Java--Making 10-integer ordering program recursive - java

I have a simple problem - I need to order 10 numbers. I had an idea how to do this recursively: Make an array of the 10 numbers, take the maximum of the ten numbers, take it out of the array, and repeat the same function with the nine numbers left. The problem was that I did not know how to implement that. I wrote the program, and it works, only it has a part that repeats all the time but with new arrays, because you cannot change the size of the array.
/* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone {
public static void main (String[] args) throws java.lang.Exception {
int[] sortedArray = new int[]{0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
Scanner input = new Scanner(System.in);
int in0 = input.nextInt();
int in1 = input.nextInt();
int in2 = input.nextInt();
int in3 = input.nextInt();
int in4 = input.nextInt();
int in5 = input.nextInt();
int in6 = input.nextInt();
int in7 = input.nextInt();
int in8 = input.nextInt();
int in9 = input.nextInt();
int[] numArray = new int[]{in0, in1, in2, in3, in4, in5, in6, in7, in8, in9};
int numArrayLength = numArray.length;
recursiveSort(numArray);
for (int i=0;i<numArrayLength;i++) {
System.out.print(numArray[i]+",");
}
sortedArray[0] = numArray[0];
System.out.println(" ");
int[] numArray2 = Arrays.copyOfRange(numArray, 1, numArrayLength);
int numArray2Length = numArray2.length;
recursiveSort(numArray2);
for (int j=0;j<numArray2Length;j++) {
System.out.print(numArray2[j]+",");
}
sortedArray[1] = numArray2[0];
System.out.println(" ");
int[] numArray3 = Arrays.copyOfRange(numArray2, 1, numArray2Length);
int numArray3Length = numArray3.length;
recursiveSort(numArray3);
for (int k=0;k<numArray3Length;k++) {
System.out.print(numArray3[k]+",");
}
sortedArray[2] = numArray3[0];
System.out.println(" ");
int[] numArray4 = Arrays.copyOfRange(numArray3, 1, numArray3Length);
int numArray4Length = numArray4.length;
recursiveSort(numArray4);
for (int k=0;k<numArray4Length;k++) {
System.out.print(numArray4[k]+",");
}
sortedArray[3] = numArray4[0];
System.out.println(" ");
int[] numArray5 = Arrays.copyOfRange(numArray4, 1, numArray4Length);
int numArray5Length = numArray5.length;
recursiveSort(numArray5);
for (int k=0;k<numArray5Length;k++) {
System.out.print(numArray5[k]+",");
}
sortedArray[4] = numArray5[0];
System.out.println(" ");
int[] numArray6 = Arrays.copyOfRange(numArray5, 1, numArray5Length);
int numArray6Length = numArray6.length;
recursiveSort(numArray6);
for (int k=0;k<numArray6Length;k++) {
System.out.print(numArray6[k]+",");
}
sortedArray[5] = numArray6[0];
System.out.println(" ");
int[] numArray7 = Arrays.copyOfRange(numArray6, 1, numArray6Length);
int numArray7Length = numArray7.length;
recursiveSort(numArray7);
for (int k=0;k<numArray7Length;k++) {
System.out.print(numArray7[k]+",");
}
sortedArray[6] = numArray7[0];
System.out.println(" ");
int[] numArray8 = Arrays.copyOfRange(numArray7, 1, numArray7Length);
int numArray8Length = numArray8.length;
recursiveSort(numArray8);
for (int k=0;k<numArray8Length;k++) {
System.out.print(numArray8[k]+",");
}
sortedArray[7] = numArray8[0];
System.out.println(" ");
int[] numArray9 = Arrays.copyOfRange(numArray8, 1, numArray8Length);
int numArray9Length = numArray9.length;
recursiveSort(numArray9);
for (int k=0;k<numArray9Length;k++) {
System.out.print(numArray9[k]+",");
}
sortedArray[8] = numArray9[0];
System.out.println(" ");
int[] numArray10 = Arrays.copyOfRange(numArray9, 1, numArray9Length);
int numArray10Length = numArray10.length;
recursiveSort(numArray10);
for (int k=0;k<numArray10Length;k++) {
System.out.print(numArray10[k]+",");
}
sortedArray[9] = numArray10[0];
System.out.println(" ");
sortedArray[2] = numArray3[0];
for (int dasdasd=0;dasdasd<sortedArray.length;dasdasd++) {
System.out.print(sortedArray[dasdasd]+",");
}
}
private static int[] recursiveSort(int numArray[]) {
int numArrayLength = numArray.length;
int maximum = 0;
for (int i=0;i<numArrayLength;i++) {
if (numArray[i] > maximum) {
maximum = numArray[i];
}
}
int indexOfMaximum = -1;
for (int j=0;j<numArrayLength;j++) {
if (numArray[j] == maximum) {
indexOfMaximum = j;
break;
}
}
int temporary = numArray[0];
numArray[0] = numArray[indexOfMaximum];
numArray[indexOfMaximum] = temporary;
return numArray;
}
}
As you can see, the
int[] numArray(n) = Arrays.copyOfRange(numArray(n-1), 1, numArray(n-1)Length);
int numArray(n)Length = numArray(n).length;
recursiveSort(numArray(n));
for (int k=0;k<numArray(n)Length;k++) {
System.out.print(numArray(n)[k]+",");
}
sortedArray[(n-1)] = numArray(n)[0];
System.out.println(" ");
constantly repeats, so there is probably a recursive solution that will work nicely. Maybe I can do something using ArrayLists because their size can change...
Any help will be appreciated!
Thank you!

I suggest a recursive routine that uses an explicit start index for the part that remains to be sorted:
private static void recursiveSort(int[] array, int start) {
if (start < array.length - 1) {
int maximum = array[start];
int maximumIndex = start;
for (int i = start + 1; i < array.length; ++i) {
if (array[i] > maximum) {
maximum = array[i];
maximumIndex = i;
}
}
if (maximumIndex != start) {
int tmp = array[start];
array[start] = array[maximumIndex];
array[maximumIndex] = tmp;
}
recursiveSort(array, start + 1);
}
}
This actually does recursion (unlike your code, which iterates calling a routine named "recursiveSort" but isn't recursive at all). The whole process would be started by calling:
recursiveSort(numArray, 0);
When it returns, the array will be sorted in descending order.
As a general heuristic, when you are struggling with how to make a method recursive, you should consider adding arguments to the method to help with the bookkeeping.

Is this homework or you just need to have the numbers ordered? Java has an easy way to do this if you use ArrayList() instead of array[]. You would just need to call Collections.sort(yourArrayList);

I suggest not trying to make your own sorting algorithm. Many smart people have already done that hard work for you.
The "recursive" sort that you were trying to implement (aka bubble sort which Ted has shown you how to truly make recursive) will work, but it is grossly inefficient. See a comparison of sorting algorithms here.
Below is a demo of the algorithm you were trying to implement compared to a shell sort, one of the fastest sorting algorithms available. The implementation I used was taken from here. Run it and you will see that shell sort is on average 7 to 8 times faster than bubble sort.
public class SortingDemo {
// Methods required for Shell sort
public static void shellSort(Comparable[] a) {
int N = a.length;
int h = 1;
while (h < N/3) h = 3*h + 1;
while (h >= 1) {
for (int i = h; i < N; i++) {
for (int j = i; j >= h && less(a[j], a[j-h]); j -= h) {
exch(a, j, j-h);
}
}
assert isHsorted(a, h);
h /= 3;
}
assert isSorted(a);
}
private static boolean less(Comparable v, Comparable w) {
return (v.compareTo(w) < 0);
}
private static void exch(Object[] a, int i, int j) {
Object swap = a[i];
a[i] = a[j];
a[j] = swap;
}
private static boolean isSorted(Comparable[] a) {
for (int i = 1; i < a.length; i++)
if (less(a[i], a[i-1])) return false;
return true;
}
private static boolean isHsorted(Comparable[] a, int h) {
for (int i = h; i < a.length; i++)
if (less(a[i], a[i-h])) return false;
return true;
}
// Method required for "recursive" sort
private static void recursiveSort(Integer[] array, int start) {
if (start < array.length - 1) {
int maximum = array[start];
int maximumIndex = start;
for (int i = start + 1; i < array.length; ++i) {
if (array[i] > maximum) {
maximum = array[i];
maximumIndex = i;
}
}
if (maximumIndex != start) {
int tmp = array[start];
array[start] = array[maximumIndex];
array[maximumIndex] = tmp;
}
recursiveSort(array, start + 1);
}
}
public static void main(String[] args) {
int desiredArraySize = 1000;
int minSizeOfNumberInArray = 0;
int maxSizeOfNumberInArray = 100;
Integer[] array = new Integer[desiredArraySize]; // Used Integer instead of int to utilize Comparable interface
for(int i = 0; i < array.length; i++) {
int randomInt = (int) Math.random() * (maxSizeOfNumberInArray - minSizeOfNumberInArray);
array[i] = randomInt;
}
long startTime = System.nanoTime();
recursiveSort(array, 0);
long endTime = System.nanoTime();
long recursiveSortTime = endTime - startTime;
System.out.println(String.format("\"Recursive\" sort completed in %d ns", recursiveSortTime));
startTime = System.nanoTime();
shellSort(array);
endTime = System.nanoTime();
long shellSortTime = endTime - startTime;
System.out.println(String.format("Shell sort completed in %d ns", shellSortTime));
System.out.println(String.format("\"Recursive\" sort took %f times longer", (float)recursiveSortTime / (float)shellSortTime));
}
}

When learning programming, both writing your own sorting algorithms and your own recursive algorithms are great exercises for solidifying your understanding of how things work. It's time well invested, even if someone's already done it better.
You noticed a pattern that repeats, and associated that with recursion. When evaluating whether recursion is a good fit, I would encourage you to tweak that thought process with the notion of "divide-and-conquer". If you're solving only one element with each recursion, then your stack will grow very deep, which should be avoided. If you can split your problem into roughly even chunks and process each chunk recursively, then recursion will be a good fit. Otherwise, a loop is already an excellent fit for repeating patterns.

Related

how to construct an array of 100 elements containing the numbers 1 -100 for which shellsort, with the increments 1 4 13 40 (The worst case)?

This is the original question
"Shell Sort worst case. Construct an array of 100 elements containing the numbers 1 through 100 for which shellsort, with the increments 1 4 13 40, uses as large a number of compares as you can find."
There are 100! permutations for an array of 100 elements, it's terrifying to go through each permutation and find which one has the maximum number of compares. Is there any smarter way to approach this problem? My approach this problem is through violence, but only randomly shuffle the array 100000000 time which is less than 100! and it take me half an hour to get the final output.
I pasted my code below. I appreciate any suggestions from you guys!
`
package ch_2_1;
import edu.princeton.cs.algs4.StdOut;
import edu.princeton.cs.algs4.StdRandom;
import java.util.Arrays;
public class exer_19
{
public static void main(String[] args)
{
// initial permutation
int[] array = new int[100];
for ( int i = 1; i < 101; i++)
{
array[i-1] = i;
}
// find the worst case and the number of compares
worst_input(array);
}
private static void worst_input(int[] array)
{
int max_count = 0;
int[] copy = new int[100];
int[] worst_case = new int[100];
for ( int i = 0; i < 100000000; i++)
{
int[] temp = generate(array);
for (int j = 0; j < 100; j++){ copy[j] = temp[j];}
Shell_sort operation = new Shell_sort();
operation.shell_sort(temp);
if (operation.compare() > max_count)
{
max_count = operation.compare();
worst_case = copy;
}
System.out.println(i);
}
for ( int s : worst_case){ System.out.print(s + " ");}
System.out.println();
System.out.println(max_count);
System.out.println();
}
private static int[] generate( int[] array)
{
StdRandom.shuffle(array);
return array;
}
private static class Shell_sort // it's necessary to create a private class to hold the shell sort method
// b/c the method must record the # of compares to sort the array, and this # count need to be returned
// to the worst_input method. Therefore, having a class to encapsulate the two methods is very helpful
{
private int count = 0;
private void shell_sort(int[] test)
{
int N = test.length;
int h = 1;
while (h < N/3) h = 3*h + 1; // 1, 4, 13, 40, 121...
while ( h > 0)
{
for ( int i = h; i < N; i++) // starting from the largest h-value th element of the array (simplified: ith element)
{
// if ith element is less than i-h th element, swap the two, continue this process until the condition is not met
for ( int j = i; j >= h && less( test[j], test[j-h]); j = j - h)
{
exchange( test, j, j-h);
count++;
}
}
// when reached the end of the array, update h value
h = h/3;
}
}
private int compare()
{
return count;
}
}
private static boolean less( int current, int previous)
{
return current < previous;
}
private static void exchange(int[] array, int cur_index, int pre_index)
{
int temp = array[pre_index];
array[pre_index] = array[cur_index];
array[cur_index] = temp;
}
}
`

Selection sort Java, a small error but can't seem to figure out

I am trying to measure the time it takes for my system to sort an Array of 50 thousand random numbers using Selection sort, however i am getting an error. The error i received is null pointed exception.
Would someone look at where i am going wrong here
import java.util.Random;
public class SelectionSort {
public static void main(String[] args) {
int arrayOne[] = null;
int arr[] = { 9, 1, 8, 5, 7, -1, 6, 0, 2, 2718 };
int arr1[] = fillArray(arrayOne);
int sortedArr[] = selectionSort(arr1);
System.out.println("Selection sort implemented below");
System.currentTimeMillis();
long start = System.currentTimeMillis();
print(sortedArr);
long elapsed = System.currentTimeMillis() - start;
System.out.println(elapsed);
}
private static int[] fillArray(int[] array) {
Random generator2 = new Random(System.currentTimeMillis());
for (int x = 0; x < 50000; x++) {
array[x] = generator2.nextInt();
}
return array;
}
private static int[] selectionSort(int[] arr) {
int minIndex, tmp;
int n = arr.length;
for (int i = 0; i < n - 1; i++) {
minIndex = i;
for (int j = i + 1; j < n; j++)
if (arr[j] < arr[minIndex])
minIndex = j;
if (minIndex != i) {
tmp = arr[i];
arr[i] = arr[minIndex];
arr[minIndex] = tmp;
}
}
return arr;
}
private static void print(int[] Array) {
// TODO prints the array
for (int i = 0; i < Array.length; i++)
System.out.print(Array[i] + " ");
System.out.println();
}
}
You're trying to fill arrayOne which is null:
int arrayOne[] = null;
You need to initialize it with sufficient capacity before you try to fill it:
int arrayOne[] = new int[50000];
When you call fillArray(arrayOne) you are passing null as the parameter. Then in the fillArray method, you are trying to put values into a null array.
By the way, if you had run through the code line by line in a debugger, you would have spotted this problem in seconds. In fact, even the exception thrown would have told you exactly which line had the problem.

error ';' expected in array defining, ; already used

public class AssignmentChapter8
{
public static void main(String[] args)
{
int randomNumbers = new int[100];
int counter = 0;
while(counter < randomNumbers.length)
{
randomNumbers[counter] = (int)(Math.random() * 25);
counter++;
}
int oddNumbers[] = new int[100];
oddNumbers[] = getOddNumbers(randomNumbers);
int evenNumbers[] = new int[100];
evenNumbers[] = getEvenNumbers(randomNumbers);
System.out.println("The odd numbers are:");
for(int k = 0; k < oddNumbers.length; k++)
System.out.print("\t" + oddNumbers[k]);
System.out.println("The even numbers are:");
for(int l = 0; l < evenNumbers.length; l++)
System.out.print("\t" + evenNumbers[l]);
}
public static int getOddNumbers(int randomNumbers)
{
int oddNumbers[] = new int[100];
int counterA = 0;
int counterB = 0;
int counter = 0;
int placeholder;
while(counter < randomNumbers.length)
{
if(randomNumbers[counterA] % 2 > 0)
{
oddNumbers[counterB] = randomNumbers[counterA];
counterB++;
}
counterA++;
counter++;
}
return oddNumbers;
}
public static int getEvenNumbers(int randomNumbers)
{
int evenNumbers[] = new int[100];
int counterA = 0;
int counterB = 0;
int counter = 0;
int placeholder;
while(counter < randomNumbers.length)
{
if(randomNumbers[counterA] % 2 > 0)
{
evenNumbers[counterB] = randomNumbers[counterA];
counterB++;
}
counterA++;
counter++;
}
return evenNumbers;
}
}
I have been trying to execute a program to sort variables in arrays, but I keep getting a ';' expected error in the line after declaration of the array where the program is supposed to retrieve an array from a function. Any help would be appreciated.
This is bad syntax (which causes the ';' expected error ):
oddNumbers[] = getOddNumbers(randomNumbers);
The brackets are not needed. You can do this:
oddNumbers = getOddNumbers(randomNumbers);
Besides that, you have plenty of errors:
int randomNumbers[] = new int[100]; // you need the brackets
Your return value in the method declaration is wrong (you are returning an array, not an int):
public static int[] getEvenNumbers(int randomNumbers)
int oddNumbers[] = new int[100];
By initializing oddNumbers become an array. when you want to assign values,
oddNumbers={elements of array}
Both left and right hand side both should arrays.
May be you should use IDE for coding then it may help you to understand some issues like this.

How can I find the smallest covering prefix of an array in Java?

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;
}
}

Radix sort in java help

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++;
}

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