays class. So in your code you cannot use methods coming from the Arrays class, you must found a workaround or implement the feature yourself.

Try this split method -
public static String[] split(String strString, String strDelimiter) {
String[] strArray;
int iOccurrences = 0;
int iIndexOfInnerString = 0;
int iIndexOfDelimiter = 0;
int iCounter = 0;
//Check for null input strings.
if (strString == null) {
throw new IllegalArgumentException("Input string cannot be null.");
}
//Check for null or empty delimiter strings.
if (strDelimiter.length() <= 0 || strDelimiter == null) {
throw new IllegalArgumentException("Delimeter cannot be null or empty.");
}
if (strString.startsWith(strDelimiter)) {
strString = strString.substring(strDelimiter.length());
}
if (!strString.endsWith(strDelimiter)) {
strString += strDelimiter;
}
while((iIndexOfDelimiter = strString.indexOf(strDelimiter,
iIndexOfInnerString)) != -1) {
iOccurrences += 1;
iIndexOfInnerString = iIndexOfDelimiter +
strDelimiter.length();
}
strArray = new String[iOccurrences];
iIndexOfInnerString = 0;
iIndexOfDelimiter = 0;
while((iIndexOfDelimiter = strString.indexOf(strDelimiter,
iIndexOfInnerString)) != -1) {
strArray[iCounter] = strString.substring(iIndexOfInnerString,iIndexOfDelimiter);
iIndexOfInnerString = iIndexOfDelimiter +
strDelimiter.length();
iCounter += 1;
}
return strArray;
}