Javascript Insert Sort

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A Java insert sort evaluates each item in a list. If an item is smaller than the previous one, the sort swaps the items. Otherwise, the items stay in the same place and the two items are compared in the list.

Computers are good at sorting the list of items. By using loops, you can search through all the items in a list and change their order until they appear in a certain way.

In programming, there are standard ways to sort lists. We call these sorting algorithms . Sort algorithms read all of the items in a list and sort them use a particular set of statements. One of the most current sorting algorithms you will come across is some sort of insertion.

In this guide, we will discuss how to implement a Java insert sort algorithm . We’ll walk through an example along the way so you can learn the logic behind what type works.

Let s start!

What does a Java insert output?

A Java insert type works like sorting cards in the hand in a deck of cards. Insert sorts check each item in a list and swap them with an item to its left. Whether an item is swapped depends on whether the item is greater or less than the previous item.

Imagine, for a moment, you are playing a card game - whist, rummy, whatever. When you order your cards, what do you do?

You’ll start from the left and check if the second card is sorted. If this card is greater than the last, it must remain in the same position. Otherwise, it must move one position in the list.

You will have spent all the cards in your hand and do this until every card appears in the correct order.

when to use sort input?

There are multiple kinds of inserts used when there are only a few left elements that need to be sorted

There are more efficient algorithms that you can use to sort lists of data, as a kind of merge. That’s why you shouldn’t always use an insert sort by default. However, kinds of integration are more effective than the kinds of bubbles and selection for positions sorting a list.

An insert is sort of a simple sorting algorithm , which means it is useful for beginners to learn.

sorting insert procedure

Viewing a deck of cards is not the most intuitive in the world. To begin with, look at an example of programming. Consider the following list:

8 6 3 9

In this list, we assume that the first item is sorted.

Our next step is to compare the second item in our list with the first. If the first element is greater than the second element, this element is placed in front of the first element.

In this case, 6 is greater than 8. This means that 6 will return a position in our list and 8 will advance a position:

6 8 3 9

Now we need to compare the third element with the elements to its left: 3 is greater than 8? No, so 8 move the position to the right:

6 3 8 9

3 is greater than 6? No, so we’re going to move number 6:

3 6 8 9

Now we need to compare if the fourth point on our list - the last element - is superior to anything else on our list. In this case, 9 is greater than all of the elements before it: 8, 6, and 3.

Here is the algorithm we used to sort the list:

We now have a sorted array. The insertion kinds are sorted on one item at a time. Let’s see how to implement this sorting algorithm in Java

How to perform an insert sort in Java

the higher of two values ‚Äã‚Äãbeing compared is inserted one position to the right each time the sort function is executed.

It’s okay to speak in theoretical terms, but how does it implement in Java? It’s a good question. Let’s write a class that will sort by insertion on a list of student grades.

Prepare the Array Library

Let’s start by importing the Array Library into our Java program. We’ll use this library to print our list to the console once we’ve sorted it:

declare a sort method

We will start by declaring a method that iterates through our list and sorts our data in ascending order >:

Let’s start by finding out how many elements in our input array. This allows us to create a loop that goes through each item in our list. We initialize a loop that loops until we get out of our list.

In our loop, we declared two variables: the key and the last one.

The Java variable "key" keeps track of the element we are currently ordering. The "last" variable keeps track of how many items should be sorted to the left of the item.

Our program will compare the value of "key" to each element to its left until we find a smaller element. This happens in our Java "while" loop.

Define a main function

When we run this code, nothing happens. Indeed, we have not yet defined our main function. We define a main function that defines a network of int (an array of numbers). This main function uses the insertSort () function we declared to sort these digits. Paste this code after declaring your entrySort Java method :

In our main method, we’ve declared a list of the numbers we want to sort. We instantiated our InsertionSort () method called sortNumbers . We use this method to sort the list of numbers in ascending order. This method changes the values ‚Äã‚Äãin our "digits" vector; we didn’t declare a separate array to store its values.

Next, we used the Arrays.toString () method to convert our array of numbers to a string. <

h2> exam complexity

insertion sort has an average case complexity of O (n ^ 2). This happens when no items are sorted.

Best case complexity occurs if an array is sorted. this produces a time complexity of O (n). indeed, the inner loop in some kind of insertion will not be executed at all in this case.

in the worst case, an insertion is made at O ‚Äã‚Äãsort (n ^ 2). This happens if an array is in ascending or descending order and you want to sort in the reverse order (ie, ascending to descending). the confro nto of each element with all the other elements.


Insertion kinds are an efficient way of sorting data. orts compares the values starting from the second in a list. If this value is greater than the one to its left, our list does not change. Otherwise, the value is moved until the element to its left is less.

You are now ready to start writing your own insertion sorting algorithm in Java! If you are looking for more Java learning resources , check out our Java Learning Guide .

Javascript Insert Sort find: Questions


Finding the index of an item in a list

5 answers

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?


Answer #1

>>> ["foo", "bar", "baz"].index("bar")

Reference: Data Structures > More on Lists

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])

Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
>>> next(g)

Most places where I once would have used index, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item"s not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

  1. Check for it first with item in my_list (clean, readable approach), or
  2. Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)


Answer #2

One thing that is really helpful in learning Python is to use the interactive help function:

>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)

 |  index(...)
 |      L.index(value, [start, [stop]]) -> integer -- return first index of value

which will often lead you to the method you are looking for.


Answer #3

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():

for i, j in enumerate(["foo", "bar", "baz"]):
    if j == "bar":

The index() function only returns the first occurrence, while enumerate() returns all occurrences.

As a list comprehension:

[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]

Here"s also another small solution with itertools.count() (which is pretty much the same approach as enumerate):

from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]

This is more efficient for larger lists than using enumerate():

$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop

Javascript Insert Sort iat: Questions


InsecurePlatformWarning: A true SSLContext object is not available. This prevents urllib3 from configuring SSL appropriately

3 answers

Tried to perform REST GET through python requests with the following code and I got error.

Code snip:

import requests
header = {"Authorization": "Bearer..."}
url = az_base_url + az_subscription_id + "/resourcegroups/Default-Networking/resources?" + az_api_version
r = requests.get(url, headers=header)


          InsecurePlatformWarning: A true SSLContext object is not available. 
          This prevents urllib3 from configuring SSL appropriately and may cause certain SSL connections to fail. 
          For more information, see

My python version is 2.7.3. I tried to install urllib3 and requests[security] as some other thread suggests, I still got the same error.

Wonder if anyone can provide some tips?


Answer #1

The docs give a fair indicator of what"s required., however requests allow us to skip a few steps:

You only need to install the security package extras (thanks @admdrew for pointing it out)

$ pip install requests[security]

or, install them directly:

$ pip install pyopenssl ndg-httpsclient pyasn1

Requests will then automatically inject pyopenssl into urllib3

If you"re on ubuntu, you may run into trouble installing pyopenssl, you"ll need these dependencies:

$ apt-get install libffi-dev libssl-dev


Dynamic instantiation from string name of a class in dynamically imported module?

3 answers

In python, I have to instantiate certain class, knowing its name in a string, but this class "lives" in a dynamically imported module. An example follows:

loader-class script:

import sys
class loader:
  def __init__(self, module_name, class_name): # both args are strings
      modul = sys.modules[module_name]
      instance = modul.class_name() # obviously this doesn"t works, here is my main problem!
    except ImportError:
       # manage import error

some-dynamically-loaded-module script:

class myName:
  # etc...

I use this arrangement to make any dynamically-loaded-module to be used by the loader-class following certain predefined behaviours in the dyn-loaded-modules...


Answer #1

You can use getattr

getattr(module, class_name)

to access the class. More complete code:

module = __import__(module_name)
class_ = getattr(module, class_name)
instance = class_()

As mentioned below, we may use importlib

import importlib
module = importlib.import_module(module_name)
class_ = getattr(module, class_name)
instance = class_()


How to get all of the immediate subdirectories in Python

3 answers

I"m trying to write a simple Python script that will copy a index.tpl to index.html in all of the subdirectories (with a few exceptions).

I"m getting bogged down by trying to get the list of subdirectories.


Answer #1

import os
def get_immediate_subdirectories(a_dir):
    return [name for name in os.listdir(a_dir)
            if os.path.isdir(os.path.join(a_dir, name))]


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