Develop Reference

Why does this algorithm not sort the last index correctly?

I am supposed to write a version of the selection sort algorithm that moves the smallest number to the front of the array while simultaneously moving the largest number to the end of the array. I understand that it is basically working from both ends in so there are two sorted sub arrays, however my code seems to misplace the last number in the array. My code:
public static void dualSelectionSort(int[]a) {
for(int i=0; i<a.length-1;i++) {
int min=i;
int max=(a.length-1)-i;
for (int j=i+1;j<a.length-i;j++) {
if(a[j]<a[min]) {
min=j;
}
if(a[j]>a[max]) {
max=j;
}
}
int temp1=a[min];
int temp2=a[max];
a[min]=a[i];
a[max]=a[(a.length-1)-i];
a[i]=temp1;
a[(a.length-1)-i]=temp2;
}
}
If given the input [5,4,3,2,1], it outputs [1,2,5,3,4] instead of [1,2,3,4,5].
What am I missing?

The problem is in the inner loop.
It is starting from i+1.
So the code for comparing max is actually comparing a[1] to a[4].
you can change it as
for (int j=i;j<a.length-i;j++)
Also as you can read in comment there can be a case where with the above change the code will still not work.
This will happen because i == max and so the existing swap approach will not work.
For example:
assume array =[6,4,5],inner loop max =0,min=1.Existing swap will make it [4,6,6].
Hence,swap should be handled differently.
if(i==max) {
swap(a,max,a.length-1-i);
swap(a,min,i);
}else {
swap(a,min,i);
swap(a,max,(a.length-1)-i);
}
The complete code is as below:
public static void dualSelectionSort(int[]a) {
for(int i=0; i<a.length-1;i++) {
int min=i;
int max=(a.length-1)-i;
for (int j=i;j<a.length-i;j++) {
if(a[j]<a[min]) {
min=j;
}
if(a[j]>a[max]) {
max=j;
}
}
if(i==max) {
swap(a,max,a.length-1-i);
swap(a,min,i);
}else {
swap(a,min,i);
swap(a,max,(a.length-1)-i);
}
}
}
public static void swap(int[] a,int i,int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}

In each iteration of the outer loop, you start with the first index (i) as the initial minimum index and the last index (a.length-1-i) as the initial maximum index. Then you compare both of these indices to all the indices in the range i+1 to a.length-i-1. But if the first index (i) actually contain the max value, you never compare it to a[max], so a[max] will never contain the correct value in the end of the inner loop.
Here's a suggested fix:
public static void dualSelectionSort(int[]a) {
for(int i = 0; i < a.length - i - 1; i++) {
if (a[i] > a[(a.length - 1) - i]) {
int temp = a[i];
a[i] = a[(a.length - 1) - i];
a[(a.length - 1) - i] = temp;
}
int min = i;
int max = (a.length - 1) - i;
for (int j = i + 1; j < (a.length -1 ) - i; j++) {
if(a[j] < a[min]) {
min = j;
}
if(a[j] > a[max]) {
max = j;
}
}
int temp1 = a[min];
int temp2 = a[max];
a[min] = a[i];
a[max] = a[(a.length - 1) - i];
a[i] = temp1;
a[(a.length - 1) - i] = temp2;
}
}
The three things I changed:
The outer loop can end much sooner than you end it - once i >= a.length-i-1 - since in each iteration you find the minimum and maximum value, so you reduce your remaining unsorted array by 2.
When initializing the min and max indices, make sure min index actually contains a value smaller the max index. If a[i] > a[(a.length-1)-i], swap them before setting min to i and max to (a.length-1)-i.
The inner loop should iterate j from i+1 to (a.length-1)-i-1.

Java Array Manipulation and Recursion

So I have spent a considerable amount of time struggling to comprehend what is wrong with my code. I have an example program that I compared mine to, which works. My code is structured differently (it's all in one method, as requested by my professor) than the example (which uses two methods). I'm supposed to create a a recursive, divide-and-conquer solution to count inversions in an int array.
I am lost on why the example program maintains the manipulations to the input array throughout the recursion, while mine does not. I know Java is pass-by-value, so I am confused why the example works. Any help with me understanding the differences in these solutions would be greatly appreciated! Thanks!
Example code with two methods - merge and invCounter:
public static long merge(int[] arr, int[] left, int[] right) {
int i = 0, j = 0, count = 0;
while (i < left.length || j < right.length) {
if (i == left.length) {
arr[i+j] = right[j];
j++;
} else if (j == right.length) {
arr[i+j] = left[i];
i++;
} else if (left[i] <= right[j]) {
arr[i+j] = left[i];
i++;
} else {
arr[i+j] = right[j];
count += left.length-i;
j++;
}
}
return count;
}
//the recursive function
public static long invCounter(int[] arr) {
int sum = 0;
if (arr.length < 2)
return 0;
int m = (arr.length + 1) / 2;
int left[] = Arrays.copyOfRange(arr, 0, m);
int right[] = Arrays.copyOfRange(arr, m, arr.length);
sum += invCounter(left);
sum += invCounter(right);
sum += merge(arr, left, right);
return sum;
}
My single-method implementation (attempt):
public static int invCounter(int ranking[]) {
int sum = 0;
int result[] = new int[ranking.length];
int resIndx = 0;
if (ranking.length < 2) {
return 0; //base case
}
//divide
int left[] = Arrays.copyOfRange(ranking, 0, ranking.length/2);
int right[] = Arrays.copyOfRange(ranking, ranking.length/2,
ranking.length);
sum += invCounter(left);
sum += invCounter(right);
int i = 0, j = 0;
while (i < left.length || j < right.length) {
if (i == left.length) {
//i empty, just add j
result[resIndx++] = right[j++];
}
else if (j == right.length) {
//j empty, just add i
result[resIndx++] = left[i++];
}
else if (right[j] < left[i]) {
//inversion
result[resIndx++] = right[j++];
sum += left.length - i;
}
else {
//no inversion
result[resIndx++] = left[i++];
}
}
ranking = Arrays.copyOf(result, result.length);
return sum;
}
Why is the example program able to maintain an updated array through the recursion while mine is not?
UPDATE (10/22/15):
So I discovered that I am able to get the correct results if I replace result with ranking and just modify this array directly. My question now though is why can't I use the result array to temporarily store the results and then copy them into the ranking (argument) array at the end? This seems to me like it would be doing the same exact thing as putting the values in earlier, however the changes to ranking aren't reflected if I change it at the end.

Your method doesn't modify the rankings parameter, instead it creates a new int array (result), and you work on it. Try directly set value on the rankings array, not on result array, or simply set the result variable to the rankings.
public static int invCounter(int ranking[]) {
int sum = 0;
int result[] = ranking;
//other code...
Edit: Or you can copy it's content, but not with Arrays.copyOf, because it first CREATES a new array and then copy into it. Use instead System.arrayCopy which copies into an EXISTING array:
System.arrayCopy(result, 0, rankings, 0, result.length();

Codility PermCheck Solution isn't working on a few data sets

Trying to solve codility lessons for practice and working on this.
Written my code in Java and tested the code on a wide range of inputs, however the code fails for extreme_min_max, single and double in the codility test results.
Assumption given:
N is an integer within the range [1..100,000].
Each element of array A is an integer within the range [1..1,000,000,000].
Explanation of my code:
1. Sort the given array.
2. Iterate over each element in the array to find the difference between every consecutive pair. If the difference is not 1, Then its not a perm hence return 0. In case there is only one element in the array, return 1.
Can anyone please help me find out the bug(s) in my code?
My code:
public int solution(int[] A)
{
if(A.length == 1)
return 1;
Arrays.sort(A);
for (int i = 0; i < A.length-1; i++)
{
long diff = Math.abs(A[i] - A[i+1]);
if(diff!=1)
return 0;
}
return 1;
}

Here is simple and better implementation which runs in O(N) time complexity and takes O(N) space complexity.
public int solution(int[] A)
{
int size = A.length;
int hashArray[] = new int[size+1];
for (int i = 0; i < size; i++)
{
if(A[i]>size)
return 0;
else
hashArray[A[i]]+=1;
}
for(int i=1;i<=size;i++)
if(hashArray[i]!=1)
return 0;
return 1;
}

Try this in C# (Score 100%) :
using System;
using System.Linq;
class Solution {
public int solution(int[] A) {
if (A.Any(x => x == 0)) { return 0; }
var orderSelect = A.OrderBy(x => x).GroupBy(x => x);
if (orderSelect.Any(x => x.Count() > 1)) { return 0; }
var res = Enumerable.Range(1, A.Length).Except(A);
return res.Any() ? 0 : 1;
}
}

Pretty simple:
Your code doesn't check this condition:
A permutation is a sequence containing each element from 1 to N once, and only once.
Ensure that the first element after sorting is 1, and everything should work.

I'm not big on Java syntax, but what you want to do here is:
Create an array temp the length of A - initialized to 0.
Go over A and do temp[A[i]]++.
Go over temp, and if any place in the array is not 1, return false.

If duplicate exists - return 0 I have implemented with 100% pass
https://codility.com/demo/results/trainingWX2E92-ASF/
public static int permCheck(int A[]){
Set<Integer> bucket = new HashSet<Integer>();
int max = 0;
int sum=0;
for(int counter=0; counter<A.length; counter++){
if(max<A[counter]) max=A[counter];
if(bucket.add(A[counter])){
sum=sum+A[counter];
}
else{
return 0;
}
}
System.out.println(max+"->"+sum);
int expectedSum = (max*(max+1))/2;
if(expectedSum==sum)return 1;
return 0;
}

Here's my first 100% code.
I can't say if it's the fastest but it seems all correct -- watch the double OR ( || ) condition.
import java.util.Arrays;
class Solution
{
public int solution(int[] A)
{
int i = 0;
int size = A.length;
if ( size > 0 && size < 100001)
{
// Sort the array ascending:
Arrays.sort(A);
// Check each element:
for(i = 0; i < size; i++)
if ( A[i] > size || A[i] != (i + 1) )
return 0;
return 1;
}
return 0;
}
}
EDIT
Actually, we need not worry about valid first element data (i.e. A[i] > 0) because, after sorting, a valid perm array must have A[0] = 1 and this is already covered by the condition A[i] = i + 1.
The upper limit for array entries (> 1,000,000,000) is restricted further by the limit on the array size itself (100,000) and we must check for conformity here as there will be a Codility test for this. So I have removed the lower limit condition on array entries.

Below code runs and gave me a 100%, the time complexity is O(n):
private static int solution(int[] A) {
int isPermutation = 1; // all permutations start at 1
int n = A.length;
Arrays.sort(A);
if (n == 0) return 0; // takes care of edge case where an empty array is passed
for (int i = 0; i < n; i++) {
if (A[i] != isPermutation) { //if current array item is not equals to permutation, return 0;
return 0;
}
isPermutation++;
}
return 1;
}

100% score with complexity O(N)
public int solution(int[] A) {
int res = 1;
if (A.length == 1 && A[0]!=1)
return 0;
int[] B = new int[A.length];
for (int j : A) {
int p = j - 1;
if (A.length > p)
B[p] = j;
}
for (int i = 0; i < B.length - 1; i++) {
if (B[i] + 1 != B[i + 1]) {
res = 0;
break;
}
}
return res;
}

Implementing a binary insertion sort using binary search in Java

I'm having trouble combining these two algorithms together. I've been asked to modify Binary Search to return the index that an element should be inserted into an array. I've been then asked to implement a Binary Insertion Sort that uses my Binary Search to sort an array of randomly generated ints.
My Binary Search works the way it's supposed to, returning the correct index whenever I test it alone. I wrote out Binary Insertion Sort to get a feel for how it works, and got that to work as well. As soon as I combine the two together, it breaks. I know I'm implementing them incorrectly together, but I'm not sure where my problem lays.
Here's what I've got:
public class Assignment3
{
public static void main(String[] args)
{
int[] binary = { 1, 7, 4, 9, 10, 2, 6, 12, 3, 8, 5 };
ModifiedBinaryInsertionSort(binary);
}
static int ModifiedBinarySearch(int[] theArray, int theElement)
{
int leftIndex = 0;
int rightIndex = theArray.length - 1;
int middleIndex = 0;
while(leftIndex <= rightIndex)
{
middleIndex = (leftIndex + rightIndex) / 2;
if (theElement == theArray[middleIndex])
return middleIndex;
else if (theElement < theArray[middleIndex])
rightIndex = middleIndex - 1;
else
leftIndex = middleIndex + 1;
}
return middleIndex - 1;
}
static void ModifiedBinaryInsertionSort(int[] theArray)
{
int i = 0;
int[] returnArray = new int[theArray.length + 1];
for(i = 0; i < theArray.length; i++)
{
returnArray[ModifiedBinarySearch(theArray, theArray[i])] = theArray[i];
}
for(i = 0; i < theArray.length; i++)
{
System.out.print(returnArray[i] + " ");
}
}
}
The return value I get for this when I run it is 1 0 0 0 0 2 0 0 3 5 12. Any suggestions?
UPDATE: updated ModifiedBinaryInsertionSort
static void ModifiedBinaryInsertionSort(int[] theArray)
{
int index = 0;
int element = 0;
int[] returnArray = new int[theArray.length];
for (int i = 1; i < theArray.lenght - 1; i++)
{
element = theArray[i];
index = ModifiedBinarySearch(theArray, 0, i, element);
returnArray[i] = element;
while (index >= 0 && theArray[index] > element)
{
theArray[index + 1] = theArray[index];
index = index - 1;
}
returnArray[index + 1] = element;
}
}

Here is my method to sort an array of integers using binary search.
It modifies the array that is passed as argument.
public static void binaryInsertionSort(int[] a) {
if (a.length < 2)
return;
for (int i = 1; i < a.length; i++) {
int lowIndex = 0;
int highIndex = i;
int b = a[i];
//while loop for binary search
while(lowIndex < highIndex) {
int middle = lowIndex + (highIndex - lowIndex)/2; //avoid int overflow
if (b >= a[middle]) {
lowIndex = middle+1;
}
else {
highIndex = middle;
}
}
//replace elements of array
System.arraycopy(a, lowIndex, a, lowIndex+1, i-lowIndex);
a[lowIndex] = b;
}
}

How an insertion sort works is, it creates a new empty array B and, for each element in the unsorted array A, it binary searches into the section of B that has been built so far (From left to right), shifts all elements to the right of the location in B it choose one right and inserts the element in. So you are building up an at-all-times sorted array in B until it is the full size of B and contains everything in A.
Two things:
One, the binary search should be able to take an int startOfArray and an int endOfArray, and it will only binary search between those two points. This allows you to make it consider only the part of array B that is actually the sorted array.
Two, before inserting, you must move all elements one to the right before inserting into the gap you've made.

I realize this is old, but the answer to the question is that, perhaps a little unintuitively, "Middleindex - 1" will not be your insertion index in all cases.
If you run through a few cases on paper the problem should become apparent.
I have an extension method that solves this problem. To apply it to your situation, you would iterate through the existing list, inserting into an empty starting list.
public static void BinaryInsert<TItem, TKey>(this IList<TItem> list, TItem item, Func<TItem, TKey> sortfFunc)
where TKey : IComparable
{
if (list == null)
throw new ArgumentNullException("list");
int min = 0;
int max = list.Count - 1;
int index = 0;
TKey insertKey = sortfFunc(item);
while (min <= max)
{
index = (max + min) >> 1;
TItem value = list[index];
TKey compKey = sortfFunc(value);
int result = compKey.CompareTo(insertKey);
if (result == 0)
break;
if (result > 0)
max = index - 1;
else
min = index + 1;
}
if (index <= 0)
index = 0;
else if (index >= list.Count)
index = list.Count;
else
if (sortfFunc(list[index]).CompareTo(insertKey) < 0)
++index;
list.Insert(index, item);
}

Dude, I think you have some serious problem with your code. Unfortunately, you are missing the fruit (logic) of this algorithm. Your divine goal here is to get the index first, insertion is a cake walk, but index needs some sweat. Please don't see this algorithm unless you gave your best and desperate for it. Never give up, you already know the logic, your goal is to find it in you. Please let me know for any mistakes, discrepancies etc. Happy coding!!
public class Insertion {
private int[] a;
int n;
int c;
public Insertion()
{
a = new int[10];
n=0;
}
int find(int key)
{
int lowerbound = 0;
int upperbound = n-1;
while(true)
{
c = (lowerbound + upperbound)/2;
if(n==0)
return 0;
if(lowerbound>=upperbound)
{
if(a[c]<key)
return c++;
else
return c;
}
if(a[c]>key && a[c-1]<key)
return c;
else if (a[c]<key && a[c+1]>key)
return c++;
else
{
if(a[c]>key)
upperbound = c-1;
else
lowerbound = c+1;
}
}
}
void insert(int key)
{
find(key);
for(int k=n;k>c;k--)
{
a[k]=a[k-1];
}
a[c]=key;
n++;
}
void display()
{
for(int i=0;i<10;i++)
{
System.out.println(a[i]);
}
}
public static void main(String[] args)
{
Insertion i=new Insertion();
i.insert(56);
i.insert(1);
i.insert(78);
i.insert(3);
i.insert(4);
i.insert(200);
i.insert(6);
i.insert(7);
i.insert(1000);
i.insert(9);
i.display();
}
}

Java - My sort is not working

I have created the following class to sort an array of strings.
public class StringSort {
private String[] hotelNames;
private int arrayLength;
public void sortHotel(String[] hotelArray) {
if (hotelArray.length <= 1) {
return;
}
this.hotelNames = hotelArray;
arrayLength = hotelArray.length;
quicksort(0, arrayLength - 1);
}
private void quicksort(int low, int high) {
int i = low, j = high;
String first = hotelNames[low];
String last = hotelNames[high];
String pivot = hotelNames[low + (high - low) / 2];
while( (first.compareTo(last)) < 0 ) { // first is less than last
while( (hotelNames[i].compareTo(pivot)) < 0 ) { // ith element is < pivot
i++;
}
while( (hotelNames[j].compareTo(pivot)) > 0) { // jth element is > pivot
j--;
}
if ( ( hotelNames[i].compareTo( hotelNames[j] )) <= 0 ) {
swap(i, j);
i++;
j--;
}
//recursive calls
if (low < j) {
quicksort(low, j);
}
if (i < high) {
quicksort(i, high);
}
}
}
private void swap(int i, int j) {
String temp = hotelNames[i];
hotelNames[i] = hotelNames[j];
hotelNames[j] = temp;
}
}
However in my main class (a class to test StringSort), when I do:
StringSort str = new StringSort();
String[] hotel1 = {"zzzz", "wwww", "dddd", "bbbbb", "bbbba", "aaaf", "aaag", "zzz"};
str.sortHotel(hotel1);
And then I have another method that prints out the array. However when it prints out, it outputs the hotel1 array as it is, unchanged. There is no 'sorting' happening, I'm not sure where I've gone wrong.

There are several problems in your implementation of quicksort:
First/last comparison. This code will made your quicksort not do anything as long as first element is less than the last element, regardless of any other order.
while( (first.compareTo(last)) < 0 ) { // first is less than last
Check before swap. This line is unnecessary:
if ( ( hotelNames[i].compareTo( hotelNames[j] )) <= 0 ) {
What you really want to do is see if the i is still less than j and bail out of the loop then. If not, then swap. After you finished with the partitioning loop, then make the recursive call, as long as there are more than two elements in each subarray.