I am trying to make an array that has different values in its cells, but for some reason it has repeating values. Where am I going wrong?
Here is my code:
package oefarray;
public class OefArray {
int[] getallenArray,differentArray;
public static void main(String[] args) {
OefArray arr = new OefArray();
arr.differentArray(10,10);
}
public void differentArray(int n, int max) {
differentArray= new int[n];
for (int i = 0; i < differentArray.length; i++) {
int value = (int) (Math.random() * max);
differentArray[i]= value;
for (int p: differentArray){
while (value == p){
value = (int) (Math.random() * max);
}
}
differentArray[i]= value;
System.out.println(differentArray[i]);
}
}
}
You're not checking whether the new generated value exists anywhere in the array, only that its value doesn't equal the current value you're examining.
differentArray= new int[n];
for (int i = 0; i < differentArray.length; i++) {
int value = 0;
while(true){
value = (int)(Math.random()*max);
boolean found = false;
for(int p: differentArray){
if(p==value){
found = true;
break;
}
}
if(!found) break;
}
differentArray[i] = value;
}
Here's an alternative to arshajii's solution which doesn't require a direct reference to ArrayList. As he pointed out, this is no more efficient than his solution. Just another way of writing it if you're not comfortable with Lists yet.
int[] nums = new int[max];
for (int i = 0; i < max; i++)
nums[i] = i;
Collections.shuffle(Arrays.asList(nums));
for (int i = 0; i < n; i++)
differentArray[i] = nums.get(i);
For a futuristic approach to this using Java 8, using IntStream.generate can produce some very terse results. This likely performs in the same performance window as the previous answer, so I make no assertion that this is more efficient. It is, however, more expressive.
public int[] differentArray(int length, int maxValue) {
if(length > maxValue) {
throw new IllegalArgumentException("The number of possible unique values is smaller than available number of slots for them.");
}
final Random random = new Random();
return IntStream.generate(() -> random.nextInt(maxValue))
.distinct()
.limit(length)
.toArray();
}
Related
I have a problem and the title actually sums it up perfectly. So i'll just go ahead and show you the code snippet.
So the methode generate, is generating an array, that is filled with numbers between 1 and 1000, including both. The length of the array is user input.
The next method, isPrime, is gonna conclude if its a prime number, so i can use those numbers with the true condition in another method. The generate method works but in isPrime i always get errors. If u can think of a better way, let me know please.
static int[] generate(int n) {
int[] arr = new int[n+1];
for(int x = 0; x <= n; x ++) {
int number = (int) (Math.random()* 999)+1;
arr[x] = number;
}
return arr;
}
static int isPrime(int p, final int q[]) {
boolean itIs = true;
//final int[] arr;
for(int r = 0; r <= p; r++) { // here it somehow states r is deadCode
for(int j = 2; j < q[r]; j++) {
if(q[r]%j == 0) {
itIs = false;
}
}
return q[r];
}
}
First, create a method to check a value is prime:
public boolean isPrime(int value) {
for (int i = 0; i < value / 2; i++) { // value / 2 is enough, doesn't need to check all values
if (value % i == 0) {
return false;
}
}
return true;
}
Then you check each value of array and put prime value to new array:
public int[] filterArray(int[] array) {
List<Integer> intList = new ArrayList<>();
for (int i = 0; i < array.length; i++) {
if (isPrime(array[i])) {
intList.add(array[i]);
}
}
Integer[] integerArray = intList.toArray(new Integer[intList.size()]);
int[] intArray = ArrayUtils.toPrimitive(integerArray);
return intArray;
}
Then you get the filtered prime array.
this is my practice before my upcoming test, I'm trying to make the user input a number. And all elements in array1 that is below the user's number, will be put in a new ArrayList.
and then I'm trying to print only the highest number in that ArrayList. If the user input is lower than all number in array1, it will return -1.
here is my code, however, when I put 920, it still returns -1, I think there's something wrong with my code to find highest number in ArrayList. Can you guys please tell me what is wrong?
static Scanner sc = new Scanner(System.in);
static int[] array1 = {900, 885, 989, 1000, 1500, 1243, 999, 915};
public static int blabla(int[] a, int b) {
Integer d = -1;
ArrayList<Integer> al = new ArrayList<Integer>();
for (int i = 0; i < array1.length; i++) { // this is to find all numbers in array1 that is below user's number, and add it to the ArrayList
if (b > array1[i]) {
al.add(array1[i]);
} // if
} // for
outerloop: // and this function below is to find maximum number in ArrayList
for (int g = (al.size()-1); g == 0; g--) {
for (int j = 0; j <=(g-1); j++) {
if (al.get(j) > al.get(g)) {
break;
}
else if(j == (g-1)) {
if (al.get(g) > al.get(j)){
d = al.get(g);
break outerloop;
}
}
} //for^2
} // for
return d;
} // priceisright
static Scanner sc = new Scanner(System.in);
static int[] array1 = {900, 885, 989, 1000, 1500, 1243, 999, 915};
public static int blabla(int[] a, int b) {
Integer d = -1;
ArrayList<Integer> al = new ArrayList<Integer>();
At this point, a1 is an empty array, so a1.length = 0, this loop never gets executed.
for (int i = 0; i < a1.length; i++) {
// this is to find all numbers in array1 that is below user's number,
// and add it to the ArrayList
if (b > a1[i]) {
al.add(a1[i]);
} // if
} // for
a1 is still empty there, the second loop won't do anything either.
// and this function below is to find maximum number in ArrayList
outerloop:
for (int g = (al.size()-1); g == 0; g--) {
for (int j = 0; j <=(g-1); j++) {
if (al.get(j) > al.get(g)) {
break;
}
else if(j == (g-1)) {
if (al.get(g) > al.get(j)){
d = al.get(g);
break outerloop;
}
}
} //for^2
} // for
return d;
} // priceisright
What about this:
// Finds the greater value in values that is below maximum.
// Returns -1 if none is found.
public static int blabla(int[] values, int maximum) {
int best_value = -1;
for (int value : values) {
if (value < maximum && value > best_value) {
best_value = value;
}
}
return best_value;
}
You can replace int[] values by List<Integer> values if your values are in an ArrayList.
You can use this
Collections.max(arrayList);
To know more about read Javadoc for Collection.max
If you want to use an ArrayList you can check fro maximum with linear time by,
public static Integer getMaximum(List<Integer> coll) {
if (coll == null) {
return null;
}
Integer i = coll.get(0);
for (int t = 1; t < coll.size(); t++) {
Integer v = coll.get(i);
if (v != null && v > i) {
i = v;
}
}
return i;
}
Or, you could change that to a SortedSet<Integer> set = new TreeSet<Integer>(); then the maximum element is always set.last();
I would simplify your two for loops by this one:
for (int g = 0; g <=(al.size()-1); g++) { //for each value in your new array
d = (al.get(g)>d)? al.get(g):d; //is current value higher than previous? if not keep old one
}
Simple way to find the max in an array or arrayList (or any collection actually) without using built in methods such as .max
Int currentMax = 0
For (int i = 0; i < al.length; i++)
{
If al[i] > currentMax
{
CurrentMax = al[i]
}
}
Answered this from my phone so I apologize for bad indents, but you get the idea :)
So I created an array with random numbers, i printed and counted the repeated numbers, now I just have to create a new array with the same numbers from the first array but without any repetitions. Can't use ArrayList by the way.
What I have is.
public static void main(String[] args) {
Random generator = new Random();
int aR[]= new int[20];
for(int i=0;i<aR.length;i++){
int number=generator.nextInt(51);
aR[i]=number;
System.out.print(aR[i]+" ");
}
System.out.println();
System.out.println();
int countRep=0;
for(int i=0;i<aR.length;i++){
for(int j=i+1;j<aR.length-1;j++){
if(aR[i]==aR[j]){
countRep++;
System.out.println(aR[i]+" "+aR[j]);
break;
}
}
}
System.out.println();
System.out.println("Repeated numbers: "+countRep);
int newaR[]= new int[aR.length - countRep];
}
Can someone help?
EDIT: Can't really use HashSet either. Also the new array needs to have the correct size.
Using Java 8 and streams you can do the following:
int[] array = new int[1024];
//fill array
int[] arrayWithoutDuplicates = Arrays.stream(array)
.distinct()
.toArray();
This will:
Turn your int[] into an IntStream.
Filter out all duplicates, so retaining distinct elements.
Save it in a new array of type int[].
Try:
Set<Integer> insertedNumbers = new HashSet<>(newaR.length);
int index = 0;
for(int i = 0 ; i < aR.length ; ++i) {
if(!insertedNumbers.contains(aR[i])) {
newaR[index++] = aR[i];
}
insertedNumbers.add(aR[i]);
}
One possible approach is to walk through the array, and for each value, compute the index at which it again occurs in the array (which is -1 if the number does not occur again). The number of values which do not occur again is the number of unique values. Then collect all values from the array for which the corresponding index is -1.
import java.util.Arrays;
import java.util.Random;
public class UniqueIntTest
{
public static void main(String[] args)
{
int array[] = createRandomArray(20, 0, 51);
System.out.println("Array " + Arrays.toString(array));
int result[] = computeUnique(array);
System.out.println("Result " + Arrays.toString(result));
}
private static int[] createRandomArray(int size, int min, int max)
{
Random random = new Random(1);
int array[] = new int[size];
for (int i = 0; i < size; i++)
{
array[i] = min + random.nextInt(max - min);
}
return array;
}
private static int[] computeUnique(int array[])
{
int indices[] = new int[array.length];
int unique = computeIndices(array, indices);
int result[] = new int[unique];
int index = 0;
for (int i = 0; i < array.length; i++)
{
if (indices[i] == -1)
{
result[index] = array[i];
index++;
}
}
return result;
}
private static int computeIndices(int array[], int indices[])
{
int unique = 0;
for (int i = 0; i < array.length; i++)
{
int value = array[i];
int index = indexOf(array, value, i + 1);
if (index == -1)
{
unique++;
}
indices[i] = index;
}
return unique;
}
private static int indexOf(int array[], int value, int offset)
{
for (int i = offset; i < array.length; i++)
{
if (array[i] == value)
{
return i;
}
}
return -1;
}
}
This sounds like a homework question, and if it is, the technique that you should pick up on is to sort the array first.
Once the array is sorted, duplicate entries will be adjacent to each other, so they are trivial to find:
int[] numbers = //obtain this however you normally would
java.util.Arrays.sort(numbers);
//find out how big the array is
int sizeWithoutDuplicates = 1; //there will be at least one entry
int lastValue = numbers[0];
//a number in the array is unique (or a first duplicate)
//if it's not equal to the number before it
for(int i = 1; i < numbers.length; i++) {
if (numbers[i] != lastValue) {
lastValue = i;
sizeWithoutDuplicates++;
}
}
//now we know how many results we have, and we can allocate the result array
int[] result = new int[sizeWithoutDuplicates];
//fill the result array
int positionInResult = 1; //there will be at least one entry
result[0] = numbers[0];
lastValue = numbers[0];
for(int i = 1; i < numbers.length; i++) {
if (numbers[i] != lastValue) {
lastValue = i;
result[positionInResult] = i;
positionInResult++;
}
}
//result contains the unique numbers
Not being able to use a list means that we have to figure out how big the array is going to be in a separate pass — if we could use an ArrayList to collect the results we would have only needed a single loop through the array of numbers.
This approach is faster (O(n log n) vs O (n^2)) than a doubly-nested loop through the array to find duplicates. Using a HashSet would be faster still, at O(n).
Find the first covering prefix of a given array.
A non-empty zero-indexed array A consisting of N integers is given. The first covering
prefix of array A is the smallest integer P such that and such that every value that
occurs in array A also occurs in sequence.
For example, the first covering prefix of array A with
A[0]=2, A[1]=2, A[2]=1, A[3]=0, A[4]=1 is 3, because sequence A[0],
A[1], A[2], A[3] equal to 2, 2, 1, 0 contains all values that occur in
array A.
My solution is
int ps ( int[] A )
{
int largestvalue=0;
int index=0;
for(each element in Array){
if(A[i]>largestvalue)
{
largestvalue=A[i];
index=i;
}
}
for(each element in Array)
{
if(A[i]==index)
index=i;
}
return index;
}
But this only works for this input, this is not a generalized solution.
Got 100% with the below.
public int ps (int[] a)
{
var length = a.Length;
var temp = new HashSet<int>();
var result = 0;
for (int i=0; i<length; i++)
{
if (!temp.Contains(a[i]))
{
temp.Add(a[i]);
result = i;
}
}
return result;
}
I would do this
int coveringPrefixIndex(final int[] arr) {
Map<Integer,Integer> indexes = new HashMap<Integer,Integer>();
// start from the back
for(int i = arr.length - 1; i >= 0; i--) {
indexes.put(arr[i],i);
}
// now find the highest value in the map
int highestIndex = 0;
for(Integer i : indexes.values()) {
if(highestIndex < i.intValue()) highestIndex = i.intValue();
}
return highestIndex;
}
Your question is from Alpha 2010 Start Challenge of Codility platform. And here is my solution which got score of 100. The idea is simple, I track an array of counters for the input array. Traversing the input array backwards, decrement the respective counter, if that counter becomes zero it means we have found the first covering prefix.
public static int solution(int[] A) {
int size = A.length;
int[] counters = new int[size];
for (int a : A)
counters[a]++;
for (int i = size - 1; i >= 0; i--) {
if (--counters[A[i]] == 0)
return i;
}
return 0;
}
here's my solution in C#:
public static int CoveringPrefix(int[] Array1)
{
// Step 1. Get length of Array1
int Array1Length = 0;
foreach (int i in Array1) Array1Length++;
// Step 2. Create a second array with the highest value of the first array as its length
int highestNum = 0;
for (int i = 0; i < Array1Length; i++)
{
if (Array1[i] > highestNum) highestNum = Array1[i];
}
highestNum++; // Make array compatible for our operation
int[] Array2 = new int[highestNum];
for (int i = 0; i < highestNum; i++) Array2[i] = 0; // Fill values with zeros
// Step 3. Final operation will determine unique values in Array1 and return the index of the highest unique value
int highestIndex = 0;
for (int i = 0; i < Array1Length; i++)
{
if (Array2[Array1[i]] < 1)
{
Array2[Array1[i]]++;
highestIndex = i;
}
}
return highestIndex;
}
100p
public static int ps(int[] a) {
Set<Integer> temp = new HashSet<Integer>();
int p = 0;
for (int i = 0; i < a.length; i++) {
if (temp.add(a[i])) {
p = i+1;
}
}
return p;
}
You can try this solution as well
import java.util.HashSet;
import java.util.Set;
class Solution {
public int ps ( int[] A ) {
Set set = new HashSet();
int index =-1;
for(int i=0;i<A.length;i++){
if(set.contains(A[i])){
if(index==-1)
index = i;
}else{
index = i;
set.add(A[i]);
}
}
return index;
}
}
Without using any Collection:
search the index of the first occurrence of each element,
the prefix is the maximum of that index. Do it backwards to finish early:
private static int prefix(int[] array) {
int max = -1;
int i = array.length - 1;
while (i > max) {
for (int j = 0; j <= i; j++) { // include i
if (array[i] == array[j]) {
if (j > max) {
max = j;
}
break;
}
}
i--;
}
return max;
}
// TEST
private static void test(int... array) {
int prefix = prefix(array);
int[] segment = Arrays.copyOf(array, prefix+1);
System.out.printf("%s = %d = %s%n", Arrays.toString(array), prefix, Arrays.toString(segment));
}
public static void main(String[] args) {
test(2, 2, 1, 0, 1);
test(2, 2, 1, 0, 4);
test(2, 0, 1, 0, 1, 2);
test(1, 1, 1);
test(1, 2, 3);
test(4);
test(); // empty array
}
This is what I tried first. I got 24%
public int ps ( int[] A ) {
int n = A.length, i = 0, r = 0,j = 0;
for (i=0;i<n;i++) {
for (j=0;j<n;j++) {
if ((long) A[i] == (long) A[j]) {
r += 1;
}
if (r == n) return i;
}
}
return -1;
}
//method must be public for codility to access
public int solution(int A[]){
Set<Integer> set = new HashSet<Integer>(A.length);
int index= A[0];
for (int i = 0; i < A.length; i++) {
if( set.contains(A[i])) continue;
index = i;
set.add(A[i]);
}
return index;
}
this got 100%, however detected time was O(N * log N) due to the HashSet.
your solutions without hashsets i don't really follow...
shortest code possible in java:
public static int solution(int A[]){
Set<Integer> set = new HashSet<Integer>(A.length);//avoid resizing
int index= -1; //value does not matter;
for (int i = 0; i < A.length; i++)
if( !set.contains(A[i])) set.add(A[index = i]); //assignment + eval
return index;
}
I got 100% with this one:
public int solution (int A[]){
int index = -1;
boolean found[] = new boolean[A.length];
for (int i = 0; i < A.length; i++)
if (!found [A[i]] ){
index = i;
found [A[i]] = true;
}
return index;
}
I used a boolean array which keeps track of the read elements.
This is what I did in Java to achieve 100% correctness and 81% performance, using a list to store and compare the values with.
It wasn't quick enough to pass random_n_log_100000 random_n_10000 or random_n_100000 tests, but it is a correct answer.
public int solution(int[] A) {
int N = A.length;
ArrayList<Integer> temp = new ArrayList<Integer>();
for(int i=0; i<N; i++){
if(!temp.contains(A[i])){
temp.add(A[i]);
}
}
for(int j=0; j<N; j++){
if(temp.contains(A[j])){
temp.remove((Object)A[j]);
}
if(temp.isEmpty()){
return j;
}
}
return -1;
}
Correctness and Performance: 100%:
import java.util.HashMap;
class Solution {
public int solution(int[] inputArray)
{
int covering;
int[] A = inputArray;
int N = A.length;
HashMap<Integer, Integer> map = new HashMap<>();
covering = 0;
for (int i = 0; i < N; i++)
{
if (map.get(A[i]) == null)
{
map.put(A[i], A[i]);
covering = i;
}
}
return covering;
}
}
Here is my Objective-C Solution to PrefixSet from Codility. 100% correctness and performance.
What can be changed to make it even more efficient? (without out using c code).
HOW IT WORKS:
Everytime I come across a number in the array I check to see if I have added it to the dictionary yet.
If it is in the dictionary then I know it is not a new number so not important in relation to the problem. If it is a new number that we haven't come across already, then I need to update the indexOftheLastPrefix to this array position and add it to the dictionary as a key.
It only used one for loop so takes just one pass. Objective-c code is quiet heavy so would like to hear of any tweaks to make this go faster. It did get 100% for performance though.
int solution(NSMutableArray *A)
{
NSUInteger arraySize = [A count];
NSUInteger indexOflastPrefix=0;
NSMutableDictionary *myDict = [[NSMutableDictionary alloc] init];
for (int i=0; i<arraySize; i++)
{
if ([myDict objectForKey:[[A objectAtIndex:i]stringValue]])
{
}
else
{
[myDict setValue:#"YES" forKey:[[A objectAtIndex:i]stringValue]];
indexOflastPrefix = i;
}
}
return indexOflastPrefix;
}
int solution(vector &A) {
// write your code in C++11 (g++ 4.8.2)
int max = 0, min = -1;
int maxindex =0,minindex = 0;
min = max =A[0];
for(unsigned int i=1;i<A.size();i++)
{
if(max < A[i] )
{
max = A[i];
maxindex =i;
}
if(min > A[i])
{
min =A[i];
minindex = i;
}
}
if(maxindex > minindex)
return maxindex;
else
return minindex;
}
fwiw: Also gets 100% on codility and it's easy to understand with only one HashMap
public static int solution(int[] A) {
// write your code in Java SE 8
int firstCoveringPrefix = 0;
//HashMap stores unique keys
HashMap hm = new HashMap();
for(int i = 0; i < A.length; i++){
if(!hm.containsKey(A[i])){
hm.put( A[i] , i );
firstCoveringPrefix = i;
}
}
return firstCoveringPrefix;
}
I was looking for the this answer in JavaScript but didn't find it so I convert the Java answer to javascript and got 93%
function solution(A) {
result=0;
temp = [];
for(i=0;i<A.length;i++){
if (!temp.includes(A[i])){
temp.push(A[i]);
result=i;
}
}
return result;
}
// you can also use imports, for example:
import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
Set<Integer> s = new HashSet<Integer>();
int index = 0;
for (int i = 0; i < A.length; i++) {
if (!s.contains(A[i])) {
s.add(A[i]);
index = i;
}
}
return index;
}
}
Hi i need some help to improve my code. I am trying to use Radixsort to sort array of 10 numbers (for example) in increasing order.
When i run the program with array of size 10 and put 10 random int numbers in like
70
309
450
279
799
192
586
609
54
657
i get this out:
450
309
192
279
54
192
586
657
54
609
Don´t see where my error is in the code.
class IntQueue
{
static class Hlekkur
{
int tala;
Hlekkur naest;
}
Hlekkur fyrsti;
Hlekkur sidasti;
int n;
public IntQueue()
{
fyrsti = sidasti = null;
}
// First number in queue.
public int first()
{
return fyrsti.tala;
}
public int get()
{
int res = fyrsti.tala;
n--;
if( fyrsti == sidasti )
fyrsti = sidasti = null;
else
fyrsti = fyrsti.naest;
return res;
}
public void put( int i )
{
Hlekkur nyr = new Hlekkur();
n++;
nyr.tala = i;
if( sidasti==null )
f yrsti = sidasti = nyr;
else
{
sidasti.naest = nyr;
sidasti = nyr;
}
}
public int count()
{
return n;
}
public static void radixSort(int [] q, int n, int d){
IntQueue [] queue = new IntQueue[n];
for (int k = 0; k < n; k++){
queue[k] = new IntQueue();
}
for (int i = d-1; i >=0; i--){
for (int j = 0; j < n; j++){
while(queue[j].count() != 0)
{
queue[j].get();
}
}
for (int index = 0; index < n; index++){
// trying to look at one of three digit to sort after.
int v=1;
int digit = (q[index]/v)%10;
v*=10;
queue[digit].put(q[index]);
}
for (int p = 0; p < n; p++){
while(queue[p].count() != 0) {
q[p] = (queue[p].get());
}
}
}
}
}
I am also thinking can I let the function take one queue as an
argument and on return that queue is in increasing order? If so how?
Please help. Sorry if my english is bad not so good in it.
Please let know if you need more details.
import java.util.Random;
public class RadTest extends IntQueue {
public static void main(String[] args)
{
int [] q = new int[10];
Random r = new Random();
int t = 0;
int size = 10;
while(t != size)
{
q[t] = (r.nextInt(1000));
t++;
}
for(int i = 0; i!= size; i++)
{
System.out.println(q[i]);
}
System.out.println("Radad: \n");
radixSort(q,size,3);
for(int i = 0; i!= size; i++)
{
System.out.println(q[i]);
}
}
}
Hope this is what you were talking about...
Thank you for your answer, I will look into it. Not looking for someone to solve the problem for me. Looking for help and Ideas how i can solve it.
in my task it says:
Implement a radix sort function for integers that sorts with queues.
The function should take one queue as an
argument and on return that queue should contain the same values in ascending
order You may assume that the values are between 0 and 999.
Can i put 100 int numbers on my queue and use radixsort function to sort it or do i need to put numbers in array and then array in radixsort function which use queues?
I understand it like i needed to put numbers in Int queue and put that queue into the function but that has not worked.
But Thank for your answers will look at them and try to solve my problem. But if you think you can help please leave comment.
This works for the test cases I tried. It's not entirely well documented, but I think that's okay. I'll leave it to you to read it, compare it to what you're currently doing, and find out why what you have might be different than mine in philosophy. There's also other things that are marked where I did them the "lazy" way, and you should do them a better way.
import java.util.*;
class Radix {
static int[] radixSort(int[] arr) {
// Bucket is only used in this method, so I declare it here
// I'm not 100% sure I recommend doing this in production code
// but it turns out, it's perfectly legal to do!
class Bucket {
private List<Integer> list = new LinkedList<Integer>();
int[] sorted;
public void add(int i) { list.add(i); sorted = null;}
public int[] getSortedArray() {
if(sorted == null) {
sorted = new int[list.size()];
int i = 0;
for(Integer val : list) {
sorted[i++] = val.intValue(); // probably could autobox, oh well
}
Arrays.sort(sorted); // use whatever method you want to sort here...
// Arrays.sort probably isn't allowed
}
return sorted;
}
}
int maxLen = 0;
for(int i : arr) {
if(i < 0) throw new IllegalArgumentException("I don't deal with negative numbers");
int len = numKeys(i);
if(len > maxLen) maxLen = len;
}
Bucket[] buckets = new Bucket[maxLen];
for(int i = 0; i < buckets.length; i++) buckets[i] = new Bucket();
for(int i : arr) buckets[numKeys(i)-1].add(i);
int[] result = new int[arr.length];
int[] posarr = new int[buckets.length]; // all int to 0
for(int i = 0; i < result.length; i++) {
// get the 'best' element, which will be the most appropriate from
// the set of earliest unused elements from each bucket
int best = -1;
int bestpos = -1;
for(int p = 0; p < posarr.length; p++) {
if(posarr[p] == buckets[p].getSortedArray().length) continue;
int oldbest = best;
best = bestOf(best, buckets[p].getSortedArray()[posarr[p]]);
if(best != oldbest) {
bestpos = p;
}
}
posarr[bestpos]++;
result[i] = best;
}
return result;
}
static int bestOf(int a, int b) {
if(a == -1) return b;
// you'll have to write this yourself :)
String as = a+"";
String bs = b+"";
if(as.compareTo(bs) < 0) return a;
return b;
}
static int numKeys(int i) {
if(i < 0) throw new IllegalArgumentException("I don't deal with negative numbers");
if(i == 0) return 1;
//return (i+"").length(); // lame method :}
int len = 0;
while(i > 0) {
len++;
i /= 10;
}
return len;
}
public static void main(String[] args) {
int[] test = {1, 6, 31, 65, 143, 316, 93, 736};
int[] res = radixSort(test);
for(int i : res) System.out.println(i);
}
}
One thing that looks strange:
for (int p = 0; p < n; p++){
while(queue[p].count() != 0) {
q[p] = (queue[p].get());
}
}
Is p supposed to be the index in q, which ranges from 0 to n-1, or in queue, which ranges from 0 to 9? It is unlikely to be both ...
Another:
for (int index = 0; index < n; index++){
// trying to look at one of three digit to sort after.
int v=1;
int digit = (q[index]/v)%10;
v*=10;
queue[digit].put(q[index]);
}
Why are you multiplying v by 10, only to overwrite it by v = 1 in the next iteration? Are you aware than v will always be one, and you will thus look at the same digit in every iteration?
Well I don't think I can help without almost posting the solution (just giving hints is more exhausting and I'm a bit tired, sorry), so I'll just contribute a nice little fuzz test so you can test your solution. How does that sound? :-)
Coming up with a good fuzztester is always a good idea if you're implementing some algorithm. While there's no 100% certainty if that runs with your implementation chances are it'll work (radix sort doesn't have any strange edge cases I'm aware of that only happen extremely rarely)
private static void fuzztest() throws Exception{
Random rnd = new Random();
int testcnt = 0;
final int NR_TESTS = 10000;
// Maximum size of array.
final int MAX_DATA_LENGTH = 1000;
// Maximum value allowed for each integer.
final int MAX_SIZE = Integer.MAX_VALUE;
while(testcnt < NR_TESTS){
int len = rnd.nextInt(MAX_DATA_LENGTH) + 1;
Integer[] array = new Integer[len];
Integer[] radix = new Integer[len];
for(int i = 0; i < len; i++){
array[i] = rnd.nextInt(MAX_SIZE);
radix[i] = new Integer(array[i]);
}
Arrays.sort(array);
sort(radix); // use your own sort function here.
for(int i = 0; i < len; i++){
if(array[i].compareTo(radix[i]) != 0){
throw new Exception("Not sorted!");
}
}
System.out.println(testcnt);
testcnt++;
}