Bucket Sort – bucket sorting. The algorithm is similar to sorting by counting, with the difference that the numbers are collected into “buckets”-ranges, then the buckets are sorted using any other, sufficiently productive, sorting algorithm, and the final chord is the expansion of the “buckets” one by one, resulting in a sorted list.

The time complexity of the algorithm is O(nk). The algorithm runs in linear time for data that obeys a uniform distribution. To put it simply, the elements must be in a certain range, without “splashes”, for example, numbers from 0.0 to 1.0. If among such numbers there are 4 or 999, then such a series, according to the laws of the yard, is no longer considered “even”.

Implementation example in Julia:

function bucketSort(numbers, bucketsCount)
buckets = Vector{Vector{Int}}()
for i in 0:bucketsCount - 1
bucket = Vector{Int}()
push!(buckets, bucket)
end
maxNumber = maximum(numbers)
for i in 0:length(numbers) - 1
bucketIndex = 1 + Int(floor(bucketsCount * numbers[1 + i] / (maxNumber + 1)))
push!(buckets[bucketIndex], numbers[1 + i])
end
for i in 0:length(buckets) - 1
bucketIndex = 1 + i
buckets[bucketIndex] = sort(buckets[bucketIndex])
end
flat = [(buckets...)...]
print(flat, "\n")
end
numbersCount = 10
maxNumber = 10
numbers = rand(1:maxNumber, numbersCount)
print(numbers,"\n")
bucketsCount = 10
bucketSort(numbers, bucketsCount)

The performance of the algorithm is also affected by the number of buckets, for more numbers it is better to take a larger number of buckets (Algorithms in a nutshell by George T. Heineman)

Radix Sort – radix sort. The algorithm is similar to counting sort in that there is no comparison of elements, instead elements are *character-by-character* grouped into *buckets* (buckets), the bucket is selected by the index of the current number-character. Time complexity – O(nd).

Works like this:

The input will be the numbers 6, 12, 44, 9

Let’s create 10 buckets of lists (0-9) into which we will add/sort numbers bit by bit.

Further:

Run a loop with counter i up to the maximum number of characters in the number

At index i from right to left we get one character for each number, if there is no character, then we consider it to be zero

The character is converted to a number

Select a bucket by index – number, put the whole number there

After finishing iterating over numbers, convert all buckets back to a list of numbers

Get numbers sorted by digit

Repeat until all digits run out

Radix Sort example in Scala:

import scala.collection.mutable.ListBuffer
import scala.util.Random.nextInt
object RadixSort {
def main(args: Array[String]) = {
var maxNumber = 200
var numbersCount = 30
var maxLength = maxNumber.toString.length() - 1
var referenceNumbers = LazyList.continually(nextInt(maxNumber + 1)).take(numbersCount).toList
var numbers = referenceNumbers
var buckets = List.fill(10)(ListBuffer[Int]())
for( i <- 0 to maxLength) { numbers.foreach( number => {
var numberString = number.toString
if (numberString.length() > i) {
var index = numberString.length() - i - 1
var character = numberString.charAt(index).toString
var characterInteger = character.toInt
buckets.apply(characterInteger) += number
}
else {
buckets.apply(0) += number
}
}
)
numbers = buckets.flatten
buckets.foreach(x => x.clear())
}
println(referenceNumbers)
println(numbers)
println(s"Validation result: ${numbers == referenceNumbers.sorted}")
}
}

The algorithm also has a version for parallel execution, for example on the GPU; there is also a variant of bit sort, which is probably very interesting and truly breathtaking!

Heapsort – heap sort. Time complexity – O(n log n), fast eh? I would call this sorting – sorting of falling stones. It seems to me that the easiest way to explain it is visually.

The input is a list of numbers, for example:
5, 0, 7, 2, 3, 9, 4

From left to right, a data structure is made – a binary tree, or as I call it – a pyramid. Pyramid elements can have a maximum of two child elements, with only one top element.

Let’s make a binary tree:
⠀⠀5
⠀0⠀7
2 3 9 4

If you look at the pyramid for a long time, you can see that these are just numbers from the array, going one after another, the number of elements in each floor is multiplied by two.

Then the fun begins, we sort the pyramid from bottom to top, using the falling pebbles method (heapify). Sorting could be started from the last floor (2 3 9 4 ), but it makes no sense because there is no floor below where one could fall.

Therefore, we start dropping elements from the penultimate floor (0 7)
⠀⠀5
⠀0⠀7
2 3 9 4

The first element to fall is selected on the right, in our case it is 7, then we look at what is under it, and below it are 9 and 4, nine is more than four, so also nine is more than seven! We drop 7 on 9, and raise 9 to place 7.
⠀⠀5
⠀0⠀9
2 3 7 4

Further, we understand that the seven has nowhere to fall below, go to the number 0, which is located on the penultimate floor on the left:
⠀⠀5
⠀0⠀9
2 3 7 4

We look at what is under it – 2 and 3, two is less than three, three is greater than zero, so we change zero and three in places:
⠀⠀5
⠀3⠀9
2 0 7 4

When you get to the end of the floor, go to the floor above and drop everything there if you can.
The result is a data structure – a heap (heap), namely max heap, because at the top is the largest element:
⠀⠀9
⠀3⠀7
2 0 5 4

If you return it to an array representation, you get a list:
[9, 3, 7, 2, 0, 5, 4]

From this we can conclude that by swapping the first and last element, we will get the first number in the final sorted position, namely 9 should be at the end of the sorted list, swap:
[4, 3, 7, 2, 0, 5, 9]

Let’s look at the binary tree:
⠀⠀4
⠀3⠀7
2 0 5 9

The result is a situation in which the lower part of the tree is sorted, you just need to drop 4 to the correct position, repeat the algorithm, but do not take into account the already sorted numbers, namely 9:
⠀⠀4
⠀3⠀7
2 0 5 9

⠀⠀7
⠀3⠀4
2 0 5 9

⠀⠀7
⠀3⠀5
2 0 4 9

It turned out that we, having dropped 4, raised the next largest number after 9 – 7. Swap the last unsorted number (4) and the largest number (7)
⠀⠀4
⠀3⠀5
2 0 7 9

It turned out that now we have two numbers in the correct final position:
4, 3, 5, 2, 0, 7, 9

Next, we repeat the sorting algorithm, ignoring those already sorted, as a result we get heap:
⠀⠀0
⠀2⠀3
4 5 7 9

Or as a list:
0, 2, 3, 4, 5, 7, 9

Implementation

The algorithm is usually divided into three functions:

Heap creation

Sifting algorithm (heapify)

Replacing the last unsorted element and the first one

A heap is created by traversing the penultimate row of the binary tree using the heapify function, from right to left to the end of the array. Then the first replacement of numbers is made in the cycle, after which the first element falls / remains in place, as a result of which the largest element falls into first place, the cycle repeats with a decrease in participants by one, because after each pass, the sorted numbers remain at the end of the list.

Heapsort example in Ruby:

DEMO = true
module Colors
BLUE = "\033[94m"
RED = "\033[31m"
STOP = "\033[0m"
end
def heapsort(rawNumbers)
numbers = rawNumbers.dup
def swap(numbers, from, to)
temp = numbers[from]
numbers[from] = numbers[to]
numbers[to] = temp
end
def heapify(numbers)
count = numbers.length()
lastParentNode = (count - 2) / 2
for start in lastParentNode.downto(0)
siftDown(numbers, start, count - 1)
start -= 1
end
if DEMO
puts "--- heapify ends ---"
end
end
def siftDown(numbers, start, rightBound)
cursor = start
printBinaryHeap(numbers, cursor, rightBound)
def calculateLhsChildIndex(cursor)
return cursor * 2 + 1
end
def calculateRhsChildIndex(cursor)
return cursor * 2 + 2
end
while calculateLhsChildIndex(cursor) <= rightBound
lhsChildIndex = calculateLhsChildIndex(cursor)
rhsChildIndex = calculateRhsChildIndex(cursor)
lhsNumber = numbers[lhsChildIndex]
biggerChildIndex = lhsChildIndex
if rhsChildIndex <= rightBound
rhsNumber = numbers[rhsChildIndex]
if lhsNumber < rhsNumber
biggerChildIndex = rhsChildIndex
end
end
if numbers[cursor] < numbers[biggerChildIndex]
swap(numbers, cursor, biggerChildIndex)
cursor = biggerChildIndex
else
break
end
printBinaryHeap(numbers, cursor, rightBound)
end
printBinaryHeap(numbers, cursor, rightBound)
end
def printBinaryHeap(numbers, nodeIndex = -1, rightBound = -1)
if DEMO == false
return
end
perLineWidth = (numbers.length() * 4).to_i
linesCount = Math.log2(numbers.length()).ceil()
xPrinterCount = 1
cursor = 0
spacing = 3
for y in (0..linesCount)
line = perLineWidth.times.map { " " }
spacing = spacing == 3 ? 4 : 3
printIndex = (perLineWidth / 2) - (spacing * xPrinterCount) / 2
for x in (0..xPrinterCount - 1)
if cursor >= numbers.length
break
end
if nodeIndex != -1 && cursor == nodeIndex
line[printIndex] = "%s%s%s" % [Colors::RED, numbers[cursor].to_s, Colors::STOP]
elsif rightBound != -1 && cursor > rightBound
line[printIndex] = "%s%s%s" % [Colors::BLUE, numbers[cursor].to_s, Colors::STOP]
else
line[printIndex] = numbers[cursor].to_s
end
cursor += 1
printIndex += spacing
end
print line.join()
xPrinterCount *= 2
print "\n"
end
end
heapify(numbers)
rightBound = numbers.length() - 1
while rightBound > 0
swap(numbers, 0, rightBound)
rightBound -= 1
siftDown(numbers, 0, rightBound)
end
return numbers
end
numbersCount = 14
maximalNumber = 10
numbers = numbersCount.times.map { Random.rand(maximalNumber) }
print numbers
print "\n---\n"
start = Time.now
sortedNumbers = heapsort(numbers)
finish = Time.now
heapSortTime = start - finish
start = Time.now
referenceSortedNumbers = numbers.sort()
finish = Time.now
referenceSortTime = start - finish
print "Reference sort: "
print referenceSortedNumbers
print "\n"
print "Reference sort time: %f\n" % referenceSortTime
print "Heap sort: "
print sortedNumbers
print "\n"
if DEMO == false
print "Heap sort time: %f\n" % heapSortTime
else
print "Disable DEMO for performance measure\n"
end
if sortedNumbers != referenceSortedNumbers
puts "Validation failed"
exit 1
else
puts "Validation success"
exit 0
end

Without visualization, this algorithm is not easy to understand, so the first thing I recommend is to write a function that will print the current view of the binary tree.

Recently, it turned out that users of modern Nvidia GPUs under Arch Linux do not need to use the bumblebee package at all, for example, for me it did not detect an external monitor when connected. I recommend removing the bumblebee package and all related packages, and installing prime using the instructions on the Arch Wiki.
Next, to launch all games on Steam and 3D applications, add prime-run, for Steam this is done like this prime-run %command% in additional launch options.
To check the correctness, you can use glxgears, prime-run glxgears. https://bbs.archlinux.org/viewtopic.php? pid=2048195#p2048195

Quicksort is a divide-and-conquer sorting algorithm. Recursively, in parts, parse the array of numbers, setting the numbers in the smaller and larger order from the selected pivot element, insert the pivot element itself into the hot-point between them. After a few recursive iterations, you’ll end up with a sorted list. Time complexity O(n^{2}).

Scheme:

We start with the fact that we get a list of elements outside, sorting boundaries. In the first step, the sort boundaries will be from start to finish.

Check that the boundaries of the beginning and end do not intersect, if this happens, then it’s time to finish

Select some element from the list, call it pivot

Shift to the right to the end to the last index, so as not to interfere

Create a counter of *smaller numbers* while equal to zero

Loop through the list from left to right, up to and including the last index where the anchor element is located

Each element is compared to the pivot

If it is less than the reference, then we swap it in places according to the index of the counter of smaller numbers. Increment the counter of smaller numbers.

When the cycle reaches the anchor element, we stop, swap the anchor element with the element by the counter of smaller numbers.

We run the algorithm separately for the left smaller part of the list, and separately for the right most part of the list.

As a result, all recursive iterations will start to stop due to the check in paragraph 2

Get sorted list

Quicksort was invented by the scientist Charles Antony Richard Hoare at Moscow State University, having learned Russian, he studied computer translation, as well as probability theory at the Kolmogorov school. In 1960, due to a political crisis, he left the Soviet Union.

An example implementation in Rust:

extern crate rand;
use rand::Rng;
fn swap(numbers: &mut [i64], from: usize, to: usize) {
let temp = numbers[from];
numbers[from] = numbers[to];
numbers[to] = temp;
}
fn quicksort(numbers: &mut [i64], left: usize, right: usize) {
if left >= right {
return
}
let length = right - left;
if length <= 1 {
return
}
let pivot_index = left + (length / 2);
let pivot = numbers[pivot_index];
let last_index = right - 1;
swap(numbers, pivot_index, last_index);
let mut less_insert_index = left;
for i in left..last_index {
if numbers[i] < pivot {
swap(numbers, i, less_insert_index);
less_insert_index += 1;
}
}
swap(numbers, last_index, less_insert_index);
quicksort(numbers, left, less_insert_index);
quicksort(numbers, less_insert_index + 1, right);
}
fn main() {
let mut numbers = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0];
let mut reference_numbers = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0];
let mut rng = rand::thread_rng();
for i in 0..numbers.len() {
numbers[i] = rng.gen_range(-10..10);
reference_numbers[i] = numbers[i];
}
reference_numbers.sort();
println!("Numbers {:?}", numbers);
let length = numbers.len();
quicksort(&mut numbers, 0, length);
println!("Numbers {:?}", numbers);
println!("Reference numbers {:?}", reference_numbers);
if numbers != reference_numbers {
println!("Validation failed");
std::process::exit(1);
}
else {
println!("Validation success!");
std::process::exit(0);
}
}

If you don't understand anything, I suggest watching a video by Rob Edwards from the San Diego State University https://www.youtube.com/watch?v=ZHVk2blR45Q it most simply, step by step, shows the essence and implementation of the algorithm.

Binary Insertion Sort is a variant of insertion sort in which the insertion position is determined using a binary search. The time complexity of the algorithm is O(n^{2})

The algorithm works like this:

Loop from zero to end of list

In the loop, a number is selected for sorting, the number is stored in a separate variable

Binary search looks for the index to insert this number against the numbers on the left

After finding the index, the numbers on the left are shifted one position to the right, starting at the insertion index. The process will erase the number to be sorted.

The previously stored number is inserted at the insertion index

At the end of the loop, the entire list will be sorted

During a binary search, it is possible that the number will not be found, but the index is not returned. Due to the peculiarities of the binary search, the number closest to the desired one will be found, then to return the index it will be necessary to compare it with the desired one, if the desired one is less, then the desired one should be at the index on the left, and if it is greater or equal, then on the right.

Source code in Go:

package main
import (
"fmt"
"math/rand"
"time"
)
const numbersCount = 20
const maximalNumber = 100
func binarySearch(numbers []int, item int, low int, high int) int {
for high > low {
center := (low + high) / 2
if numbers[center] < item { low = center + 1 } else if numbers[center] > item {
high = center - 1
} else {
return center
}
}
if numbers[low] < item {
return low + 1
} else {
return low
}
}
func main() {
rand.Seed(time.Now().Unix())
var numbers [numbersCount]int
for i := 0; i < numbersCount; i++ {
numbers[i] = rand.Intn(maximalNumber)
}
fmt.Println(numbers)
for i := 1; i < len(numbers); i++ { searchAreaLastIndex := i - 1 insertNumber := numbers[i] insertIndex := binarySearch(numbers[:], insertNumber, 0, searchAreaLastIndex) for x := searchAreaLastIndex; x >= insertIndex; x-- {
numbers[x+1] = numbers[x]
}
numbers[insertIndex] = insertNumber
}
fmt.Println(numbers)
}

Shell Sort is a variant of sorting by inserts with preliminary combing of an array of numbers.

We need to remember how insertion sort works:

The loop starts from zero to the end of the loop, thus the array is divided into two parts

For the left side, the second loop is started, comparing elements from right to left, the smaller element on the right is omitted until there is a smaller element on the left

At the end of both cycles, we get a sorted list

Once in a while, computer scientist Donald Shell wondered how to improve the insertion sort algorithm. He came up with the idea to preliminarily go through the array with two cycles, but at a certain distance, gradually reducing the “comb” until it turns into a regular insertion sort algorithm. Everything is really so simple, no pitfalls, we add another one to the two cycles from above, in which we gradually reduce the size of the “comb”. The only thing that will need to be done is to check the distance when comparing so that it does not go beyond the array.

A really interesting topic is the choice of sequence for changing the length of the comparison at each iteration of the first loop. It is interesting for the reason that the performance of the algorithm depends on it.

Different people were engaged in calculating the ideal distance, so, apparently, this topic was interesting to them. Couldn’t they just fire up Ruby, call the fastest sort() algorithm?

In general, these strange people wrote dissertations on the topic of calculating the distance / gap of the “comb” for the Shell algorithm. I just used the results of their work and checked 5 types of sequences, Hibbard, Knuth-Pratt, Ciura, Sedgewick.

from typing import List
import time
import random
from functools import reduce
import math
DEMO_MODE = False
if input("Demo Mode Y/N? ").upper() == "Y":
DEMO_MODE = True
class Colors:
BLUE = '\033[94m'
RED = '\033[31m'
END = '\033[0m'
def swap(list, lhs, rhs):
list[lhs], list[rhs] = list[rhs], list[lhs]
return list
def colorPrintoutStep(numbers: List[int], lhs: int, rhs: int):
for index, number in enumerate(numbers):
if index == lhs:
print(f"{Colors.BLUE}", end = "")
elif index == rhs:
print(f"{Colors.RED}", end = "")
print(f"{number},", end = "")
if index == lhs or index == rhs:
print(f"{Colors.END}", end = "")
if index == lhs or index == rhs:
print(f"{Colors.END}", end = "")
print("\n")
input(">")
def ShellSortLoop(numbers: List[int], distanceSequence: List[int]):
distanceSequenceIterator = reversed(distanceSequence)
while distance:= next(distanceSequenceIterator, None):
for sortArea in range(0, len(numbers)):
for rhs in reversed(range(distance, sortArea + 1)):
lhs = rhs - distance
if DEMO_MODE:
print(f"Distance: {distance}")
colorPrintoutStep(numbers, lhs, rhs)
if numbers[lhs] > numbers[rhs]:
swap(numbers, lhs, rhs)
else:
break
def ShellSort(numbers: List[int]):
global ShellSequence
ShellSortLoop(numbers, ShellSequence)
def HibbardSort(numbers: List[int]):
global HibbardSequence
ShellSortLoop(numbers, HibbardSequence)
def ShellPlusKnuttPrattSort(numbers: List[int]):
global KnuttPrattSequence
ShellSortLoop(numbers, KnuttPrattSequence)
def ShellPlusCiuraSort(numbers: List[int]):
global CiuraSequence
ShellSortLoop(numbers, CiuraSequence)
def ShellPlusSedgewickSort(numbers: List[int]):
global SedgewickSequence
ShellSortLoop(numbers, SedgewickSequence)
def insertionSort(numbers: List[int]):
global insertionSortDistanceSequence
ShellSortLoop(numbers, insertionSortDistanceSequence)
def defaultSort(numbers: List[int]):
numbers.sort()
def measureExecution(inputNumbers: List[int], algorithmName: str, algorithm):
if DEMO_MODE:
print(f"{algorithmName} started")
numbers = inputNumbers.copy()
startTime = time.perf_counter()
algorithm(numbers)
endTime = time.perf_counter()
print(f"{algorithmName} performance: {endTime - startTime}")
def sortedNumbersAsString(inputNumbers: List[int], algorithm) -> str:
numbers = inputNumbers.copy()
algorithm(numbers)
return str(numbers)
if DEMO_MODE:
maximalNumber = 10
numbersCount = 10
else:
maximalNumber = 10
numbersCount = random.randint(10000, 20000)
randomNumbers = [random.randrange(1, maximalNumber) for i in range(numbersCount)]
ShellSequenceGenerator = lambda n: reduce(lambda x, _: x + [int(x[-1]/2)], range(int(math.log(numbersCount, 2))), [int(numbersCount / 2)])
ShellSequence = ShellSequenceGenerator(randomNumbers)
ShellSequence.reverse()
ShellSequence.pop()
HibbardSequence = [
0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095,
8191, 16383, 32767, 65535, 131071, 262143, 524287, 1048575,
2097151, 4194303, 8388607, 16777215, 33554431, 67108863, 134217727,
268435455, 536870911, 1073741823, 2147483647, 4294967295, 8589934591
]
KnuttPrattSequence = [
1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573, 265720,
797161, 2391484, 7174453, 21523360, 64570081, 193710244, 581130733,
1743392200, 5230176601, 15690529804, 47071589413
]
CiuraSequence = [
1, 4, 10, 23, 57, 132, 301, 701, 1750, 4376,
10941, 27353, 68383, 170958, 427396, 1068491,
2671228, 6678071, 16695178, 41737946, 104344866,
260862166, 652155416, 1630388541
]
SedgewickSequence = [
1, 5, 19, 41, 109, 209, 505, 929, 2161, 3905,
8929, 16001, 36289, 64769, 146305, 260609, 587521,
1045505, 2354689, 4188161, 9427969, 16764929, 37730305,
67084289, 150958081, 268386305, 603906049, 1073643521,
2415771649, 4294770689, 9663381505, 17179475969
]
insertionSortDistanceSequence = [1]
algorithms = {
"Default Python Sort": defaultSort,
"Shell Sort": ShellSort,
"Shell + Hibbard" : HibbardSort,
"Shell + Prat, Knutt": ShellPlusKnuttPrattSort,
"Shell + Ciura Sort": ShellPlusCiuraSort,
"Shell + Sedgewick Sort": ShellPlusSedgewickSort,
"Insertion Sort": insertionSort
}
for name, algorithm in algorithms.items():
measureExecution(randomNumbers, name, algorithm)
reference = sortedNumbersAsString(randomNumbers, defaultSort)
for name, algorithm in algorithms.items():
if sortedNumbersAsString(randomNumbers, algorithm) != reference:
print("Sorting validation failed")
exit(1)
print("Sorting validation success")
exit(0)

In my implementation, for a random set, the fastest numbers are the Sedgwick and Hibbard gaps.

mypy

I would also like to mention the static typing analyzer for Python 3 – mypy. Helps to cope with the problems inherent in languages with dynamic typing, namely, it eliminates the possibility of sticking something where it is not necessary.

As experienced programmers say, “static typing is not needed when you have a team of professionals”, someday we will all become professionals, we will write code in complete unity and understanding with machines, but for now you can use similar utilities and languages with static typing.

Double Selection Sort – a subspecies of sorting by selection, it seems like it should be accelerated twice. The vanilla algorithm double loops through the list of numbers, finds the minimum number, and swaps it with the current digit pointed to by the loop at the level above. Double selection sort, on the other hand, looks for the minimum and maximum number, then it replaces the two digits indicated by the loop at the level above – two numbers on the left and on the right. All this orgy ends when the cursors of numbers to replace meet in the middle of the list, as a result, sorted numbers are obtained to the left and right of the visual center. The time complexity of the algorithm is similar to Selection Sort – O(n2), but supposedly there is a 30% speedup.

Corner case

Already at this stage, you can imagine the moment of collision, when the number of the left cursor (the minimum number) will point to the maximum number in the list, then the minimum number is permuted, the permutation of the maximum number immediately breaks down. Therefore, all implementations of the algorithm contain a check of such cases, replacement of indices with correct ones. In my implementation, one check was enough:

if (leftCursor == maximalNumberIndex) {
maximalNumberIndex = minimalNumberIndex;
}

Cito implementation

Cito is a lib language, a translator language. You can write on it for C, C++, C#, Java, JavaScript, Python, Swift, TypeScript, OpenCL C, while knowing absolutely nothing about these languages. The source code in the Cito language is translated into the source code in the supported languages, then you can use it as a library, or directly by correcting the generated code by hand. A kind of Write once – translate to anything.
Double Selection Sort – Cito:

public class DoubleSelectionSort
{
public static int[] sort(int[]# numbers, int length)
{
int[]# sortedNumbers = new int[length];
for (int i = 0; i < length; i++) {
sortedNumbers[i] = numbers[i];
}
for (int leftCursor = 0; leftCursor < length / 2; leftCursor++) {
int minimalNumberIndex = leftCursor;
int minimalNumber = sortedNumbers[leftCursor];
int rightCursor = length - (leftCursor + 1);
int maximalNumberIndex = rightCursor;
int maximalNumber = sortedNumbers[maximalNumberIndex];
for (int cursor = leftCursor; cursor <= rightCursor; cursor++) { int cursorNumber = sortedNumbers[cursor]; if (minimalNumber > cursorNumber) {
minimalNumber = cursorNumber;
minimalNumberIndex = cursor;
}
if (maximalNumber < cursorNumber) {
maximalNumber = cursorNumber;
maximalNumberIndex = cursor;
}
}
if (leftCursor == maximalNumberIndex) {
maximalNumberIndex = minimalNumberIndex;
}
int fromNumber = sortedNumbers[leftCursor];
int toNumber = sortedNumbers[minimalNumberIndex];
sortedNumbers[minimalNumberIndex] = fromNumber;
sortedNumbers[leftCursor] = toNumber;
fromNumber = sortedNumbers[rightCursor];
toNumber = sortedNumbers[maximalNumberIndex];
sortedNumbers[maximalNumberIndex] = fromNumber;
sortedNumbers[rightCursor] = toNumber;
}
return sortedNumbers;
}
}

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Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. It is mandatory to procure user consent prior to running these cookies on your website.