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I am trying to understand the SelectionSort for arrays.
If I understand it correctly, it compares the element[minimum] with the next elements element[i] and swaps positions if element[i] < element[minimum]. Then it repeats this with element[i + 1] until the list is sorted.
Now when I try it with the following code, I get an incorrect sort and can't figure out what I'm doing wrong.
public static void SelectionSort1(int[] list){
for (int i = 0; i < (list.length - 1); i++){
int min = i;
for (int j = i + 1; j<list.length; j++){
if(list[j] < list[min]){
min = j;
}
int smallernumber = list[i];
list[i] = list[min];
list[min] = smallernumber;
}
}
}
public static void main (String [] args){
int[] list = {5,4,2,1,3};
System.out.println("Before Sort" + Arrays.toString(list));
SelectionSort1(list);
System.out.println("After Sort " + Arrays.toString(list));
}
My output is:
Before Sort[5, 4, 2, 1, 3]
After Sort [2, 3, 1, 4, 5]
Can anyone point me in the right direction?
Your understanding is almost correct, it compares the element[minimum] with the next elements element[j] and if element[j] < element[minimum] then the minimum index is equal to j. Then it repeats until j reaches the end of the array. After it swaps the index at element[i] with element[minimum].
What this is doing is swapping element[i] with the smallest number to the right of index i. Then moving index i to the next spot. Repeating this gives you a sorted array.
TLDR:
Move the swapping outside of the second for loop. It should be:
public static void SelectionSort1(int[] list){
for (int i = 0; i < (list.length - 1); i++){
int min = i;
for (int j = i + 1; j<list.length; j++){
if(list[j] < list[min]){
min = j;
}
}
int smallernumber = list[i];
list[i] = list[min];
list[min] = smallernumber;
}
}
I want to find the K smallest elements in an array, and I was able to basically sort my array into a min-heap structure but I am still getting the wrong output.
Here are my inputs:
arr = [9,4,7,1,-2,6,5]
k = 3
Here's the Code:
public static int[] findKSmallest(int[] arr, int k) {
int[] result = new int[k];
int heapSize = arr.length;
// Write - Your - Code
for (int i = (heapSize - 1) / 2; i >= 0; i--) {
minHeap(arr, i, heapSize);
}
for (int j = 0; j < k; j++) {
result[j] = arr[j];
}
return result;
}
public static void minHeap(int[] arr, int index, int heapSize) {
int smallest = index;
while (smallest < heapSize / 2) {
int left = (2 * index) + 1; // 1 more than half the index
int right = (2 * index) + 2; // 2 more than half the index
if (left < heapSize && arr[left] < arr[index]) {
smallest = left;
}
if (right < heapSize && arr[right] < arr[smallest]) {
smallest = right;
}
if (smallest != index) {
int temp = arr[index];
arr[index] = arr[smallest];
arr[smallest] = temp;
index = smallest;
} else {
break;
}
}
}
Here is my expected output:
[-2,1,4]
Though my output is:
[-2,1,5]
Please I would like to know where I went wrong.
After you build your minHeap you have to extract element and appropriately adjust tree. You simply took elements from array by index, which is not how heap works.
I slightly modified your minHeap() method to use recursion and you should check arr[smallest] < arr[left] not the orther way around.
public static int[] findKSmallest(int[] arr, int k) {
int[] result = new int[k];
int heapSize = arr.length;
// Write - Your - Code
for (int i = (heapSize - 1) / 2; i >= 0; i--) {
minHeap(arr, heapSize, i);
}
// extract elements from heap
for (int i = heapSize - 1; i > 0; i--) {
int temp = arr[0];
arr[0] = arr[i];
arr[i] = temp;
minHeap(arr, 0, i);
}
for (int j = 0; j < k; j++) {
result[j] = arr[j];
}
return result;
}
public static void minHeap(int[] arr, int index, int heapSize) {
int smallest = index;
int left = (2 * index) + 1; // 1 more than half the index
int right = (2 * index) + 2; // 2 more than half the index
if (left < heapSize && arr[smallest] < arr[left]) {
smallest = left;
}
if (right < heapSize && arr[smallest] < arr[right]) {
smallest = right;
}
if (smallest != index) {
int temp = arr[index];
arr[index] = arr[smallest];
arr[smallest] = temp;
minHeap(arr, smallest, heapSize);
}
}
Hope this helps. As expected the result is:
[-2, 1, 4]
You can use Arrays.stream​(int[]) method:
int[] arr = {9, 4, 7, 1, -2, 6, 5};
int k = 3;
int[] minHeap = Arrays.stream(arr).sorted().limit(k).toArray();
System.out.println(Arrays.toString(minHeap)); // [-2, 1, 4]
Since Java 8 we can do Sorting stream
Alternative code:
Arrays.stream(arr).sorted().boxed().collect(Collectors.toList()).subList(0, k);
where:
int[] arr = {9,4,7,1,-2,6,5};
int k = 3; // toIndex
Alternative code in context and testbench:
public static void main(String[] args) {
int[] arr = {9, 4, 7, 1, -2, 6, 5};
int k = 3; // toIndex
List<Integer> listResult = Arrays.stream(arr)
.sorted().boxed().collect(Collectors.toList()).subList(0, k);
// print out list result
System.out.println("Output of list result: " + listResult);
int[] arrayResult = listResult.stream().mapToInt(i -> i).toArray();
// print out array result
List<String> arrayResultAsString = Arrays.stream(arrayResult)
.boxed().map(i -> String.valueOf(i)).collect(Collectors.toList());
System.out.println("Output of array result: " + arrayResultAsString);
}
Output:
Output of list result: [-2, 1, 4]
Output of array result: [-2, 1, 4]
If K is relatively small to the array size my recommendation is to use selection sort, so it may be a design pattern Strategy - depending on K to array size relation with for example two procedures: selection sort for small K and quicksort for huge K. Instead of playing with design patterns good may be also simple if statement e.g. selection<=30%<quick
I was asked to write my own implementation to remove duplicated values in an array. Here is what I have created. But after tests with 1,000,000 elements it took very long time to finish. Is there something that I can do to improve my algorithm or any bugs to remove ?
I need to write my own implementation - not to use Set, HashSet etc. Or any other tools such as iterators. Simply an array to remove duplicates.
public static int[] removeDuplicates(int[] arr) {
int end = arr.length;
for (int i = 0; i < end; i++) {
for (int j = i + 1; j < end; j++) {
if (arr[i] == arr[j]) {
int shiftLeft = j;
for (int k = j+1; k < end; k++, shiftLeft++) {
arr[shiftLeft] = arr[k];
}
end--;
j--;
}
}
}
int[] whitelist = new int[end];
for(int i = 0; i < end; i++){
whitelist[i] = arr[i];
}
return whitelist;
}
you can take the help of Set collection
int end = arr.length;
Set<Integer> set = new HashSet<Integer>();
for(int i = 0; i < end; i++){
set.add(arr[i]);
}
now if you will iterate through this set, it will contain only unique values. Iterating code is like this :
Iterator it = set.iterator();
while(it.hasNext()) {
System.out.println(it.next());
}
If you are allowed to use Java 8 streams:
Arrays.stream(arr).distinct().toArray();
Note: I am assuming the array is sorted.
Code:
int[] input = new int[]{1, 1, 3, 7, 7, 8, 9, 9, 9, 10};
int current = input[0];
boolean found = false;
for (int i = 0; i < input.length; i++) {
if (current == input[i] && !found) {
found = true;
} else if (current != input[i]) {
System.out.print(" " + current);
current = input[i];
found = false;
}
}
System.out.print(" " + current);
output:
1 3 7 8 9 10
Slight modification to the original code itself, by removing the innermost for loop.
public static int[] removeDuplicates(int[] arr){
int end = arr.length;
for (int i = 0; i < end; i++) {
for (int j = i + 1; j < end; j++) {
if (arr[i] == arr[j]) {
/*int shiftLeft = j;
for (int k = j+1; k < end; k++, shiftLeft++) {
arr[shiftLeft] = arr[k];
}*/
arr[j] = arr[end-1];
end--;
j--;
}
}
}
int[] whitelist = new int[end];
/*for(int i = 0; i < end; i++){
whitelist[i] = arr[i];
}*/
System.arraycopy(arr, 0, whitelist, 0, end);
return whitelist;
}
There exists many solution of this problem.
The sort approach
You sort your array and resolve only unique items
The set approach
You declare a HashSet where you put all item then you have only unique ones.
You create a boolean array that represent the items all ready returned, (this depend on your data in the array).
If you deal with large amount of data i would pick the 1. solution. As you do not allocate additional memory and sorting is quite fast. For small set of data the complexity would be n^2 but for large i will be n log n.
Since you can assume the range is between 0-1000 there is a very simple and efficient solution
//Throws an exception if values are not in the range of 0-1000
public static int[] removeDuplicates(int[] arr) {
boolean[] set = new boolean[1001]; //values must default to false
int totalItems = 0;
for (int i = 0; i < arr.length; ++i) {
if (!set[arr[i]]) {
set[arr[i]] = true;
totalItems++;
}
}
int[] ret = new int[totalItems];
int c = 0;
for (int i = 0; i < set.length; ++i) {
if (set[i]) {
ret[c++] = i;
}
}
return ret;
}
This runs in linear time O(n). Caveat: the returned array is sorted so if that is illegal then this answer is invalid.
class Demo
{
public static void main(String[] args)
{
int a[]={3,2,1,4,2,1};
System.out.print("Before Sorting:");
for (int i=0;i<a.length; i++ )
{
System.out.print(a[i]+"\t");
}
System.out.print ("\nAfter Sorting:");
//sorting the elements
for(int i=0;i<a.length;i++)
{
for(int j=i;j<a.length;j++)
{
if(a[i]>a[j])
{
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
//After sorting
for(int i=0;i<a.length;i++)
{
System.out.print(a[i]+"\t");
}
System.out.print("\nAfter removing duplicates:");
int b=0;
a[b]=a[0];
for(int i=0;i<a.length;i++)
{
if (a[b]!=a[i])
{
b++;
a[b]=a[i];
}
}
for (int i=0;i<=b;i++ )
{
System.out.print(a[i]+"\t");
}
}
}
OUTPUT:Before Sortng:3 2 1 4 2 1 After Sorting:1 1 2 2 3 4
Removing Duplicates:1 2 3 4
Since this question is still getting a lot of attention, I decided to answer it by copying this answer from Code Review.SE:
You're following the same philosophy as the bubble sort, which is
very, very, very slow. Have you tried this?:
Sort your unordered array with quicksort. Quicksort is much faster
than bubble sort (I know, you are not sorting, but the algorithm you
follow is almost the same as bubble sort to traverse the array).
Then start removing duplicates (repeated values will be next to each
other). In a for loop you could have two indices: source and
destination. (On each loop you copy source to destination unless they
are the same, and increment both by 1). Every time you find a
duplicate you increment source (and don't perform the copy).
#morgano
import java.util.Arrays;
public class Practice {
public static void main(String[] args) {
int a[] = { 1, 3, 3, 4, 2, 1, 5, 6, 7, 7, 8, 10 };
Arrays.sort(a);
int j = 0;
for (int i = 0; i < a.length - 1; i++) {
if (a[i] != a[i + 1]) {
a[j] = a[i];
j++;
}
}
a[j] = a[a.length - 1];
for (int i = 0; i <= j; i++) {
System.out.println(a[i]);
}
}
}
**This is the most simplest way**
What if you create two boolean arrays: 1 for negative values and 1 for positive values and init it all on false.
Then you cycle thorugh the input array and lookup in the arrays if you've encoutered the value already.
If not, you add it to the output array and mark it as already used.
package com.pari.practice;
import java.util.HashSet;
import java.util.Iterator;
import com.pari.sort.Sort;
public class RemoveDuplicates {
/**
* brute force- o(N square)
*
* #param input
* #return
*/
public static int[] removeDups(int[] input){
boolean[] isSame = new boolean[input.length];
int sameNums = 0;
for( int i = 0; i < input.length; i++ ){
for( int j = i+1; j < input.length; j++){
if( input[j] == input[i] ){ //compare same
isSame[j] = true;
sameNums++;
}
}
}
//compact the array into the result.
int[] result = new int[input.length-sameNums];
int count = 0;
for( int i = 0; i < input.length; i++ ){
if( isSame[i] == true) {
continue;
}
else{
result[count] = input[i];
count++;
}
}
return result;
}
/**
* set - o(N)
* does not guarantee order of elements returned - set property
*
* #param input
* #return
*/
public static int[] removeDups1(int[] input){
HashSet myset = new HashSet();
for( int i = 0; i < input.length; i++ ){
myset.add(input[i]);
}
//compact the array into the result.
int[] result = new int[myset.size()];
Iterator setitr = myset.iterator();
int count = 0;
while( setitr.hasNext() ){
result[count] = (int) setitr.next();
count++;
}
return result;
}
/**
* quicksort - o(Nlogn)
*
* #param input
* #return
*/
public static int[] removeDups2(int[] input){
Sort st = new Sort();
st.quickSort(input, 0, input.length-1); //input is sorted
//compact the array into the result.
int[] intermediateResult = new int[input.length];
int count = 0;
int prev = Integer.MIN_VALUE;
for( int i = 0; i < input.length; i++ ){
if( input[i] != prev ){
intermediateResult[count] = input[i];
count++;
}
prev = input[i];
}
int[] result = new int[count];
System.arraycopy(intermediateResult, 0, result, 0, count);
return result;
}
public static void printArray(int[] input){
for( int i = 0; i < input.length; i++ ){
System.out.print(input[i] + " ");
}
}
public static void main(String[] args){
int[] input = {5,6,8,0,1,2,5,9,11,0};
RemoveDuplicates.printArray(RemoveDuplicates.removeDups(input));
System.out.println();
RemoveDuplicates.printArray(RemoveDuplicates.removeDups1(input));
System.out.println();
RemoveDuplicates.printArray(RemoveDuplicates.removeDups2(input));
}
}
Output:
5 6 8 0 1 2 9 11
0 1 2 5 6 8 9 11
0 1 2 5 6 8 9 11
I have just written the above code for trying out. thanks.
public static int[] removeDuplicates(int[] arr){
HashSet<Integer> set = new HashSet<>();
final int len = arr.length;
//changed end to len
for(int i = 0; i < len; i++){
set.add(arr[i]);
}
int[] whitelist = new int[set.size()];
int i = 0;
for (Iterator<Integer> it = set.iterator(); it.hasNext();) {
whitelist[i++] = it.next();
}
return whitelist;
}
Runs in O(N) time instead of your O(N^3) time
Not a big fun of updating user input, however considering your constraints...
public int[] removeDup(int[] nums) {
Arrays.sort(nums);
int x = 0;
for (int i = 0; i < nums.length; i++) {
if (i == 0 || nums[i] != nums[i - 1]) {
nums[x++] = nums[i];
}
}
return Arrays.copyOf(nums, x);
}
Array sort can be easily replaced with any nlog(n) algorithm.
This is simple way to sort the elements in the array
public class DublicatesRemove {
public static void main(String args[]) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter size of the array");
int l = Integer.parseInt(br.readLine());
int[] a = new int[l];
// insert elements in the array logic
for (int i = 0; i < l; i++)
{
System.out.println("enter a element");
int el = Integer.parseInt(br.readLine());
a[i] = el;
}
// sorting elements in the array logic
for (int i = 0; i < l; i++)
{
for (int j = 0; j < l - 1; j++)
{
if (a[j] > a[j + 1])
{
int temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
// remove duplicate elements logic
int b = 0;
a[b] = a[0];
for (int i = 1; i < l; i++)
{
if (a[b] != a[i])
{
b++;
a[b]=a[i];
}
}
for(int i=0;i<=b;i++)
{
System.out.println(a[i]);
}
}
}
Okay, so you cannot use Set or other collections. One solution I don't see here so far is one based on the use of a Bloom filter, which essentially is an array of bits, so perhaps that passes your requirements.
The Bloom filter is a lovely and very handy technique, fast and space-efficient, that can be used to do a quick check of the existence of an element in a set without storing the set itself or the elements. It has a (typically small) false positive rate, but no false negative rate. In other words, for your question, if a Bloom filter tells you that an element hasn't been seen so far, you can be sure it hasn't. But if it says that an element has been seen, you actually need to check. This still saves a lot of time if there aren't too many duplicates in your list (for those, there is no looping to do, except in the small probability case of a false positive --you typically chose this rate based on how much space you are willing to give to the Bloom filter (rule of thumb: less than 10 bits per unique element for a false positive rate of 1%).
There are many implementations of Bloom filters, see e.g. here or here, so I won't repeat that in this answer. Let us just assume the api described in that last reference, in particular, the description of put(E e):
true if the Bloom filter's bits changed as a result of this operation. If the bits changed, this is definitely the first time object has been added to the filter. If the bits haven't changed, this might be the first time object has been added to the filter. (...)
An implementation using such a Bloom filter would then be:
public static int[] removeDuplicates(int[] arr) {
ArrayList<Integer> out = new ArrayList<>();
int n = arr.length;
BloomFilter<Integer> bf = new BloomFilter<>(...); // decide how many bits and how many hash functions to use (compromise between space and false positive rate)
for (int e : arr) {
boolean might_contain = !bf.put(e);
boolean found = false;
if (might_contain) {
// check if false positive
for (int u : out) {
if (u == e) {
found = true;
break;
}
}
}
if (!found) {
out.add(e);
}
}
return out.stream().mapToInt(i -> i).toArray();
}
Obviously, if you can alter the incoming array in place, then there is no need for an ArrayList: at the end, when you know the actual number of unique elements, just arraycopy() those.
For a sorted Array, just check the next index:
//sorted data!
public static int[] distinct(int[] arr) {
int[] temp = new int[arr.length];
int count = 0;
for (int i = 0; i < arr.length; i++) {
int current = arr[i];
if(count > 0 )
if(temp[count - 1] == current)
continue;
temp[count] = current;
count++;
}
int[] whitelist = new int[count];
System.arraycopy(temp, 0, whitelist, 0, count);
return whitelist;
}
You need to sort your array then then loop and remove duplicates. As you cannot use other tools you need to write be code yourself.
You can easily find examples of quicksort in Java on the internet (on which this example is based).
public static void main(String[] args) throws Exception {
final int[] original = new int[]{1, 1, 2, 8, 9, 8, 4, 7, 4, 9, 1};
System.out.println(Arrays.toString(original));
quicksort(original);
System.out.println(Arrays.toString(original));
final int[] unqiue = new int[original.length];
int prev = original[0];
unqiue[0] = prev;
int count = 1;
for (int i = 1; i < original.length; ++i) {
if (original[i] != prev) {
unqiue[count++] = original[i];
}
prev = original[i];
}
System.out.println(Arrays.toString(unqiue));
final int[] compressed = new int[count];
System.arraycopy(unqiue, 0, compressed, 0, count);
System.out.println(Arrays.toString(compressed));
}
private static void quicksort(final int[] values) {
if (values.length == 0) {
return;
}
quicksort(values, 0, values.length - 1);
}
private static void quicksort(final int[] values, final int low, final int high) {
int i = low, j = high;
int pivot = values[low + (high - low) / 2];
while (i <= j) {
while (values[i] < pivot) {
i++;
}
while (values[j] > pivot) {
j--;
}
if (i <= j) {
swap(values, i, j);
i++;
j--;
}
}
if (low < j) {
quicksort(values, low, j);
}
if (i < high) {
quicksort(values, i, high);
}
}
private static void swap(final int[] values, final int i, final int j) {
final int temp = values[i];
values[i] = values[j];
values[j] = temp;
}
So the process runs in 3 steps.
Sort the array - O(nlgn)
Remove duplicates - O(n)
Compact the array - O(n)
So this improves significantly on your O(n^3) approach.
Output:
[1, 1, 2, 8, 9, 8, 4, 7, 4, 9, 1]
[1, 1, 1, 2, 4, 4, 7, 8, 8, 9, 9]
[1, 2, 4, 7, 8, 9, 0, 0, 0, 0, 0]
[1, 2, 4, 7, 8, 9]
EDIT
OP states values inside array doesn't matter really. But I can assume that range is between 0-1000. This is a classic case where an O(n) sort can be used.
We create an array of size range +1, in this case 1001. We then loop over the data and increment the values on each index corresponding to the datapoint.
We can then compact the resulting array, dropping values the have not been incremented. This makes the values unique as we ignore the count.
public static void main(String[] args) throws Exception {
final int[] original = new int[]{1, 1, 2, 8, 9, 8, 4, 7, 4, 9, 1, 1000, 1000};
System.out.println(Arrays.toString(original));
final int[] buckets = new int[1001];
for (final int i : original) {
buckets[i]++;
}
final int[] unique = new int[original.length];
int count = 0;
for (int i = 0; i < buckets.length; ++i) {
if (buckets[i] > 0) {
unique[count++] = i;
}
}
final int[] compressed = new int[count];
System.arraycopy(unique, 0, compressed, 0, count);
System.out.println(Arrays.toString(compressed));
}
Output:
[1, 1, 2, 8, 9, 8, 4, 7, 4, 9, 1, 1000, 1000]
[1, 2, 4, 7, 8, 9, 1000]
public static void main(String args[]) {
int[] intarray = {1,2,3,4,5,1,2,3,4,5,1,2,3,4,5};
Set<Integer> set = new HashSet<Integer>();
for(int i : intarray) {
set.add(i);
}
Iterator<Integer> setitr = set.iterator();
for(int pos=0; pos < intarray.length; pos ++) {
if(pos < set.size()) {
intarray[pos] =setitr.next();
} else {
intarray[pos]= 0;
}
}
for(int i: intarray)
System.out.println(i);
}
I know this is kinda dead but I just wrote this for my own use. It's more or less the same as adding to a hashset and then pulling all the elements out of it. It should run in O(nlogn) worst case.
public static int[] removeDuplicates(int[] numbers) {
Entry[] entries = new Entry[numbers.length];
int size = 0;
for (int i = 0 ; i < numbers.length ; i++) {
int nextVal = numbers[i];
int index = nextVal % entries.length;
Entry e = entries[index];
if (e == null) {
entries[index] = new Entry(nextVal);
size++;
} else {
if(e.insert(nextVal)) {
size++;
}
}
}
int[] result = new int[size];
int index = 0;
for (int i = 0 ; i < entries.length ; i++) {
Entry current = entries[i];
while (current != null) {
result[i++] = current.value;
current = current.next;
}
}
return result;
}
public static class Entry {
int value;
Entry next;
Entry(int value) {
this.value = value;
}
public boolean insert(int newVal) {
Entry current = this;
Entry prev = null;
while (current != null) {
if (current.value == newVal) {
return false;
} else if(current.next != null) {
prev = current;
current = next;
}
}
prev.next = new Entry(value);
return true;
}
}
int tempvar=0; //Variable for the final array without any duplicates
int whilecount=0; //variable for while loop
while(whilecount<(nsprtable*2)-1) //nsprtable can be any number
{
//to check whether the next value is idential in case of sorted array
if(temparray[whilecount]!=temparray[whilecount+1])
{
finalarray[tempvar]=temparray[whilecount];
tempvar++;
whilecount=whilecount+1;
}
else if (temparray[whilecount]==temparray[whilecount+1])
{
finalarray[tempvar]=temparray[whilecount];
tempvar++;
whilecount=whilecount+2;
}
}
Hope this helps or solves the purpose.
package javaa;
public class UniqueElementinAnArray
{
public static void main(String[] args)
{
int[] a = {10,10,10,10,10,100};
int[] output = new int[a.length];
int count = 0;
int num = 0;
//Iterate over an array
for(int i=0; i<a.length; i++)
{
num=a[i];
boolean flag = check(output,num);
if(flag==false)
{
output[count]=num;
++count;
}
}
//print the all the elements from an array except zero's (0)
for (int i : output)
{
if(i!=0 )
System.out.print(i+" ");
}
}
/***
* If a next number from an array is already exists in unique array then return true else false
* #param arr Unique number array. Initially this array is an empty.
* #param num Number to be search in unique array. Whether it is duplicate or unique.
* #return true: If a number is already exists in an array else false
*/
public static boolean check(int[] arr, int num)
{
boolean flag = false;
for(int i=0;i<arr.length; i++)
{
if(arr[i]==num)
{
flag = true;
break;
}
}
return flag;
}
}
public static int[] removeDuplicates(int[] arr) {
int end = arr.length;
HashSet<Integer> set = new HashSet<Integer>(end);
for(int i = 0 ; i < end ; i++){
set.add(arr[i]);
}
return set.toArray();
}
You can use an auxiliary array (temp) which in indexes are numbers of main array. So the time complexity will be liner and O(n). As we want to do it without using any library, we define another array (unique) to push non-duplicate elements:
var num = [2,4,9,4,1,2,24,12,4];
let temp = [];
let unique = [];
let j = 0;
for (let i = 0; i < num.length; i++){
if (temp[num[i]] !== 1){
temp[num[i]] = 1;
unique[j++] = num[i];
}
}
console.log(unique);
If you are looking to remove duplicates using the same array and also keeping the time complexity of O(n). Then this should do the trick. Also, would only work if the array is sorted.
function removeDuplicates_sorted(arr){
let j = 0;
for(let x = 0; x < arr.length - 1; x++){
if(arr[x] != arr[x + 1]){
arr[j++] = arr[x];
}
}
arr[j++] = arr[arr.length - 1];
arr.length = j;
return arr;
}
Here is for an unsorted array, its O(n) but uses more space complexity then the sorted.
function removeDuplicates_unsorted(arr){
let map = {};
let j = 0;
for(var numbers of arr){
if(!map[numbers]){
map[numbers] = 1;
arr[j++] = numbers;
}
}
arr.length = j;
return arr;
}
Note to other readers who desire to use the Set method of solving this problem: If original ordering must be preserved, do not use HashSet as in the top result. HashSet does not guarantee the preservation of the original order, so LinkedHashSet should be used instead-this keeps track of the order in which the elements were inserted into the set and returns them in that order.
This is an interview question.
public class Test4 {
public static void main(String[] args) {
int a[] = {1, 2, 2, 3, 3, 3, 6,6,6,6,6,66,7,65};
int newlength = lengthofarraywithoutduplicates(a);
for(int i = 0 ; i < newlength ;i++) {
System.out.println(a[i]);
}//for
}//main
private static int lengthofarraywithoutduplicates(int[] a) {
int count = 1 ;
for (int i = 1; i < a.length; i++) {
int ch = a[i];
if(ch != a[i-1]) {
a[count++] = ch;
}//if
}//for
return count;
}//fix
}//end1
But, it's always better to use Stream :
int[] a = {1, 2, 2, 3, 3, 3, 6,6,6,6,6,66,7,65};
int[] array = Arrays.stream(a).distinct().toArray();
System.out.println(Arrays.toString(array));//[1, 2, 3, 6, 66, 7, 65]
How about this one, only for the sorted Array of numbers, to print the Array without duplicates, without using Set or other Collections, just an Array:
public static int[] removeDuplicates(int[] array) {
int[] nums = new int[array.length];
int addedNumber = 0;
int j = 0;
for(int i=0; i < array.length; i++) {
if (addedNumber != array[i]) {
nums[j] = array[i];
j++;
addedNumber = nums[j-1];
}
}
return Arrays.copyOf(nums, j);
}
An array of 1040 duplicated numbers processed in 33020 nanoseconds(0.033020 millisec).
public static void main(String[] args) {
Integer[] intArray = { 1, 1, 1, 2, 4, 2, 3, 5, 3, 6, 7, 3, 4, 5 };
Integer[] finalArray = removeDuplicates(intArray);
System.err.println(Arrays.asList(finalArray));
}
private static Integer[] removeDuplicates(Integer[] intArray) {
int count = 0;
Integer[] interimArray = new Integer[intArray.length];
for (int i = 0; i < intArray.length; i++) {
boolean exists = false;
for (int j = 0; j < interimArray.length; j++) {
if (interimArray[j]!=null && interimArray[j] == intArray[i]) {
exists = true;
}
}
if (!exists) {
interimArray[count] = intArray[i];
count++;
}
}
final Integer[] finalArray = new Integer[count];
System.arraycopy(interimArray, 0, finalArray, 0, count);
return finalArray;
}
I feel Android Killer's idea is great, but I just wondered if we can leverage HashMap. So I did a little experiment. And I found HashMap seems faster than HashSet.
Here is code:
int[] input = new int[1000000];
for (int i = 0; i < input.length; i++) {
Random random = new Random();
input[i] = random.nextInt(200000);
}
long startTime1 = new Date().getTime();
System.out.println("Set start time:" + startTime1);
Set<Integer> resultSet = new HashSet<Integer>();
for (int i = 0; i < input.length; i++) {
resultSet.add(input[i]);
}
long endTime1 = new Date().getTime();
System.out.println("Set end time:"+ endTime1);
System.out.println("result of set:" + (endTime1 - startTime1));
System.out.println("number of Set:" + resultSet.size() + "\n");
long startTime2 = new Date().getTime();
System.out.println("Map start time:" + startTime1);
Map<Integer, Integer> resultMap = new HashMap<Integer, Integer>();
for (int i = 0; i < input.length; i++) {
if (!resultMap.containsKey(input[i]))
resultMap.put(input[i], input[i]);
}
long endTime2 = new Date().getTime();
System.out.println("Map end Time:" + endTime2);
System.out.println("result of Map:" + (endTime2 - startTime2));
System.out.println("number of Map:" + resultMap.size());
Here is result:
Set start time:1441960583837
Set end time:1441960583917
result of set:80
number of Set:198652
Map start time:1441960583837
Map end Time:1441960583983
result of Map:66
number of Map:198652
This is not using Set, Map, List or any extra collection, only two arrays:
package arrays.duplicates;
import java.lang.reflect.Array;
import java.util.Arrays;
public class ArrayDuplicatesRemover<T> {
public static <T> T[] removeDuplicates(T[] input, Class<T> clazz) {
T[] output = (T[]) Array.newInstance(clazz, 0);
for (T t : input) {
if (!inArray(t, output)) {
output = Arrays.copyOf(output, output.length + 1);
output[output.length - 1] = t;
}
}
return output;
}
private static <T> boolean inArray(T search, T[] array) {
for (T element : array) {
if (element.equals(search)) {
return true;
}
}
return false;
}
}
And the main to test it
package arrays.duplicates;
import java.util.Arrays;
public class TestArrayDuplicates {
public static void main(String[] args) {
Integer[] array = {1, 1, 2, 2, 3, 3, 3, 3, 4};
testArrayDuplicatesRemover(array);
}
private static void testArrayDuplicatesRemover(Integer[] array) {
final Integer[] expectedResult = {1, 2, 3, 4};
Integer[] arrayWithoutDuplicates = ArrayDuplicatesRemover.removeDuplicates(array, Integer.class);
System.out.println("Array without duplicates is supposed to be: " + Arrays.toString(expectedResult));
System.out.println("Array without duplicates currently is: " + Arrays.toString(arrayWithoutDuplicates));
System.out.println("Is test passed ok?: " + (Arrays.equals(arrayWithoutDuplicates, expectedResult) ? "YES" : "NO"));
}
}
And the output:
Array without duplicates is supposed to be: [1, 2, 3, 4]
Array without duplicates currently is: [1, 2, 3, 4]
Is test passed ok?: YES
I was asked to write my own implementation to remove duplicated values in an array. Here is what I have created. But after tests with 1,000,000 elements it took very long time to finish. Is there something that I can do to improve my algorithm or any bugs to remove ?
I need to write my own implementation - not to use Set, HashSet etc. Or any other tools such as iterators. Simply an array to remove duplicates.
public static int[] removeDuplicates(int[] arr) {
int end = arr.length;
for (int i = 0; i < end; i++) {
for (int j = i + 1; j < end; j++) {
if (arr[i] == arr[j]) {
int shiftLeft = j;
for (int k = j+1; k < end; k++, shiftLeft++) {
arr[shiftLeft] = arr[k];
}
end--;
j--;
}
}
}
int[] whitelist = new int[end];
for(int i = 0; i < end; i++){
whitelist[i] = arr[i];
}
return whitelist;
}
you can take the help of Set collection
int end = arr.length;
Set<Integer> set = new HashSet<Integer>();
for(int i = 0; i < end; i++){
set.add(arr[i]);
}
now if you will iterate through this set, it will contain only unique values. Iterating code is like this :
Iterator it = set.iterator();
while(it.hasNext()) {
System.out.println(it.next());
}
If you are allowed to use Java 8 streams:
Arrays.stream(arr).distinct().toArray();
Note: I am assuming the array is sorted.
Code:
int[] input = new int[]{1, 1, 3, 7, 7, 8, 9, 9, 9, 10};
int current = input[0];
boolean found = false;
for (int i = 0; i < input.length; i++) {
if (current == input[i] && !found) {
found = true;
} else if (current != input[i]) {
System.out.print(" " + current);
current = input[i];
found = false;
}
}
System.out.print(" " + current);
output:
1 3 7 8 9 10
Slight modification to the original code itself, by removing the innermost for loop.
public static int[] removeDuplicates(int[] arr){
int end = arr.length;
for (int i = 0; i < end; i++) {
for (int j = i + 1; j < end; j++) {
if (arr[i] == arr[j]) {
/*int shiftLeft = j;
for (int k = j+1; k < end; k++, shiftLeft++) {
arr[shiftLeft] = arr[k];
}*/
arr[j] = arr[end-1];
end--;
j--;
}
}
}
int[] whitelist = new int[end];
/*for(int i = 0; i < end; i++){
whitelist[i] = arr[i];
}*/
System.arraycopy(arr, 0, whitelist, 0, end);
return whitelist;
}
There exists many solution of this problem.
The sort approach
You sort your array and resolve only unique items
The set approach
You declare a HashSet where you put all item then you have only unique ones.
You create a boolean array that represent the items all ready returned, (this depend on your data in the array).
If you deal with large amount of data i would pick the 1. solution. As you do not allocate additional memory and sorting is quite fast. For small set of data the complexity would be n^2 but for large i will be n log n.
Since you can assume the range is between 0-1000 there is a very simple and efficient solution
//Throws an exception if values are not in the range of 0-1000
public static int[] removeDuplicates(int[] arr) {
boolean[] set = new boolean[1001]; //values must default to false
int totalItems = 0;
for (int i = 0; i < arr.length; ++i) {
if (!set[arr[i]]) {
set[arr[i]] = true;
totalItems++;
}
}
int[] ret = new int[totalItems];
int c = 0;
for (int i = 0; i < set.length; ++i) {
if (set[i]) {
ret[c++] = i;
}
}
return ret;
}
This runs in linear time O(n). Caveat: the returned array is sorted so if that is illegal then this answer is invalid.
class Demo
{
public static void main(String[] args)
{
int a[]={3,2,1,4,2,1};
System.out.print("Before Sorting:");
for (int i=0;i<a.length; i++ )
{
System.out.print(a[i]+"\t");
}
System.out.print ("\nAfter Sorting:");
//sorting the elements
for(int i=0;i<a.length;i++)
{
for(int j=i;j<a.length;j++)
{
if(a[i]>a[j])
{
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
//After sorting
for(int i=0;i<a.length;i++)
{
System.out.print(a[i]+"\t");
}
System.out.print("\nAfter removing duplicates:");
int b=0;
a[b]=a[0];
for(int i=0;i<a.length;i++)
{
if (a[b]!=a[i])
{
b++;
a[b]=a[i];
}
}
for (int i=0;i<=b;i++ )
{
System.out.print(a[i]+"\t");
}
}
}
OUTPUT:Before Sortng:3 2 1 4 2 1 After Sorting:1 1 2 2 3 4
Removing Duplicates:1 2 3 4
Since this question is still getting a lot of attention, I decided to answer it by copying this answer from Code Review.SE:
You're following the same philosophy as the bubble sort, which is
very, very, very slow. Have you tried this?:
Sort your unordered array with quicksort. Quicksort is much faster
than bubble sort (I know, you are not sorting, but the algorithm you
follow is almost the same as bubble sort to traverse the array).
Then start removing duplicates (repeated values will be next to each
other). In a for loop you could have two indices: source and
destination. (On each loop you copy source to destination unless they
are the same, and increment both by 1). Every time you find a
duplicate you increment source (and don't perform the copy).
#morgano
import java.util.Arrays;
public class Practice {
public static void main(String[] args) {
int a[] = { 1, 3, 3, 4, 2, 1, 5, 6, 7, 7, 8, 10 };
Arrays.sort(a);
int j = 0;
for (int i = 0; i < a.length - 1; i++) {
if (a[i] != a[i + 1]) {
a[j] = a[i];
j++;
}
}
a[j] = a[a.length - 1];
for (int i = 0; i <= j; i++) {
System.out.println(a[i]);
}
}
}
**This is the most simplest way**
What if you create two boolean arrays: 1 for negative values and 1 for positive values and init it all on false.
Then you cycle thorugh the input array and lookup in the arrays if you've encoutered the value already.
If not, you add it to the output array and mark it as already used.
package com.pari.practice;
import java.util.HashSet;
import java.util.Iterator;
import com.pari.sort.Sort;
public class RemoveDuplicates {
/**
* brute force- o(N square)
*
* #param input
* #return
*/
public static int[] removeDups(int[] input){
boolean[] isSame = new boolean[input.length];
int sameNums = 0;
for( int i = 0; i < input.length; i++ ){
for( int j = i+1; j < input.length; j++){
if( input[j] == input[i] ){ //compare same
isSame[j] = true;
sameNums++;
}
}
}
//compact the array into the result.
int[] result = new int[input.length-sameNums];
int count = 0;
for( int i = 0; i < input.length; i++ ){
if( isSame[i] == true) {
continue;
}
else{
result[count] = input[i];
count++;
}
}
return result;
}
/**
* set - o(N)
* does not guarantee order of elements returned - set property
*
* #param input
* #return
*/
public static int[] removeDups1(int[] input){
HashSet myset = new HashSet();
for( int i = 0; i < input.length; i++ ){
myset.add(input[i]);
}
//compact the array into the result.
int[] result = new int[myset.size()];
Iterator setitr = myset.iterator();
int count = 0;
while( setitr.hasNext() ){
result[count] = (int) setitr.next();
count++;
}
return result;
}
/**
* quicksort - o(Nlogn)
*
* #param input
* #return
*/
public static int[] removeDups2(int[] input){
Sort st = new Sort();
st.quickSort(input, 0, input.length-1); //input is sorted
//compact the array into the result.
int[] intermediateResult = new int[input.length];
int count = 0;
int prev = Integer.MIN_VALUE;
for( int i = 0; i < input.length; i++ ){
if( input[i] != prev ){
intermediateResult[count] = input[i];
count++;
}
prev = input[i];
}
int[] result = new int[count];
System.arraycopy(intermediateResult, 0, result, 0, count);
return result;
}
public static void printArray(int[] input){
for( int i = 0; i < input.length; i++ ){
System.out.print(input[i] + " ");
}
}
public static void main(String[] args){
int[] input = {5,6,8,0,1,2,5,9,11,0};
RemoveDuplicates.printArray(RemoveDuplicates.removeDups(input));
System.out.println();
RemoveDuplicates.printArray(RemoveDuplicates.removeDups1(input));
System.out.println();
RemoveDuplicates.printArray(RemoveDuplicates.removeDups2(input));
}
}
Output:
5 6 8 0 1 2 9 11
0 1 2 5 6 8 9 11
0 1 2 5 6 8 9 11
I have just written the above code for trying out. thanks.
public static int[] removeDuplicates(int[] arr){
HashSet<Integer> set = new HashSet<>();
final int len = arr.length;
//changed end to len
for(int i = 0; i < len; i++){
set.add(arr[i]);
}
int[] whitelist = new int[set.size()];
int i = 0;
for (Iterator<Integer> it = set.iterator(); it.hasNext();) {
whitelist[i++] = it.next();
}
return whitelist;
}
Runs in O(N) time instead of your O(N^3) time
Not a big fun of updating user input, however considering your constraints...
public int[] removeDup(int[] nums) {
Arrays.sort(nums);
int x = 0;
for (int i = 0; i < nums.length; i++) {
if (i == 0 || nums[i] != nums[i - 1]) {
nums[x++] = nums[i];
}
}
return Arrays.copyOf(nums, x);
}
Array sort can be easily replaced with any nlog(n) algorithm.
This is simple way to sort the elements in the array
public class DublicatesRemove {
public static void main(String args[]) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter size of the array");
int l = Integer.parseInt(br.readLine());
int[] a = new int[l];
// insert elements in the array logic
for (int i = 0; i < l; i++)
{
System.out.println("enter a element");
int el = Integer.parseInt(br.readLine());
a[i] = el;
}
// sorting elements in the array logic
for (int i = 0; i < l; i++)
{
for (int j = 0; j < l - 1; j++)
{
if (a[j] > a[j + 1])
{
int temp = a[j];
a[j] = a[j + 1];
a[j + 1] = temp;
}
}
}
// remove duplicate elements logic
int b = 0;
a[b] = a[0];
for (int i = 1; i < l; i++)
{
if (a[b] != a[i])
{
b++;
a[b]=a[i];
}
}
for(int i=0;i<=b;i++)
{
System.out.println(a[i]);
}
}
}
Okay, so you cannot use Set or other collections. One solution I don't see here so far is one based on the use of a Bloom filter, which essentially is an array of bits, so perhaps that passes your requirements.
The Bloom filter is a lovely and very handy technique, fast and space-efficient, that can be used to do a quick check of the existence of an element in a set without storing the set itself or the elements. It has a (typically small) false positive rate, but no false negative rate. In other words, for your question, if a Bloom filter tells you that an element hasn't been seen so far, you can be sure it hasn't. But if it says that an element has been seen, you actually need to check. This still saves a lot of time if there aren't too many duplicates in your list (for those, there is no looping to do, except in the small probability case of a false positive --you typically chose this rate based on how much space you are willing to give to the Bloom filter (rule of thumb: less than 10 bits per unique element for a false positive rate of 1%).
There are many implementations of Bloom filters, see e.g. here or here, so I won't repeat that in this answer. Let us just assume the api described in that last reference, in particular, the description of put(E e):
true if the Bloom filter's bits changed as a result of this operation. If the bits changed, this is definitely the first time object has been added to the filter. If the bits haven't changed, this might be the first time object has been added to the filter. (...)
An implementation using such a Bloom filter would then be:
public static int[] removeDuplicates(int[] arr) {
ArrayList<Integer> out = new ArrayList<>();
int n = arr.length;
BloomFilter<Integer> bf = new BloomFilter<>(...); // decide how many bits and how many hash functions to use (compromise between space and false positive rate)
for (int e : arr) {
boolean might_contain = !bf.put(e);
boolean found = false;
if (might_contain) {
// check if false positive
for (int u : out) {
if (u == e) {
found = true;
break;
}
}
}
if (!found) {
out.add(e);
}
}
return out.stream().mapToInt(i -> i).toArray();
}
Obviously, if you can alter the incoming array in place, then there is no need for an ArrayList: at the end, when you know the actual number of unique elements, just arraycopy() those.
For a sorted Array, just check the next index:
//sorted data!
public static int[] distinct(int[] arr) {
int[] temp = new int[arr.length];
int count = 0;
for (int i = 0; i < arr.length; i++) {
int current = arr[i];
if(count > 0 )
if(temp[count - 1] == current)
continue;
temp[count] = current;
count++;
}
int[] whitelist = new int[count];
System.arraycopy(temp, 0, whitelist, 0, count);
return whitelist;
}
You need to sort your array then then loop and remove duplicates. As you cannot use other tools you need to write be code yourself.
You can easily find examples of quicksort in Java on the internet (on which this example is based).
public static void main(String[] args) throws Exception {
final int[] original = new int[]{1, 1, 2, 8, 9, 8, 4, 7, 4, 9, 1};
System.out.println(Arrays.toString(original));
quicksort(original);
System.out.println(Arrays.toString(original));
final int[] unqiue = new int[original.length];
int prev = original[0];
unqiue[0] = prev;
int count = 1;
for (int i = 1; i < original.length; ++i) {
if (original[i] != prev) {
unqiue[count++] = original[i];
}
prev = original[i];
}
System.out.println(Arrays.toString(unqiue));
final int[] compressed = new int[count];
System.arraycopy(unqiue, 0, compressed, 0, count);
System.out.println(Arrays.toString(compressed));
}
private static void quicksort(final int[] values) {
if (values.length == 0) {
return;
}
quicksort(values, 0, values.length - 1);
}
private static void quicksort(final int[] values, final int low, final int high) {
int i = low, j = high;
int pivot = values[low + (high - low) / 2];
while (i <= j) {
while (values[i] < pivot) {
i++;
}
while (values[j] > pivot) {
j--;
}
if (i <= j) {
swap(values, i, j);
i++;
j--;
}
}
if (low < j) {
quicksort(values, low, j);
}
if (i < high) {
quicksort(values, i, high);
}
}
private static void swap(final int[] values, final int i, final int j) {
final int temp = values[i];
values[i] = values[j];
values[j] = temp;
}
So the process runs in 3 steps.
Sort the array - O(nlgn)
Remove duplicates - O(n)
Compact the array - O(n)
So this improves significantly on your O(n^3) approach.
Output:
[1, 1, 2, 8, 9, 8, 4, 7, 4, 9, 1]
[1, 1, 1, 2, 4, 4, 7, 8, 8, 9, 9]
[1, 2, 4, 7, 8, 9, 0, 0, 0, 0, 0]
[1, 2, 4, 7, 8, 9]
EDIT
OP states values inside array doesn't matter really. But I can assume that range is between 0-1000. This is a classic case where an O(n) sort can be used.
We create an array of size range +1, in this case 1001. We then loop over the data and increment the values on each index corresponding to the datapoint.
We can then compact the resulting array, dropping values the have not been incremented. This makes the values unique as we ignore the count.
public static void main(String[] args) throws Exception {
final int[] original = new int[]{1, 1, 2, 8, 9, 8, 4, 7, 4, 9, 1, 1000, 1000};
System.out.println(Arrays.toString(original));
final int[] buckets = new int[1001];
for (final int i : original) {
buckets[i]++;
}
final int[] unique = new int[original.length];
int count = 0;
for (int i = 0; i < buckets.length; ++i) {
if (buckets[i] > 0) {
unique[count++] = i;
}
}
final int[] compressed = new int[count];
System.arraycopy(unique, 0, compressed, 0, count);
System.out.println(Arrays.toString(compressed));
}
Output:
[1, 1, 2, 8, 9, 8, 4, 7, 4, 9, 1, 1000, 1000]
[1, 2, 4, 7, 8, 9, 1000]
public static void main(String args[]) {
int[] intarray = {1,2,3,4,5,1,2,3,4,5,1,2,3,4,5};
Set<Integer> set = new HashSet<Integer>();
for(int i : intarray) {
set.add(i);
}
Iterator<Integer> setitr = set.iterator();
for(int pos=0; pos < intarray.length; pos ++) {
if(pos < set.size()) {
intarray[pos] =setitr.next();
} else {
intarray[pos]= 0;
}
}
for(int i: intarray)
System.out.println(i);
}
I know this is kinda dead but I just wrote this for my own use. It's more or less the same as adding to a hashset and then pulling all the elements out of it. It should run in O(nlogn) worst case.
public static int[] removeDuplicates(int[] numbers) {
Entry[] entries = new Entry[numbers.length];
int size = 0;
for (int i = 0 ; i < numbers.length ; i++) {
int nextVal = numbers[i];
int index = nextVal % entries.length;
Entry e = entries[index];
if (e == null) {
entries[index] = new Entry(nextVal);
size++;
} else {
if(e.insert(nextVal)) {
size++;
}
}
}
int[] result = new int[size];
int index = 0;
for (int i = 0 ; i < entries.length ; i++) {
Entry current = entries[i];
while (current != null) {
result[i++] = current.value;
current = current.next;
}
}
return result;
}
public static class Entry {
int value;
Entry next;
Entry(int value) {
this.value = value;
}
public boolean insert(int newVal) {
Entry current = this;
Entry prev = null;
while (current != null) {
if (current.value == newVal) {
return false;
} else if(current.next != null) {
prev = current;
current = next;
}
}
prev.next = new Entry(value);
return true;
}
}
int tempvar=0; //Variable for the final array without any duplicates
int whilecount=0; //variable for while loop
while(whilecount<(nsprtable*2)-1) //nsprtable can be any number
{
//to check whether the next value is idential in case of sorted array
if(temparray[whilecount]!=temparray[whilecount+1])
{
finalarray[tempvar]=temparray[whilecount];
tempvar++;
whilecount=whilecount+1;
}
else if (temparray[whilecount]==temparray[whilecount+1])
{
finalarray[tempvar]=temparray[whilecount];
tempvar++;
whilecount=whilecount+2;
}
}
Hope this helps or solves the purpose.
package javaa;
public class UniqueElementinAnArray
{
public static void main(String[] args)
{
int[] a = {10,10,10,10,10,100};
int[] output = new int[a.length];
int count = 0;
int num = 0;
//Iterate over an array
for(int i=0; i<a.length; i++)
{
num=a[i];
boolean flag = check(output,num);
if(flag==false)
{
output[count]=num;
++count;
}
}
//print the all the elements from an array except zero's (0)
for (int i : output)
{
if(i!=0 )
System.out.print(i+" ");
}
}
/***
* If a next number from an array is already exists in unique array then return true else false
* #param arr Unique number array. Initially this array is an empty.
* #param num Number to be search in unique array. Whether it is duplicate or unique.
* #return true: If a number is already exists in an array else false
*/
public static boolean check(int[] arr, int num)
{
boolean flag = false;
for(int i=0;i<arr.length; i++)
{
if(arr[i]==num)
{
flag = true;
break;
}
}
return flag;
}
}
public static int[] removeDuplicates(int[] arr) {
int end = arr.length;
HashSet<Integer> set = new HashSet<Integer>(end);
for(int i = 0 ; i < end ; i++){
set.add(arr[i]);
}
return set.toArray();
}
You can use an auxiliary array (temp) which in indexes are numbers of main array. So the time complexity will be liner and O(n). As we want to do it without using any library, we define another array (unique) to push non-duplicate elements:
var num = [2,4,9,4,1,2,24,12,4];
let temp = [];
let unique = [];
let j = 0;
for (let i = 0; i < num.length; i++){
if (temp[num[i]] !== 1){
temp[num[i]] = 1;
unique[j++] = num[i];
}
}
console.log(unique);
If you are looking to remove duplicates using the same array and also keeping the time complexity of O(n). Then this should do the trick. Also, would only work if the array is sorted.
function removeDuplicates_sorted(arr){
let j = 0;
for(let x = 0; x < arr.length - 1; x++){
if(arr[x] != arr[x + 1]){
arr[j++] = arr[x];
}
}
arr[j++] = arr[arr.length - 1];
arr.length = j;
return arr;
}
Here is for an unsorted array, its O(n) but uses more space complexity then the sorted.
function removeDuplicates_unsorted(arr){
let map = {};
let j = 0;
for(var numbers of arr){
if(!map[numbers]){
map[numbers] = 1;
arr[j++] = numbers;
}
}
arr.length = j;
return arr;
}
Note to other readers who desire to use the Set method of solving this problem: If original ordering must be preserved, do not use HashSet as in the top result. HashSet does not guarantee the preservation of the original order, so LinkedHashSet should be used instead-this keeps track of the order in which the elements were inserted into the set and returns them in that order.
This is an interview question.
public class Test4 {
public static void main(String[] args) {
int a[] = {1, 2, 2, 3, 3, 3, 6,6,6,6,6,66,7,65};
int newlength = lengthofarraywithoutduplicates(a);
for(int i = 0 ; i < newlength ;i++) {
System.out.println(a[i]);
}//for
}//main
private static int lengthofarraywithoutduplicates(int[] a) {
int count = 1 ;
for (int i = 1; i < a.length; i++) {
int ch = a[i];
if(ch != a[i-1]) {
a[count++] = ch;
}//if
}//for
return count;
}//fix
}//end1
But, it's always better to use Stream :
int[] a = {1, 2, 2, 3, 3, 3, 6,6,6,6,6,66,7,65};
int[] array = Arrays.stream(a).distinct().toArray();
System.out.println(Arrays.toString(array));//[1, 2, 3, 6, 66, 7, 65]
How about this one, only for the sorted Array of numbers, to print the Array without duplicates, without using Set or other Collections, just an Array:
public static int[] removeDuplicates(int[] array) {
int[] nums = new int[array.length];
int addedNumber = 0;
int j = 0;
for(int i=0; i < array.length; i++) {
if (addedNumber != array[i]) {
nums[j] = array[i];
j++;
addedNumber = nums[j-1];
}
}
return Arrays.copyOf(nums, j);
}
An array of 1040 duplicated numbers processed in 33020 nanoseconds(0.033020 millisec).
public static void main(String[] args) {
Integer[] intArray = { 1, 1, 1, 2, 4, 2, 3, 5, 3, 6, 7, 3, 4, 5 };
Integer[] finalArray = removeDuplicates(intArray);
System.err.println(Arrays.asList(finalArray));
}
private static Integer[] removeDuplicates(Integer[] intArray) {
int count = 0;
Integer[] interimArray = new Integer[intArray.length];
for (int i = 0; i < intArray.length; i++) {
boolean exists = false;
for (int j = 0; j < interimArray.length; j++) {
if (interimArray[j]!=null && interimArray[j] == intArray[i]) {
exists = true;
}
}
if (!exists) {
interimArray[count] = intArray[i];
count++;
}
}
final Integer[] finalArray = new Integer[count];
System.arraycopy(interimArray, 0, finalArray, 0, count);
return finalArray;
}
I feel Android Killer's idea is great, but I just wondered if we can leverage HashMap. So I did a little experiment. And I found HashMap seems faster than HashSet.
Here is code:
int[] input = new int[1000000];
for (int i = 0; i < input.length; i++) {
Random random = new Random();
input[i] = random.nextInt(200000);
}
long startTime1 = new Date().getTime();
System.out.println("Set start time:" + startTime1);
Set<Integer> resultSet = new HashSet<Integer>();
for (int i = 0; i < input.length; i++) {
resultSet.add(input[i]);
}
long endTime1 = new Date().getTime();
System.out.println("Set end time:"+ endTime1);
System.out.println("result of set:" + (endTime1 - startTime1));
System.out.println("number of Set:" + resultSet.size() + "\n");
long startTime2 = new Date().getTime();
System.out.println("Map start time:" + startTime1);
Map<Integer, Integer> resultMap = new HashMap<Integer, Integer>();
for (int i = 0; i < input.length; i++) {
if (!resultMap.containsKey(input[i]))
resultMap.put(input[i], input[i]);
}
long endTime2 = new Date().getTime();
System.out.println("Map end Time:" + endTime2);
System.out.println("result of Map:" + (endTime2 - startTime2));
System.out.println("number of Map:" + resultMap.size());
Here is result:
Set start time:1441960583837
Set end time:1441960583917
result of set:80
number of Set:198652
Map start time:1441960583837
Map end Time:1441960583983
result of Map:66
number of Map:198652
This is not using Set, Map, List or any extra collection, only two arrays:
package arrays.duplicates;
import java.lang.reflect.Array;
import java.util.Arrays;
public class ArrayDuplicatesRemover<T> {
public static <T> T[] removeDuplicates(T[] input, Class<T> clazz) {
T[] output = (T[]) Array.newInstance(clazz, 0);
for (T t : input) {
if (!inArray(t, output)) {
output = Arrays.copyOf(output, output.length + 1);
output[output.length - 1] = t;
}
}
return output;
}
private static <T> boolean inArray(T search, T[] array) {
for (T element : array) {
if (element.equals(search)) {
return true;
}
}
return false;
}
}
And the main to test it
package arrays.duplicates;
import java.util.Arrays;
public class TestArrayDuplicates {
public static void main(String[] args) {
Integer[] array = {1, 1, 2, 2, 3, 3, 3, 3, 4};
testArrayDuplicatesRemover(array);
}
private static void testArrayDuplicatesRemover(Integer[] array) {
final Integer[] expectedResult = {1, 2, 3, 4};
Integer[] arrayWithoutDuplicates = ArrayDuplicatesRemover.removeDuplicates(array, Integer.class);
System.out.println("Array without duplicates is supposed to be: " + Arrays.toString(expectedResult));
System.out.println("Array without duplicates currently is: " + Arrays.toString(arrayWithoutDuplicates));
System.out.println("Is test passed ok?: " + (Arrays.equals(arrayWithoutDuplicates, expectedResult) ? "YES" : "NO"));
}
}
And the output:
Array without duplicates is supposed to be: [1, 2, 3, 4]
Array without duplicates currently is: [1, 2, 3, 4]
Is test passed ok?: YES
I tried to make an implementation of bubble sort, but I am not sure whether it is correct or not. If you can give it a look and if it is a bubble sort and can be done in better way please don't be shy. Here is the code:
package Exercises;
import java.util.*;
public class BubbleSort_6_18
{
public static void main(String[] args)
{
Random generator = new Random();
int[] list = new int[11];
for(int i=0; i<list.length; i++)
{
list[i] = generator.nextInt(10);
}
System.out.println("Original Random array: ");
printArray(list);
bubbleSort(list);
System.out.println("\nAfter bubble sort: ");
printArray(list);
}
public static void bubbleSort(int[] list)
{
for(int i=0; i<list.length; i++)
{
for(int j=i + 1; j<list.length; j++)
{
if(list[i] > list[j])
{
int temp = list[i];
list[i] = list[j];
list[j] = temp;
}
}
}
}
public static void printArray(int[] list)
{
for(int i=0; i<list.length; i++)
{
System.out.print(list[i] + ", ");
}
}
}
private static int [] bublesort (int[] list , int length) {
boolean swap = true;
int temp;
while(swap){
swap = false;
for(int i = 0;i < list.length-1; i++){
if(list[i] > list[i+1]){
temp = list[i];
list[i] = list[i+1];
list[i+1] = temp;
swap = true;
}
}
}
return list;
}
Mohammod Hossain implementation is quite good but he does alot of unecessary iterations, sadly he didnt accept my edit and i can't comment due to reputations points so here is how it should look like:
public void sort(int[] array) {
int temp = 0;
boolean swap = true;
int range = array.length - 1;
while (swap) {
swap = false;
for (int i = 0; i < range; i++) {
if (array[i] > array[i + 1]) {
temp = array[i];
array[i] = array[i + 1];
array[i + 1] = temp;
swap = true;
}
}
range--;
}
}
This is the calssical implementation for bubble sort and it seems to be OK. There are several optimizations that can be done, but the overall idea is the same. Here are some ideas:
If there is an iteration of the outer cycle when no swap is performed in the inner cycle, then break, no use to continue
On each iteration of the outer cycle swap the direction of the inner one - do it once left to right and then do it once right to left(this helps avoid elements moving slowly towards the right end).
{
System.out.println("The Elments Before Sorting:");
for(i=0;i<a.length;i++)
{
System.out.print(a[i]+"\t");
}
for(i=1;i<=a.length-1;i++)
{
for(j=0;j<=a.length-i-1;j++)
{
if((a[j])>(a[j+1]))
{
temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
}
System.out.println("The Elements After Sorting:");
for(i=0;i<a.length;i++)
{
System.out.println(a[i]+"\t");
}
}
}
Short Answer: This is definitely NOT Bubble sort. It is a variant of Selection sort (a less efficient variant than the commonly known one).
It might be helpful to see a visualization of how they work on VisuAlgo
Why this is not bubble sort?
Because you loop over the array and compare each element to each other element on its right. if the right element is smaller you swap. Thus, at the end of the first outer loop iteration you will have the smallest element on the left most position and you have done N swaps in the worst case (think of a reverse-ordered array).
If you think about it, you did not really need to do all these swaps, you could have searched for the minimum value on the right then after you find it you swap. This is simply the idea of Selection sort, you select the min of remaining unsorted elements and put it in its correct position.
How does bubble sort look like then?
In bubble sort you always compare two adjacent elements and bubble the larger one to the right. At the end of the first iteration of the outer loop, you would have the largest element on the right-most position. The swap flag stops the outer loop when the array is already sorted.
void bubbleSort(int[] arr) {
boolean swap = true;
for(int i = arr.length - 1; i > 0 && swap; i--) {
swap = false;
// for the unsorted part of the array, bubble the largest element to the right.
for (int j = 0; j < i; j++) {
if (arr[j] > arr[j+1]) {
// swap
int temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
swap = true;
}
}
}
}
Yes it seems to be Bubble sort swapping the elements
Bubble sort
void bubbleSort(int arr[])
{
int n = arr.length;
for (int i = 0; i < n-1; i++)
for (int j = 0; j < n-i-1; j++)
if (arr[j] > arr[j+1])
{
// swap temp and arr[i]
int temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
It will give in worst case O(n^2) and even if array is sorted.
I think you got the idea of bubble sort by looking at your code:
Bubble sort usually works like the following:
Assume aNumber is some random number:
for (int i = 0; i < aNumber; i++)
{
for(int j = 0; j < aNumber; j++)
//Doing something with i and j, usually running it as a loop for 2D array
//array[i][j] will give you a complete sort.
}
How bubble sort works is it iterates through every single possible spot of the array. i x j times
The down side to this is, it will take square the number of times to sort something. Not very efficient, but it does get the work done in the easiest way.
You can loop over the array until no more elements are swapped
When you put the element at the last position you know it's the largest, so you can recuce the inner loop by 1
A bubblesort version with while loops from my first undergraduate year ("the BlueJ era").
public static void bubbleSort()
{
int[] r = randomArrayDisplayer();
int i = r.length;
while(i!=0){
int j = 0;
while(j!=i-1){
if(r[j+1]<r[j]){
swap(r,j,j+1);
}
j++;
}
i--;
}
}
private static void swap(int[] r, int u, int v)
{
int value = r[u];
r[u] = r[v];
r[v] = value;
arrayDisplayer(r);
}
My advice is to display every step in order to be sure of the correct behaviour.
public class BubbleSort {
public static void main(String[] args) {
int arr[] = {64, 34, 25, 12, 22, 11, 90};
BubbleSort client=new BubbleSort();
int[] result=client.bubbleSort(arr);
for(int i:result)
{
System.out.println(i);
}
}
public int[] bubbleSort(int[] arr)
{
int n=arr.length;
for(int i=0;i<n;i++)
{
for(int j=0;j<n-i-1;j++)
if(arr[j]>arr[j+1])
swap(arr,j,j+1);
}
return arr;
}
private int[] swap(int[] arr, int i, int j) {
int temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
return arr;
}
}
Above code is looks like implementation Selection sort , it's not a bubble sort.
Please find below code for bubble sort.
Class BubbleSort {
public static void main(String []args) {
int n, c, d, swap;
Scanner in = new Scanner(System.in);
System.out.println("Input number of integers to sort");
n = in.nextInt();
int array[] = new int[n];
System.out.println("Enter " + n + " integers");
for (c = 0; c < n; c++)
array[c] = in.nextInt();
for (c = 0; c < ( n - 1 ); c++) {
for (d = 0; d < n - c - 1; d++) {
if (array[d] > array[d+1]) /* For descending order use < */
{
swap = array[d];
array[d] = array[d+1];
array[d+1] = swap;
}
}
}
System.out.println("Sorted list of numbers");
for (c = 0; c < n; c++)
System.out.println(array[c]);
}
}
/*
Implementation of Bubble sort using Java
*/
import java.util.Arrays;
import java.util.Scanner;
public class BubbleSort {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
System.out.println("Enter the number of elements of array");
int n = in.nextInt();
int []a = new int[n];
System.out.println("Enter the integer array");
for(int i=0; i<a.length; i++)
{
a[i]=in.nextInt();
}
System.out.println("UnSorted array: "+ Arrays.toString(a));
for(int i=0; i<n; i++)
{
for(int j=1; j<n; j++)
{
if(a[j-1]>a[j])
{
int temp = a[j-1];
a[j-1]=a[j];
a[j]=temp;
}
}
}
System.out.println("Sorted array: "+ Arrays.toString(a));
}
}
/*
****************************************
Time Complexity: O(n*n)
Space Complexity: O(1)
****************************************
*/
class BubbleSort {
public static void main(String[] args) {
int a[] = {5,4,3,2,1};
int length = a.length - 1;
for (int i = 0 ; i < length ; i++) {
for (int j = 0 ; j < length-i ; j++) {
if (a[j] > a[j+1]) {
int swap = a[j];
a[j] = a[j+1];
a[j+1] = swap;
}
}
}
for (int x : a) {
System.out.println(x);
}
}
}
int[] nums = new int[] { 6, 3, 2, 1, 7, 10, 9 };
for(int i = nums.Length-1; i>=0; i--)
for(int j = 0; j<i; j++)
{
int temp = 0;
if( nums[j] < nums[j + 1])
{
temp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = temp;
}
}
package com.examplehub.sorts;
public class BubbleSort implements Sort {
/**
* BubbleSort algorithm implements.
*
* #param numbers the numbers to be sorted.
*/
public void sort(int[] numbers) {
for (int i = 0; i < numbers.length - 1; ++i) {
boolean swapped = false;
for (int j = 0; j < numbers.length - 1 - i; ++j) {
if (numbers[j] > numbers[j + 1]) {
int temp = numbers[j];
numbers[j] = numbers[j + 1];
numbers[j + 1] = temp;
swapped = true;
}
}
if (!swapped) {
break;
}
}
}
/**
* Generic BubbleSort algorithm implements.
*
* #param array the array to be sorted.
* #param <T> the class of the objects in the array.
*/
public <T extends Comparable<T>> void sort(T[] array) {
for (int i = 0; i < array.length - 1; ++i) {
boolean swapped = false;
for (int j = 0; j < array.length - 1 - i; ++j) {
if (array[j].compareTo(array[j + 1]) > 0) {
T temp = array[j];
array[j] = array[j + 1];
array[j + 1] = temp;
swapped = true;
}
}
if (!swapped) {
break;
}
}
}
}
source from
function bubbleSort(arr,n) {
if (n == 1) // Base case
return;
// One pass of bubble sort. After
// this pass, the largest element
// is moved (or bubbled) to end. and count++
for (let i = 0; i <n-1; i++){
if (arr[i] > arr[i + 1])
{
let temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp;
}
}
// Largest element is fixed,
// recur for remaining array
console.log("Bubble sort Steps ", arr, " Bubble sort array length reduce every recusrion ", n);
bubbleSort(arr, n - 1);
}
let arr1 = [64, 3400, 251, 12, 220, 11, 125]
bubbleSort(arr1, arr1.length);
console.log("Sorted array : ", arr1);
Here's an implementation for the bubble sort algorithm using Stack:
static void bubbleSort(int[] elements) {
Stack<Integer> primaryStack = new Stack<>();
Stack<Integer> secondaryStack = new Stack<>();
int lastIndex = elements.length - 1;
for (int element : elements) {
primaryStack.push(element);
} // Now all the input elements are in primaryStack
// Do the bubble sorting
for (int i = 0; i < elements.length; i++) {
if (i % 2 == 0) sort(elements, i, primaryStack, secondaryStack, lastIndex);
else sort(elements, i, secondaryStack, primaryStack, lastIndex);
}
}
private static void sort(int[] elements, int i, Stack<Integer> stackA, Stack<Integer> stackB, int lastIndex) {
while (!stackA.isEmpty()) { // Move an element from stack A to stack B
int element = stackA.pop();
if (stackB.isEmpty() || element >= stackB.peek()) { // Don't swap, just push
stackB.push(element);
} else { // Swap, then push
int temp = stackB.pop();
stackB.push(element);
stackB.push(temp);
}
}
elements[lastIndex - i] = stackB.pop();
}