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Remove a Column from a 2D Array

I have little problem with removing col from a 2d array.
The goal is to remove from every row specific index "2" and give it back.
Its how it need to be
I made it but got little problem with 0 at the end.
Its how I got
private static void removeEntry(int[][] workArray, int col) {
int row = workArray.length;
//largest row count
int max = 0;
int tempNum = 0;
for (int[] ints : workArray) {
tempNum = 0;
for (int j = 0; j < ints.length; j++) {
tempNum++;
if (tempNum > max) {
max = tempNum;
}
}
}
int [][] newArray = new int[row][max];
for(int i = 0; i < row; i++) {
for(int j = 0; j < max; j++) {
if(j < col && j < workArray[i].length) {
newArray[i][j] = workArray[i][j];
}else if (j == col) {
// Do nothing
} else if (j > col && j < workArray[i].length) {
newArray[i][j - 1] = workArray[i][j];
}
}
}
for (int i = 0; i < workArray.length; i++) {
for (int j = 0; j < workArray[i].length; j++) {
workArray[i][j] = newArray[i][j];
}
}
Then I tried to remove 0 but don't work
int remIndex = 0;
for (int i = 0; i < workArray.length; i++) {
for (int j = remIndex; j < workArray[i].length-1; j++) {
if(workArray[i][j] == remIndex){
workArray[i][j] = workArray[i][j + 1];
}
}
}

An array is a container of data that occupies a contiguous block of memory, its size should be defined when the array is being instantiated and can't be changed.
If you need an array of smaller or greater length, then you need to create a new one and copy all previously added elements that should be retained.
There's also no such thing in Java as "2D" arrays, we can create a nested array, i.e. an array containing other arrays.
And similarly to how we can reassign an integer value at a particular index in a plain array int[], we can change a reference in the array of arrays, i.e. we can make it point to another (newly created) array.
And that's what is required according to do in your assignment since the method is void. I.e. the given array needs to be changed by replacing all of its "rows", that greater or equal in length than the given column to remove col, with a new array that will retain all the previous element except for the one at index col.
private static void removeEntry(int[][] workArray, int col) {
for (int row = 0; row < workArray.length; row++) {
if (workArray[row].length <= col) { // no need to change anything
continue;
}
int newLength = workArray[row].length - 1;
int[] newRow = new int[newLength]; // creating a new row shorter by 1
for (int oldCol = 0, newCol = 0; oldCol < workArray[row].length; oldCol++, newCol++) {
if (oldCol == col) { // skipping the target column
newCol--; // newCol would be incremented automatically at the end of the iteration, but we want it to remain the same
continue;
}
newRow[newCol] = workArray[row][oldCol];
}
workArray[row] = newRow; // reassigning the row
}
}
main()
public static void main(String[] args) {
int[][] testArr =
{{1, 2},
{1, 2, 3},
{1, 2, 3, 4}};
removeEntry(testArr, 2);
// printing the testArr
for (int[] arr: testArr) {
System.out.println(Arrays.toString(arr));
}
}
Output:
[1, 2]
[1, 2]
[1, 2, 4]
A link to the Online Demo
If you've made the method to be void by mistake, you are required to return a new array, i.e. the return type int[] (check the requirements of your assignment carefully). Then a minor change needs to be applied to the logic explained and implemented above: every array should be replaced with a new one and then placed into the newly created resulting array.
Note Arrays.copyOf() allows creating a duplicate of the hole array, and System.arraycopy() can help you to copy the elements in the given range from one array into another, but because you are working on an assignment I suggest you to do it manually with loops because you are expected to demonstrate the knowledge on how to manipulate with arrays, not the knowledge of special utility features (unless otherwise specified in the assignment).
private static int[][] removeEntry(int[][] workArray, int col) {
int[][] result = new int[workArray.length][];
for (int row = 0; row < workArray.length; row++) {
int newLength = col < workArray[row].length ? workArray[row].length - 1 : workArray[row].length;
int[] newRow = new int[newLength];
for (int oldCol = 0, newCol = 0; oldCol < workArray[row].length; oldCol++, newCol++) {
if (oldCol == col) {
newCol--;
continue;
}
newRow[newCol] = workArray[row][oldCol];
}
result[row] = newRow; // reassigning the row
}
return result;
}
main()
public static void main(String[] args) {
int[][] testArr =
{{1, 2},
{1, 2, 3},
{1, 2, 3, 4}};
int[][] newArr = removeEntry(testArr, 2);
// printing the testArr
for (int[] arr: newArr) {
System.out.println(Arrays.toString(arr));
}
}
Output:
[1, 2]
[1, 2]
[1, 2, 4]
A link to the Online Demo

Here is my approach on this. I used System.arraycopy() to copy the array up to the removed column and from directly after the removed column to the end. This way, we remove the column completely from the array.
private static int[][] removeEntry(int[][] workArray, int col) {
int[][] resultArray = new int[workArray.length][];
int index = 0;
for (int[] row : workArray) {
if (row.length - 1 < col) {
resultArray[index] = row;
index++;
continue;
}
int[] arrayCopy = new int[row.length - 1];
System.arraycopy(row, 0, arrayCopy, 0, col);
System.arraycopy(row, col + 1, arrayCopy, col, row.length - col - 1);
resultArray[index] = arrayCopy;
index++;
}
return resultArray;
}

Arrays are of fixed size. At the time of initializing array, a default value is stored in it. Here, 0 is stored as default. So I would suggest you to not initialize "col" as "max" but rather do it in for loop like this :-
int [][] newArray = new int[row][];
for(int i = 0; i < row; i++) {
if(col <= workArray[i].length){
newArray[i] = new int[workArray[i].length-1]; //initialize it here
}
for(int j = 0; j < workArray[i].length; j++) {
if(j < col) {
newArray[i][j] = workArray[i][j];
}else if (j == col) {
// Do nothing
} else if (j > col) {
newArray[i][j - 1] = workArray[i][j];
}
}
}
return newArray;
You are also supposed to return modified array as we can't changed the size of array and thus we had to create a new one ( newArray ). So do change the method definition as :-
private static int[][] removeEntry(int[][] workArray, int col)
Finally, whole method would look like :-
private static int[][] removeEntry(int[][] workArray, int col) {
int row = workArray.length;
//largest row count
int [][] newArray = new int[row][];
for(int i = 0; i < row; i++) {
if(col <= workArray[i].length){
newArray[i] = new int[workArray[i].length-1]; //initialize it here
}
for(int j = 0; j < workArray[i].length; j++) {
if(j < col) {
newArray[i][j] = workArray[i][j];
}else if (j == col) {
// Do nothing
} else if (j > col) {
newArray[i][j - 1] = workArray[i][j];
}
}
}
return newArray;
}
and you can use it like :-
workArray = removeEntry(workArray, 2);

Checking if two int arrays have duplicate elements, and extract one of the duplicate elements from them

I'm trying to write a method, union(), that will return an int array, and it takes two int array parameters and check if they are sets, or in other words have duplicates between them. I wrote another method, isSet(), it takes one array argument and check if the array is a set. The problem is I want to check if the two arrays in the union method have duplicates between them, if they do, I want to extract one of the duplicates and put it in the unionArray[] int array. This is what I tried so far.
public int[] union(int[] array1, int[] array2){
int count = 0;
if (isSet(array1) && isSet(array2)){
for (int i = 0; i < array1.length; i++){
for (int j = 0; j < array2.length; j++){
if (array1[i] == array2[j]){
System.out.println(array2[j]);
count ++;
}
}
}
}
int[] array3 = new int[array2.length - count];
int[] unionArray = new int[array1.length + array3.length];
int elementOfUnion = 0;
for (int i = 0; i< array1.length; i++){
unionArray[i] = array1[i];
elementOfUnion = i + 1 ;
}
int index = 0;
for (int i = elementOfUnion; i < unionArray.length; i++){
unionArray[i] = array3[index];
index++;
}
return unionArray;
}
public boolean isSet(int[] array){
boolean duplicates = true;
for (int i = 0; i < array.length; i++){
for(int n = i+1; n < array.length; n++){
if (array[i] == array[n])
duplicates = false;
}
}
return duplicates;
}
What I was trying to do is to use all of array1 elements in the unionArray, check if array2 has any duplicates with array1, and then move all the non-duplicate elements from array2 to a new array3, and concatenate array3 to unionArray.

It will be much easier to do it with Collection API or Stream API. However, you have mentioned that you want to do it purely using arrays and without importing any class, it will require a few lengthy (although simple) processing units. The most important theories that drive the logic is how (given below) a union is calculated:
n(A U B) = n(A) + n(B) - n(A ∩ B)
and
n(Only A) = n(A) - n(A ∩ B)
n(Only B) = n(B) - n(A ∩ B)
A high-level summary of this solution is depicted with the following diagram:
Rest of the logic has been very clearly mentioned through comments in the code itself.
public class Main {
public static void main(String[] args) {
// Test
display(union(new int[] { 1, 2, 3, 4 }, new int[] { 3, 4, 5, 6 }));
display(union(new int[] { 1, 2, 3 }, new int[] { 4, 5, 6 }));
display(union(new int[] { 1, 2, 3, 4 }, new int[] { 1, 2, 3, 4 }));
display(union(new int[] { 1, 2, 3, 4 }, new int[] { 3, 4 }));
display(union(new int[] { 1, 2, 3, 4 }, new int[] { 4, 5 }));
display(union(new int[] { 1, 2, 3, 4, 5, 6 }, new int[] { 7, 8 }));
}
public static int[] union(int[] array1, int[] array2) {
// Create an array of the length equal to that of the smaller of the two array
// parameters
int[] intersection = new int[array1.length <= array2.length ? array1.length : array2.length];
int count = 0;
// Put the duplicate elements into intersection[]
for (int i = 0; i < array1.length; i++) {
for (int j = 0; j < array2.length; j++) {
if (array1[i] == array2[j]) {
intersection[count++] = array1[i];
}
}
}
// Create int []union of the length as per the n(A U B) = n(A) + n(B) - n(A ∩ B)
int[] union = new int[array1.length + array2.length - count];
// Copy array1[] minus intersection[] into union[]
int lastIndex = copySourceOnly(array1, intersection, union, count, 0);
// Copy array2[] minus intersection[] into union[]
lastIndex = copySourceOnly(array2, intersection, union, count, lastIndex);
// Copy intersection[] into union[]
for (int i = 0; i < count; i++) {
union[lastIndex + i] = intersection[i];
}
return union;
}
static int copySourceOnly(int[] source, int[] exclude, int[] target, int count, int startWith) {
int j, lastIndex = startWith;
for (int i = 0; i < source.length; i++) {
// Check if source[i] is present in intersection[]
for (j = 0; j < count; j++) {
if (source[i] == exclude[j]) {
break;
}
}
// If j has reached count, it means `break;` was not executed i.e. source[i] is
// not present in intersection[]
if (j == count) {
target[lastIndex++] = source[i];
}
}
return lastIndex;
}
static void display(int arr[]) {
System.out.print("[");
for (int i = 0; i < arr.length; i++) {
System.out.print(i < arr.length - 1 ? arr[i] + ", " : arr[i]);
}
System.out.println("]");
}
}
Output:
[1, 2, 5, 6, 3, 4]
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4]
[1, 2, 3, 4]
[1, 2, 3, 5, 4]
[1, 2, 3, 4, 5, 6, 7, 8]

Using Java's streams could make this quite simpler:
public int[] union(int[] array1, int[] array2) {
return Stream.of(array1, array2).flatMapToInt(Arrays::stream).distinct().toArray();
}

Even with all the restrictions of only using arrays, you can simplify your code a lot. No need to check for sets. Just :
allocate an array to store all elements of the union (i.e., int[] tmp_union ), which at worst will be all elements from both arrays array1 and array2.
iterate over the elements of array1 and compared them against the elements from tmp_union array, add them into the tmp_union array only if they were not yet added to that array.
Repeat 2) for the array2.
During this process keep track of the number of elements added to the tmp_union array so far (i.e., added_so_far). In the end, copy the elements from the tmp_union array into a new array (i.e., unionArray) with space allocated just for the union elements. The code would look something like:
public static int[] union(int[] array1, int[] array2){
int[] tmp_union = new int[array1.length + array2.length];
int added_so_far = add_unique(array1, tmp_union, 0);
added_so_far = add_unique(array2, tmp_union, added_so_far);
return copyArray(tmp_union, added_so_far);
}
private static int[] copyArray(int[] ori, int size) {
int[] dest = new int[size];
for(int i = 0; i < size; i++)
dest[i] = ori[i];
return dest;
}
private static int add_unique(int[] array, int[] union, int added_so_far) {
for (int element : array)
if (!is_present(union, added_so_far, element))
union[added_so_far++] = element;
return added_so_far;
}
private static boolean is_present(int[] union, int added_so_far, int element) {
for (int z = 0; z < added_so_far; z++)
if (element == union[z])
return true;
return false;
}

Sorting list of numbers

I am trying to sort a list of numbers from smallest to the biggest and print it. I've tried two things:
1.
public class Sorter {
public static void main(String[] args) {
int[] numbers = {1, 3, 8, 2, 5, -2, 0, 7, 15};
int[] sorted = new int[numbers.length];
for (int a = 0; a < numbers.length; a++) {
int check = 0;
for (int b = 0; b < numbers.length; b++) {
if (numbers[a] < numbers[b]) {
check++;
}
}
sorted[check] = numbers[a];
}
for (int c = numbers.length - 1; c >= 0; c--) {
System.out.print(sorted[c] + ", ");
}
}
}
and this thing works, but won't work with repeated values, so I tried this other thing
public class Sortertwo {
public static void main(String[] args) {
int[] numinput = {3, 2, 1, 4, 7, 3, 17, 5, 2, 2, -2, -4};
int[] numsorted = new int[numinput.length];
int n = 0;
for (; n < numinput.length; ) {
for (int b = 0; b < numinput.length; b++) {
int check = 0;
for (int c = 0; c < numinput.length; c++) {
if (numinput[b] <= numinput[c]) {
check++;
}
}
if (check >= (numinput.length - n) && numinput[b] != 0) {
numsorted[n] = numinput[b];
numinput[b] = 0;
n++;
}
if (n >= (numinput.length)) {
break;
}
}
}
for (int g = 0; g < numinput.length; g++) {
System.out.print(numsorted[g] + ", ");
}
}
}
Where it relies on the thing that once the number from the first array is used (the smallest one is found), it has to be ignored when the program goes through the array next time around.
I tried to assign it like null value, but it doesn't work, so I assigned it to zero and then ignore it, which is a problem, because the list cant have a zero in it.
Is there any like better way to go about it? Thanks.

You can always use:
Arrays.sort(numbers);

If you want to use your first method then change this:
if (numbers[a] < numbers[b])
{
check++;
}
to:
if (numbers[a] <= numbers[b])
{
check++;
}

Unless this is homework, using Arrays.sort as the comments suggest, should be the way to go
import java.util.Arrays;
public class S {
public static void main(String ... args) {
int[] numbers = {1, 3, 8, 2, 5, -2, 0, 7, 15};
Arrays.sort(numbers);
System.out.println(Arrays.toString(numbers));
}
}
Prints:
[-2, 0, 1, 2, 3, 5, 7, 8, 15]

Add up each element in two uneven arrays in Java

Im Swedish so maybe I did give the wrong title.
I have two arrays of different size:
{2, 5, 10, 13}
{5, 7, 5, 22, 44, 75}
I want to add each element and put it in a third array.
So the result should be {7, 12, 15, 25, 44, 75}
I have manage to done some code.
I get an exeption of out of bounds.
I think the problem is that I can´t add a non existing element.
But how can I solve it?
public static void main(String[] args) {
int[] samling = {1, 2, 4, 3, 8};
int[] samling2 = {1, 2, 4, 3, 8, 8, 3};
int[] svar = concateArrays(samling, samling2);
for(int i=0; i < svar.length; i++)
System.out.println("Ny Array " + svar[i]);
}
public static int[] concateArrays(int[] samling, int[] samling2)
{
int sum = samling.length + samling2.length;
int[] total = new int[sum];
for(int i=0; i < total.length; i++){
//if (samling2.length != 0) // || samling.length != 0)
total[i] = samling[i] + samling2[i];
}
return total;
}

The length of the output array shouldn't be the sum of lengths of the input arrays, it should be the length of the longer input array. And before accessing an element of either input array, you must check the current index i is a valid index of that array.
public static int[] concateArrays(int[] samling, int[] samling2)
{
int[] total = new int[Math.max(samling.length,samling2.length)];
for(int i=0; i < total.length; i++) {
total[i] = (i < samling.length ? samling[i] : 0) +
(i < samling2.length ? samling2[i] : 0);
}
return total;
}

You can use one loop to loop over both arrays simultaneously.
You first need to check which array is the longest, and make a new array:
int[] arr = new int[longest];
Then you need to walk over the array. In this example, I assume the latter array is always the longest.
for (int i = 0; i < samling2.length; i++) {
int totalValue = samling[i];
if (i < samling.length) {
totalValue += samling2[i];
}
arr[i] = totalValue;
}

You can make an array of length equal to max of the two arrays you have. Then you can add the elements from both arrays if the specific index exist in both arrays else simply copy the index from the array which contain that index.
See the following code to get a better picture
import java.io.*;
class GFG {
public static void main(String[] args) {
int[] samling = {1, 2, 4, 3, 8};
int[] samling2 = {1, 2, 4, 3, 8, 8, 3};
int[] svar = concateArrays(samling, samling2);
System.out.println("Ny Array :");
for(int i=0; i < svar.length; i++)
System.out.print(svar[i] + " ");
}
public static int[] concateArrays(int[] samling, int[] samling2)
{
int len = 0;
if (samling.length > samling2.length)
len = samling.length;
else
len = samling2.length;
int[] total = new int[len];
for(int i=0; i < len; i++){
if (i >= samling2.length) {
total[i] = samling[i];
}else if( i >= samling.length) {
total[i] = samling2[i];
}else{
total[i] = samling2[i] + samling[i];
}
}
return total;
}
}

Another variant, with no need for a conditional operation in each iteration of the copying loop:
public static int[] concateArrays(int[] samling, int[] samling2)
{
// max length of the two arrays
int maxLen = Math.max(samling.length,samling2.length);
// decide which of the inputs is shorter and which is longer
int[] shorter = maxLen == samling.length ? samling2 : samling;
int[] longer = maxLen == samling.length ? samling : samling2;
int[] total = new int[maxLen];
// add both as long as there are elements in the shorter
for(int i=0; i < shorter.length; i++) {
total[i] = shorter[i] + longer[i];
}
// copy the remainder of the longer
for(int i=shorter.length; i < longer.length; i++) {
total[i] = longer[i];
}
return total;
}

You can try this :
public static int[] concateArrays(int[] samling, int[] samling2)
{
int minLength = samling.length;
int[] array = null;
if (samling2.length > samling.length) {
array = samling2;
} else {
minLength = samling2.length;
array = samling;
}
for (int i = 0; i < minLength; i++) {
array[i] = samling[i] + samling2[i];
}
return array;
}

How to get unique items from an array?

I am Java beginner, I found a few topics regarding this theme, but none of them worked for me.
I have an array like this:
int[] numbers = {1, 1, 2, 1, 3, 4, 5};
and I would need to get this output:
1, 2, 3, 4, 5
Every item from that array just once.
But how to get it?

The simpliest solution without writing your own algorithm:
Integer[] numbers = {1, 1, 2, 1, 3, 4, 5};
Set<Integer> uniqKeys = new TreeSet<Integer>();
uniqKeys.addAll(Arrays.asList(numbers));
System.out.println("uniqKeys: " + uniqKeys);
Set interface guarantee uniqueness of values. TreeSet additionally sorts this values.

You can use a Set<Integer> and save lot of time since it holds unique elements. If you aren't allowed to use any class from Java Collections, sort the array and count the unique elements. You can sort the array manually or use Arrays#sort.
I'll post the Set<Integer> code:
int[] numbers = {1, 1, 2, 1, 3, 4, 5};
Set<Integer> setUniqueNumbers = new LinkedHashSet<Integer>();
for(int x : numbers) {
setUniqueNumbers.add(x);
}
for(Integer x : setUniqueNumbers) {
System.out.println(x);
}
Note that I prefer to use LinkedHashSet as Set implementation since it maintains the order of how the elements were inserted. This means, if your array was {2 , 1 , 2} then the output will be 2, 1 and not 1, 2.

In Java 8:
final int[] expected = { 1, 2, 3, 4, 5 };
final int[] numbers = { 1, 1, 2, 1, 3, 4, 5 };
final int[] distinct = Arrays.stream(numbers)
.distinct()
.toArray();
Assert.assertArrayEquals(Arrays.toString(distinct), expected, distinct);
final int[] unorderedNumbers = { 5, 1, 2, 1, 4, 3, 5 };
final int[] distinctOrdered = Arrays.stream(unorderedNumbers)
.sorted()
.distinct()
.toArray();
Assert.assertArrayEquals(Arrays.toString(distinctOrdered), expected, distinctOrdered);

//Running total of distinct integers found
int distinctIntegers = 0;
for (int j = 0; j < array.length; j++)
{
//Get the next integer to check
int thisInt = array[j];
//Check if we've seen it before (by checking all array indexes below j)
boolean seenThisIntBefore = false;
for (int i = 0; i < j; i++)
{
if (thisInt == array[i])
{
seenThisIntBefore = true;
}
}
//If we have not seen the integer before, increment the running total of distinct integers
if (!seenThisIntBefore)
{
distinctIntegers++;
}
}

Below code will print unique integers have a look:
printUniqueInteger(new int[]{1, 1, 2, 1, 3, 4, 5});
static void printUniqueInteger(int array[]){
HashMap<Integer, String> map = new HashMap();
for(int i = 0; i < array.length; i++){
map.put(array[i], "test");
}
for(Integer key : map.keySet()){
System.out.println(key);
}
}

Simple Hashing will be far efficient and faster than any Java inbuilt function:
public class Main
{
static int HASH[];
public static void main(String[] args)
{
int[] numbers = {1, 1, 2, 1, 3, 4, 5};
HASH=new int[100000];
for(int i=0;i<numbers.length;i++)
{
if(HASH[numbers[i]]==0)
{
System.out.print(numbers[i]+",");
HASH[numbers[i]]=1;
}
}
}
}
Time Complexity: O(N), where N=numbers.length
DEMO

public class Practice {
public static void main(String[] args) {
List<Integer> list = new LinkedList<>(Arrays.asList(3,7,3,-1,2,3,7,2,15,15));
countUnique(list);
}
public static void countUnique(List<Integer> list){
Collections.sort(list);
Set<Integer> uniqueNumbers = new HashSet<Integer>(list);
System.out.println(uniqueNumbers.size());
}
}

In JAVA8, you can simply use
stream()
and
distinct()
to get unique elements.
intArray = Arrays.stream(intArray).distinct().toArray();

There is an easier way to get a distinct list:
Integer[] intArray = {1,2,3,0,0,2,4,0,2,5,2};
List<Integer> intList = Arrays.asList(intArray); //To List
intList = new ArrayList<>(new LinkedHashSet<>(intList)); //Distinct
Collections.sort(intList); //Optional Sort
intArray = intList.toArray(new Integer[0]); //Back to array
Outputs:
1 2 3 0 0 2 4 0 2 5 2 //Array
1 2 3 0 0 2 4 0 2 5 2 //List
1 2 3 0 4 5 //Distinct List
0 1 2 3 4 5 //Distinct Sorted List
0 1 2 3 4 5 //Distinct Sorted Array
See jDoodle Example

You could do it like this:
int[] numbers = {1, 1, 2, 1, 3, 4, 5};
ArrayList<Integer> store = new ArrayList<Integer>(); // so the size can vary
for (int n = 0; n < numbers.length; n++){
if (!store.contains(numbers[n])){ // if numbers[n] is not in store, then add it
store.add(numbers[n]);
}
}
numbers = new int[store.size()];
for (int n = 0; n < store.size(); n++){
numbers[n] = store.get(n);
}
Integer and int can be (almost) used interchangeably. This piece of code takes your array "numbers" and changes it so that all duplicate numbers are lost. If you want to sort it, you can add Collections.sort(store); before numbers = new int[store.size()]

I don't know if you've solved your issue yet, but my code would be:
int[] numbers = {1, 1, 2, 1, 3, 4, 5};
int x = numbers.length;
int[] unique = new int[x];
int p = 0;
for(int i = 0; i < x; i++)
{
int temp = numbers[i];
int b = 0;
for(int y = 0; y < x; y++)
{
if(unique[y] != temp)
{
b++;
}
}
if(b == x)
{
unique[p] = temp;
p++;
}
}
for(int a = 0; a < p; a++)
{
System.out.print(unique[a]);
if(a < p-1)
{
System.out.print(", ");
}
}

String s1[]= {"hello","hi","j2ee","j2ee","sql","jdbc","hello","jdbc","hybernet","j2ee"};
int c=0;
for(int i=0;i<s1.length;i++)
{
for(int j=i+1;j<s1.length;j++)
{
if(s1[i]==(s1[j]) )
{
c++;
}
}
if(c==0)
{
System.out.println(s1[i]);
}
else
{
c=0;
}
}
}
}

To find out unique data:
public class Uniquedata
{
public static void main(String[] args)
{
int c=0;
String s1[]={"hello","hi","j2ee","j2ee","sql","jdbc","hello","jdbc","hybernet","j2ee","hello","hello","hybernet"};
for(int i=0;i<s1.length;i++)
{
for(int j=i+1;j<s1.length;j++)
{
if(s1[i]==(s1[j]) )
{
c++;
s1[j]="";
}}
if(c==0)
{
System.out.println(s1[i]);
}
else
{
s1[i]="";
c=0;
}
}
}
}

you can use
Object[] array = new HashSet<>(Arrays.asList(numbers)).toArray();

Here is my piece of code using counting sort (partially)
Output is a sorted array consiting of unique elements
void findUniqueElementsInArray(int arr[]) {
int[] count = new int[256];
int outputArrayLength = 0;
for (int i = 0; i < arr.length; i++) {
if (count[arr[i]] < 1) {
count[arr[i]] = count[arr[i]] + 1;
outputArrayLength++;
}
}
for (int i = 1; i < 256; i++) {
count[i] = count[i] + count[i - 1];
}
int[] sortedArray = new int[outputArrayLength];
for (int i = 0; i < arr.length; i++) {
sortedArray[count[arr[i]] - 1] = arr[i];
}
for (int i = 0; i < sortedArray.length; i++) {
System.out.println(sortedArray[i]);
}
}
Reference - discovered this solution while
trying to solve a problem from HackerEarth

If you are a Java programmer, I recommend you to use this.
It will work.
public class DistinctElementsInArray {
//Print all distinct elements in a given array without any duplication
public static void printDistinct(int arr[], int n) {
// Pick all elements one by one
for (int i = 0; i < n; i++) {
// Check if the picked element is already existed
int j;
for (j = 0; j < i; j++)
if (arr[i] == arr[j])
break;
// If not printed earlier, then print it
if (i == j)
System.out.print(arr[i] + " ");
}
}
public static void main(String[] args) {
int array[] = { 4, 5, 9, 5, 4, 6, 6, 5, 4, 10, 6, 4, 5, 3, 8, 4, 8, 3 };
// 4 - 5 5 - 4 9 - 1 6 - 3 10 - 1 3 - 2 8 - 2
int arrayLength = array.length;
printDistinct(array, arrayLength);
}
}

public class DistinctArray {
public static void main(String[] args) {
int num[]={1,2,5,4,1,2,3,5};
for(int i =0;i<num.length;i++)
{
boolean isDistinct=false;
for(int j=0;j<i;j++)
{
if(num[j]==num[i])
{
isDistinct=true;
break;
}
}
if(!isDistinct)
{
System.out.print(num[i]+" ");
}
}
}
}