I'm trying to make a program where I count the amount of comparisons quicksort makes but its not 100% accurate. how would I make this comparison counter accurate?
The quicksort takes in an array of n random integers.
Code I have so far:
public int QuickSort(int[] list, int from, int to) {
int icount = 0;
if (from >= to) {
return icount;
}
int p = from;
int i = from;
int j = to;
while (i <= j) {
if (list[i] <= list[p]) {
i++;
icount++;
} else if (list[j] >= list[p]) {
j--;
icount++;
} else {
swap(list, i, j);
i++;
j--;
icount++;
}
}
if (p < j) { // place the pivot in its correct position
swap(list, j, p);
p = j;
icount++;
} else if (p > i) {
swap(list, i, p);
p = i;
icount++;
}
icount += QuickSort(list, from, p - 1);
icount += QuickSort(list, p + 1, to);
return icount;
}
Is there anything wrong with the current code? or is it not possible to be completely accurate with random integers?
I going to do searching the value in the array, did I need to create a method to handle it? For example, the array logged 32,21,13,44,22, and I going to find 22 of the comparison. How can I implement this?
public class binarySearch {
public static void main(String [] args) {
int i = binarySearch(0, new int[]{32,21,13,44,22});
System.out.println("Iterations: " + i);
}
public static int binarySearch(int key, int[] array) {
int left = 0;
int mid;
int right = array.length - 1;
int i = 0;
while (left <= right) {
mid = (left + right) / 2;
int comp = Integer.compare(key, array[mid]);
i++;
if (comp < 0) {
right = mid - 1;
} else if (comp > 0) {
left = mid + 1;
} else {
break; // success
}
}
return i;
}
}
My final answer is here. May help you all in the future.
public static int binarySearch(int key, int[] array) {
int left = 0;
int mid;
int right = array.length - 1;
int i = 0;
while (left <= right) {
mid = (left + right) / 2;
int comp = Integer.compare(key, array[mid]);
i++;
if (comp < 0) {
right = mid - 1;
} else if (comp > 0) {
left = mid + 1;
} else {
break; // success
}
}
return i;
}
If you have shuffled array, all you can do is go through an array and find your number.
BinarySearch works only with sorted array. I think your solution could look like this:
public static int binarySearch(int[] arr, int key) {
Arrays.sort(arr);
return Arrays.binarySearch(arr, key);
}
I am working on a homework task where I am supposed to make a function that will do a binary insertion sort, but my function does not seem to work properly.
Here I have tried to combine a binary search function with a insertion sort function (it is specified in the homework task that it needs to be in the form of a function: insertionSort(int[] array, int lo, int hi))
public static void insertionSort(int[] array, int lo, int hi){
int mid;
int pos;
for (int i = 1; i < array.length; i++) {
int x= array[i];
while (lo < hi) {
mid = lo + (hi -lo)/2;
if (x == array[mid]) {
pos = mid;
}
if (x > array[mid]) {
lo = mid+1;
}
else if (x < array[mid]) {
hi = mid-1;
}
}
pos = lo;
for (int j = i; j > pos; j--) {
array[j] = array[j-1];
}
array[pos] = x;
}
}
If I try to run it with the list {2,5,1,8,3}, the output will be
2 5 1 3 1 (if lo < hi and if lo > hi)
2 5 3 8 5 (if lo==hi)
What I am expecting though, is a sorted list...
Any idea of what I am doing wrong?
Just to give you a possible idea:
public static void insertionSort(int[] array) {
if (array.length <= 1) {
return;
}
// Start with an initially sorted part.
int loSorted = array.length - 1;
//int hiSorted = array.length;
while (loSorted > 0) {
// Take one from the array
int x = array[0];
// Where in the sorted part to insert?
int insertI = insertPosition(array, loSorted);
// Insert x at insertI
...
--loSorted;
}
}
whenever I need binary search, my function looks the following way:
public static void binarySearch(int arr[], int first, int last, int key){
int mid = (first + last)/2;
while( first <= last ){
if ( arr[mid] < key ){
first = mid + 1;
}else if ( arr[mid] == key ){
System.out.println("Element is found at index: " + mid);
break;
}else{
last = mid - 1;
}
mid = (first + last)/2;
}
if ( first > last ){
System.out.println("Element is not found!");
}
}
In your main method the call looks like:
public static void main(String[] args) {
int arr[] = {10,20,30,40,50};
int key = 30;
int last=arr.length-1;
binarySearch(arr,0,last,key);
}
I hope I was able to help you!
Thank you for your input. I changed the function a little bit, and it seems to be working now
` public static void insertionSort(int[] array, int lo, int hi){
int mid;
int pos;
for (int i = 1; i < array.length; i++) {
int j = i -1;
int x = array[i];
while (lo <= hi) {
mid = lo + (hi -lo)/2;
if (x == array[mid]) {
pos = mid;
break;
}
if (x > array[mid]) {
lo = mid+1;
}
else if (x < array[mid]) {
hi = mid-1;
}
}
while (j >= 0 && array[j] > x) {
array[j + 1] = array[j];
j = j - 1;
}
array[j + 1] = x;
}
}
the problem seemed to lay in the last part, where I was trying to move the elements into their right positions. The function is probably not perfect tho, so constructive criticism is welcome :)
Can I get some help please? I have tried many methods to get this to work i got the array sorted and to print but after that my binary search function doesnt want to run and give me right results. It always gives me -1. Any help?
public class BinarySearch {
public static final int NOT_FOUND = -1;
public static int binarySearch(double[] a, double key) {
int low = 0;
int high = a.length -1;
int mid;
while (low<=high) {
mid = (low+high) /2;
if (mid > key)
high = mid -1;
else if (mid < key)
low = mid +1;
else
return mid;
}
return NOT_FOUND;
}
public static void main(String[] args) {
double key = 10.5, index;
double a[] ={10,5,4,10.5,30.5};
int i;
int l = a.length;
int j;
System.out.println("The array currently looks like");
for (i=0; i<a.length; i++)
System.out.println(a[i]);
System.out.println("The array after sorting looks like");
for (j=1; j < l; j++) {
for (i=0; i < l-j; i++) {
if (a[i] > a[i+1]) {
double temp = a[i];
a[i] = a[i+1];
a[i+1] = temp;
}
}
}
for (i=0;i < l;i++) {
System.out.println(a[i]);
}
System.out.println("Found " + key + " at " + binarySearch(double a[], key));
}
}
you are not actually comparing with the array values. in
while (low <= high) {
mid = (low + high) / 2;
if (mid > key) {
high = mid - 1;
} else if (mid < key) {
low = mid + 1;
} else {
return mid;
}
}
Instead use this section
while (low <= high) {
mid = (low + high) / 2;
if (a[mid] > key) {
high = mid - 1;
} else if (a[mid] < key) {
low = mid + 1;
} else {
return mid;
}
}
You were correct to find the indexes, but what you were doing is that you were just comparing index number with your key, which is obviously incorrect. When you write a[mid] you will actually compare your key with the number which is at index mid.
Also the last line of code is giving compile error, it should be
System.out.println("Found " + key + " at " + binarySearch(a, key));
Here
if (mid > key)
high = mid -1;
else if (mid < key)
low = mid +1;
else
return mid;
You're comparing index to a value (key) in array. You should instead compare it to a[mid]
And,
System.out.println("Found " + key + " at " + binarySearch(double a[], key));
Should be
System.out.println("Found " + key + " at " + binarySearch(a, key));
public static double binarySearch(double[] a, double key) {
if (a.length == 0) {
return -1;
}
int low = 0;
int high = a.length-1;
while(low <= high) {
int middle = (low+high) /2;
if (b> a[middle]){
low = middle +1;
} else if (b< a[middle]){
high = middle -1;
} else { // The element has been found
return a[middle];
}
}
return -1;
}
int binarySearch(int list[], int lowIndex, int highIndex, int find)
{
if (highIndex>=lowIndex)
{
int mid = lowIndex + (highIndex - lowIndex)/2;
// If the element is present at the
// middle itself
if (list[mid] == find)
return mid;
// If element is smaller than mid, then
// it can only be present in left subarray
if (list[mid] > find)
return binarySearch(list, lowIndex, mid-1, find);
// Else the element can only be present
// in right subarray
return binarySearch(list, mid+1, highIndex, find);
}
// We reach here when element is not present
// in array
return -1;
}
I somehow find the iterative version not quite easy to read, recursion makes it nice and easy :-)
public class BinarySearch {
private static int binarySearchMain(int key, int[] arr, int start, int end) {
int middle = (end-start+1)/2 + start; //get index of the middle element of a particular array portion
if (arr[middle] == key) {
return middle;
}
if (key < arr[middle] && middle > 0) {
return binarySearchMain(key, arr, start, middle-1); //recurse lower half
}
if (key > arr[middle] && middle < arr.length-1) {
return binarySearchMain(key, arr, middle+1, end); //recurse higher half
}
return Integer.MAX_VALUE;
}
public static int binarySearch(int key, int[] arr) { //entry point here
return binarySearchMain(key, arr, 0, arr.length-1);
}
}
Here is a solution without heap. The same thing can be done in an array.
If we need to find 'k' largest numbers, we take an array of size 'k' populated with first k items from the main data source. Now, keep on reading an item, and place it in the result array, if it has a place.
public static void largestkNumbers() {
int k = 4; // find 4 largest numbers
int[] arr = {4,90,7,10,-5,34,98,1,2};
int[] result = new int[k];
//initial formation of elems
for (int i = 0; i < k; ++i) {
result[i] = arr[i];
}
Arrays.sort(result);
for ( int i = k; i < arr.length; ++i ) {
int index = binarySearch(result, arr[i]);
if (index > 0) {
// insert arr[i] at result[index] and remove result[0]
insertInBetweenArray(result, index, arr[i]);
}
}
}
public static void insertInBetweenArray(int[] arr, int index, int num) {
// insert num at arr[index] and remove arr[0]
for ( int i = 0 ; i < index; ++i ) {
arr[i] = arr[i+1];
}
arr[index-1] = num;
}
public static int binarySearch(int[] arr, int num) {
int lo = 0;
int hi = arr.length - 1;
int mid = -1;
while( lo <= hi ) {
mid = (lo+hi)/2;
if ( arr[mid] > num ) {
hi = mid-1;
} else if ( arr[mid] < num ) {
lo = mid+1;
} else {
return mid;
}
}
return mid;
}
int BinSearch(int[] array, int size, int value)
{
if(size == 0) return -1;
if(array[size-1] == value) return size-1;
if(array[0] == value) return 0;
if(size % 2 == 0) {
if(array[size-1] == value) return size-1;
BinSearch(array,size-1,value);
}
else
{
if(array[size/2] == value) return (size/2);
else if(array[size/2] > value) return BinSearch(array, (size/2)+1, value);
else if(array[size/2] < value) return (size/2)+BinSearch(array+size/2, size/2, value);
}
}
or
Binary Search in Array
/**
* Find whether 67 is a prime no
* Domain consists 25 of prime numbers
* Binary Search
*/
int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};
int min = 0,
mid,
max = primes.length,
key = 67,
count= 0;
boolean isFound = false;
while (!isFound) {
if (count < 6) {
mid = (min + max) / 2;
if (primes[mid] == key) {
isFound = true;
System.out.println("Found prime at: " + mid);
} else if (primes[mid] < key) {
min = mid + 1;
isFound = false;
} else if (primes[mid] > key) {
max = mid - 1;
isFound = false;
}
count++;
} else {
System.out.println("No such number");
isFound = true;
}
}
/**
HOPE YOU LIKE IT
A.K.A Binary Search
Take number array of 10 elements, input a number a check whether the number
is present:
**/
package array;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.IOException;
class BinaryS
{
public static void main(String args[]) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter a number: ");
int n=Integer.parseInt(br.readLine());
int a[]={10,20,30,40,50,60,70,80,90,100};
int upper=a.length-1,lower=0,mid;
boolean found=false;
int pos=0;
while(lower<=upper)
{
mid=(upper+lower)/2;
if(n<a[mid])upper=mid-1;
else if(n>a[mid])lower=mid+1;
else
{
found=true;
pos=mid;
break;
}
}
if(found)System.out.println(n+" found at index "+pos);
else System.out.println(n+" not found in array");
}
}
Well I know I am posting this answer much later.
But according to me its always better to check boundary condition at first.
That will make your algorithm more efficient.
public static int binarySearch(int[] array, int element){
if(array == null || array.length == 0){ // validate array
return -1;
}else if(element<array[0] || element > array[array.length-1]){ // validate value our of range that to be search
return -1;
}else if(element == array[0]){ // if element present at very first element of array
return 0;
}else if(element == array[array.length-1]){ // if element present at very last element of array
return array.length-1;
}
int start = 0;
int end = array.length-1;
while (start<=end){
int midIndex = start + ((end-start)/2); // calculate midIndex
if(element < array[midIndex]){ // focus on left side of midIndex
end = midIndex-1;
}else if(element > array[midIndex]){// focus on right side of midIndex
start = midIndex+1;
}else {
return midIndex; // You are in luck :)
}
}
return -1; // better luck next time :(
}
static int binarySearchAlgorithm() {
// Array should be in sorted order. Mandatory requirement
int[] a = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int lowIndex = 0;
int valueToFind = 8;
int highIndex = a.length - 1;
while (lowIndex <= highIndex) {
//Finding the midIndex;
int midIndex = (highIndex + lowIndex) / 2;
// Checking if midIndex value of array contains the value to be find.
if (a[midIndex] == valueToFind) {
return midIndex;
}
// Checking the mid Index value is less than the value to be find.
else if (a[midIndex] < valueToFind) {
// If Yes, changing the lowIndex value to midIndex value + 1;
lowIndex = midIndex + 1;
} else if (a[midIndex] > valueToFind) {
// If Yes, changing the highIndex value to midIndex value - 1;
highIndex = midIndex - 1;
} else {
return -1;
}
}
return -1;
}
I'm reviewing algorithm stuff and stuck in a simple quick sort algorithm implementation in java
import java.util.Arrays;
import java.util.Random;
public class QuickSort {
int arr[];
boolean randomPick;
Random rand = null;
public QuickSort(int[] inputArray) {
arr = inputArray;
randomPick = false;
}
public QuickSort(int[] inputArray, boolean random) {
arr = inputArray;
if (random) {
randomPick = true;
rand = new Random(System.currentTimeMillis());
}
else {
randomPick = false;
}
}
public int[] sort() {
int start = 0;
int end = arr.length - 1;
try {
_sort(start, end);
}
catch (StackOverflowError e) {
System.out.println("StackOverflow: " + Arrays.toString(arr));
}
return arr;
}
private void _sort(int start, int end) {
int i = start;
int j = end;
int pivotLoc;
if (!randomPick) {
pivotLoc = (j + i) / 2;
}
else {
pivotLoc = rand.nextInt(j - i) + i;
}
int pivot = arr[pivotLoc];
while (i < j) { // swapping numbers around the pivot
while (arr[i] < pivot) {
i++;
}
while (arr[j] > pivot) {
j--;
}
if (i < j) {
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
i++;
j--;
}
}
if (i - start > 1) {
_sort(start, i);
}
if (end - i > 1) {
_sort(i, end);
}
}
}
The code can either pick the middle number as the pivot or randomly pick a number as the pivot and it sorts the array by calling _sort recurrently. It works in the first case but fails in the second.
The situation is: when it reaches a subarray as this {3,5,5}, i==start==0 and j==end==2. Finally 5 and 5 are swapped, i becomes 2 and j becomes 1 (i++ and j--). Then since i - start>1 it will call _sort over and over and eventually evoke the stackoverflow error. However the error is supposed to happen in the first case (fixed pivot), which hasn't happened so far...
I don't know what's wrong with my code. I know it's not a pleasure to read code written by others but any help?
You've made a small mistake in your recursion condition, both the start and end compares are against i, the start should be against j
You have:
if (i - start > 1) {
_sort(start, i);
}
if (end - i > 1) {
_sort(i, end);
}
Needs to be:
if (j > start) {
_sort(start, j);
}
if (end > i) {
_sort(i, end);
}
And your i and j comparison needs to allow equals:
// here ----v
if (i <= j) {
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
i++;
j--;
}