Related
Basically I need to find the 2nd maximum number in a array.
Suppose the size of the array is 5 and elements are user-inputed.
Here's my solution:
class Main {
public static void main(String[] args) {
int q = 0;
Scanner sc = new Scanner(System.in);
int[] arr = new int[5];
for (int i = 0; i < 5; i++) {
arr[i] = sc.nextInt();
}
System.out.println(getSecondLargest(arr, 5));
}
public static int getSecondLargest(int[] a, int total) {
int temp;
for (int i = 0; i < total; i++) {
for (int j = i + 1; j < total; j++) {
if (a[i] == a[j]) {
return a[i];
}
if (a[i] > a[j]) {
temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
return a[total - 2];
}
}
Everything works but it fails when the input has the multiple duplicates values of the maximum number. For eg - 5 5 5 4 3 it needs to give 4 as output but it returns 5 as output.
I also tried to simplify code since size is already mentioned:
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int q = 0;
Scanner sc = new Scanner(System.in);
int[] arr = new int[5];
for (int i = 0; i < 5; i++) {
arr[i] = sc.nextInt();
}
print2largest(arr, 5);
}
static void print2largest(int arr[], 5) {
Arrays.sort(arr);
System.out.println(arr[3]);
}
}
But for the above simplified code to work perfectly no duplicate values must be present in the array.
How to get 2nd maximum element in case of there are multiple duplicate values.
You can use Java-Stream to do this:
int[] arr = new int[] {3,5, 9, 7, 4, 12};
int secondLargest = Arrays.stream(arr)
.boxed()
.distinct() // remove duplicates
.sorted(Comparator.comparingInt(value -> (int) value).reversed())
.skip(1) // skip the first largest
.findFirst()
.get();
System.out.println(secondLargest);
Output:
9
You can use a TreeSet with reverse comparing:
static void print2largest(int arr[]) {
// Creates a TreeSet where, each time an element is added, will order it
// with a reverse comparator (so the largest values will come first).
// Since this is a set, it'll ignore duplicated elements.
final Set<Integer> set = new TreeSet<>(Collections.reverseOrder());
// Add all values from the input array to this set
for (int value : arr) set.add(value);
// Transforms the set into a list so we can get the second largest element
final List<Integer> sortedValues = new ArrayList(set);
// Returns the second largest element from this set
System.out.println(sortedValues.get(1));
}
You can do two functions.(Simple algorithm)
-First function:
Remove duplicates.
-Second function:
Sort a given array, and return the element located in `[array.length-2]`.
Remember to check if `array.length >= 2`, else,( `array.length = 1` ), return
array[0].
We can find the second largest number in an array in java by sorting the array and returning the 2nd largest number.
public class SecondLargestInArrayExample {
public static int getSecondLargest(int[] a, int total) {
int temp;
for (int i = 0; i < total; i++) {
for (int j = i + 1; j < total; j++) {
if (a[i] > a[j]) {
temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
return a[total - 2];
}
public static void main(String args[]) {
int a[] = {1, 2, 5, 6, 3, 2};
int b[] = {44, 66, 99, 77, 33, 22, 55};
System.out.println("Second Largest: " + getSecondLargest(a, 6));
System.out.println("Second Largest: " + getSecondLargest(b, 7));
}
}
Output:
Second Largest: 5
Second Largest: 77
Hello I am having difficulty implementing a counting sort method in java. I believe the problem comes from the last two loops I have in the method. I am getting an ArrayIndexOutOfBounds exception : 8. I believe this comes from my second to last for loop when at index 5 the value is 8 but I am not sure how to resolve this. Any help is appreciated. Thank you!
In my code k is the highest value in the input array.
Code:
public static void main(String[] args) {
int [] arrayOne = {0,1,1,3,4,5,3,0};
int [] output = Arrays.copyOf(arrayOne, arrayOne.length);
System.out.println(Arrays.toString(arrayOne));
countingSort(arrayOne, output, 5);
System.out.println(Arrays.toString(output));
}
public static void countingSort(int[] input, int[] output , int k){
int [] temp = Arrays.copyOf(input, k+1);
for (int i = 0; i <= k; i++){
temp[i] = 0;
}
for (int j = 0; j <= input.length - 1; j++){
temp[input[j]] = temp[input[j]] + 1;
}
for (int i = 1; i <= k; i++){
temp[i] = temp[i] + temp[i-1];
}
for (int j = input.length; j >= 1; j--){
output[temp[input[j]]] = input[j];
temp[input[j]] = temp[input[j]] - 1;
}
}
The problem is in the first loop because the array temp lenght is 6 and you are doing 7 interations in there.
So at the end of the for it is trying to do temp[6]=0 and the last position of your array is temp[5].
To fix this change your first loop to:
for (int i = 0; i < k; i++){
In the last loop you will get the same exception cause input[8] doesn't exist.
import java.util.Arrays;
public class CountingSort {
public static void main(String[] args) {
int[] input = {0,1,1,3,4,5,3,0};
int[] output = new int[input.length];
int k = 5; // k is the largest number in the input array
System.out.println("before sorting:");
System.out.println(Arrays.toString(input));
output = countingSort(input, output, k);
System.out.println("after sorting:");
System.out.println(Arrays.toString(output));
}
public static int[] countingSort(int[] input, int[] output, int k) {
int counter[] = new int[k + 1];
for (int i : input) { counter[i]++; }
int ndx = 0;
for (int i = 0; i < counter.length; i++) {
while (0 < counter[i]) {
output[ndx++] = i;
counter[i]--;
}
}
return output;
}
}
Above code is adapted from: http://www.java67.com/2017/06/counting-sort-in-java-example.html
this may help but try using the Arraya.sort() method.
e.g:
//A Java program to sort an array of integers in ascending order.
// A sample Java program to sort an array of integers
// using Arrays.sort(). It by default sorts in
// ascending order
import java.util.Arrays;
public class SortExample
{
public static void main(String[] args)
{
// Our arr contains 8 elements
int[] arr = {13, 7, 6, 45, 21, 9, 101, 102};
Arrays.sort(arr);
System.out.printf("Modified arr[] : %s",
Arrays.toString(arr));
}
}
example is a snippet from https://www.geeksforgeeks.org/arrays-sort-in-java-with-examples/
As per algorithm following implementation, I have prepared for the count sort technique
public static int[] countSort(int elements[]) {
int[] sorted = new int[elements.length+1];
int[] range = new int[getMax(elements)+1];
for(int i=0;i<range.length;i++) {
range[i] = getCount(i, elements);
try {
range[i] = range[i]+range[i-1];
}catch(ArrayIndexOutOfBoundsException ae) {
continue;
}
}
for(int i=0;i<elements.length;i++) {
sorted[range[elements[i]]] = elements[i];
range[elements[i]] = range[elements[i]]-1;
}
return sorted;
}
public static int getCount(int value,int[] elements) {
int count = 0;
for(int element:elements) {
if(element==value) count++;
}
return count;
}
public static int getMax(int elements[]) {
int max = elements[0];
for(int i=0;i<elements.length;i++) {
if(max<elements[i]) {
max = elements[i];
}
}
return max;
}
Please review and let me know if any feedback and it is more helpful.
Note :
Non-negative no won't support in the above implementation.
don't use 0th index of the sorted array.
I'm writing a code that ask at the user to insert the numbers of the array and then write each numbers, do the same thing in another array, and at the end compare the first array with the second array and print out the bubble sort of all numbers, so a kind of bubble sort for the first and second array togheter. I wrote this below, but I don t know how to compare with one method the two different arrays.
public static void main(String[] args) {
public static int[] macello(int[]A){
for(int i=0; i<A.length-1; i++){
for(int j=0; j<A.length-1-i;j++){
if(A[j]>A[j+1]){
int temp = A[j+1];
A[j+1]= A[j];
A[j] = temp;
}
}
}
return A;
}
public static void printArray2(int[]A){
for(int i = 0; i<A.length; i++){
System.out.print(A[i]+",");
}
}
Scanner scan = new Scanner(System.in);
System.out.println("Insert the capacity's array1: ");
int n = scan.nextInt();
int[]numbers1 = {n};
for(int i=0; i<n; i++){
System.out.println("Insert the value of each numbers: ");
int j =0;
numbers1[j] = scan.nextInt();
j++;
}
System.out.println("Insert the capacity's array2: ");
int m = scan.nextInt();
int[]numbers2 = {m};
for(int i=0; i<m; i++){
System.out.println("Insert the value of each numbers: ");
int j=0;
numbers2[j] = scan.nextInt();
j++;
}
macello(Arrays.equals(numbers1,numbers2));
printArray2(Arrays.equals(numbers1,numbers2));
}
}
You mention in the comments that you've already solved bubblesort. So I'm going to assume you have a method with the signature void bubbleSort(int[] arr).
Your code shows you understand how to acquire an array from the user, so we don't need to handle that.
Now what you're describing is bubbleSorting these two arrays. To do this, you need one -big- array that holds them both.
int combinedLength = array1.length + array2.length;
int[] combined = new int[combinedLength];
for(int i = 0; i < array1.length; i++) {
combined[i] = array1[i];
}
for(int i = 0; i < array2.length; i++) {
combined[array1.length + i] = array2[i];
}
// now you can bubbleSort
bubbleSort(combined);
arrayPrint(combined);
Ideally you wrap that logic in a merge method - this particular method leverages the Arrays and System classes to do some of the lifting for you. Obviously you could use the "naive" logic above if you want.
int[] merge(int[] a , int[] b) {
int[] c = Arrays.copyOf(a, a.length + b.length);
System.arraycopy(b,0,c,a.length,b.length);
return c;
}
If you also make a method that acquires an array, like this:
public int[] acquireArray(Scanner sc) {
System.out.println("Length? ");
int len = sc.nextInt();
int[] arr = new int[len];
for(int i = 0; i < len; i++) {
System.out.println("Enter element " + (i+1) + ":");
arr[i] = sc.nextInt();
}
return arr;
}
Then your code becomes very, very clean:
Scanner sc = new Scanner(System.in);
int[] a = acquireArray(sc);
int[] b = acquireArray(sc);
int[] c = merge(a,b);
bubbleSort(c);
arrayPrint(c);
I made a driver to test each of these ideas out to make sure they all work. I am a bit concerned, though, because you mention recursion. As you can see in this driver, there is no recursion here. Also be aware that I take a number of shortcuts that are probably not allowed (such as System.arraycopy, Arrays.copyOf, and Arrays.toString). I just wanted to validate the various pieces of functionality. The message uses 1 indexing because that's what most people think in. If you enter 5 elements, they'll be 1-5. You and I know Java stores them 0 indexed, 0-4. It's just a matter of taste and UX.
import java.util.*;
public class BubbleSort {
public static void main(String...args) {
Scanner sc = new Scanner(System.in);
int[] a = acquireArray(sc);
int[] b = acquireArray(sc);
int[] c = merge(a,b);
bubbleSort(c);
printArray(c);
}
public static int[] acquireArray(Scanner sc) {
System.out.println("Length? ");
int len = sc.nextInt();
int[] arr = new int[len];
for(int i = 0; i < len; i++) {
System.out.println("Enter element " + (i+1) + ":");
arr[i] = sc.nextInt();
}
return arr;
}
public static int[] merge(int[] a , int[] b) {
int[] c = Arrays.copyOf(a, a.length + b.length);
System.arraycopy(b,0,c,a.length,b.length);
return c;
}
public static void bubbleSort(int[] a) {
boolean swapped = true;
int j = 0;
while(swapped) {
swapped = false;
j++;
for(int i = 0; i < a.length - j; i++) {
if(a[i] > a[i+1]) {
int t = a[i];
a[i] = a[i+1];
a[i+1] = t;
swapped = true;
}
}
}
}
public static void printArray(int[] a) {
System.out.println(Arrays.toString(a));
}
}
And here's what I get when I run it
C:\files\j>java BubbleSort
Length?
5
Enter element 1:
1
Enter element 2:
5
Enter element 3:
3
Enter element 4:
9
Enter element 5:
7
Length?
5
Enter element 1:
2
Enter element 2:
6
Enter element 3:
4
Enter element 4:
0
Enter element 5:
8
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
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