Most Difficult Javascript Interview Questions

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Information technology is one of the most profitable and rewarding areas of business. But before you get that awesome CS gig, you need to go through the application process. That’s why we’ve put together this guide - to help you prepare for the most common IT interview questions.

With the growing tech industry, computer science has become a popular career choice for many students. Common roles for which computer science graduates apply are: computer engineer, computer engineer, software development engineer, computer software developer, and application engineer.

Computer Science is one of the most profitable graduates ; s through engineering, mathematics, science and business. As a computer science graduate, you can earn an average salary of $ 50 - $ 130,000 per year . In this area, you will also have great potential for growth as salaries for IT professionals are known to increase dramatically from year to year.

Computer technical interview questions

A person speaking in an interview.
Being good at a technical interview can be the difference between getting hired or fired .

To help you with your interview process, we’ve compiled some tips and tricks that you can use to make your interview a success. Here is a list of some common IT questions you can ask. help to get good results at your next interview.

in a computer maintenance, your answers to general questions and techniques will be an important factor in your selection for the position. the Most questions will be based on coding, programming languages ‚Äã‚Äã, operating systems, hardware and software computer and other subjects related to computer systems.

These are the types of q uestions that establish basic knowledge. For example, what is a class / superclass? What is an object? What is a file? Linked list, class variable, multiple inheritance, base class, input and output, conversion constructor, object oriented programming, etc.

Let’s take a look at some technical interview questions and answer the questions above.

Which programming languages ‚Äã‚Äãdo you prefer and why?

This is a very simple question and you should know that there is no right answer. If the position you are applying for requires knowledge of a certain language, you can mention a few things you like about that particular language. Or you can mention your preferred language and justify your preference. For example:

JavaScript is my favorite language because it is easy to use and my main choice for final back-development. Before JavaScript websites were static, however after JS the web equation changed. Now with JS we have features that help build user friendly and vibrant websites. Here are a few reasons why JS is my favorite programming language to work with.

What is a class and a superclass?

The answer to this question is similar to the others should be at the beginning of the language. You can answer this question as follows:

A class defines the project from which the class objects are created example, the characteristics of an object and with which methods and variables it is associated. A super class, on the other hand, refers to the considered direct class.

What is the difference between Process and Thread?

Although a process and a thread may seem conceptually similar, the two are independent sequences of execution. You can answer this question as follows:

Process is a running program. The thread is a segment of a process, which means that a process can have multiple threads.

Compared to a thread, the process takes longer to complete. The process also takes longer to create while the thread takes less time. The process is isolated while the threads share memory.

What is a constructor?

In object oriented programming, a constructor is a method of a class or structure and initializes an object of that type. Your answer might look like this:

Most of the CS interview questions as discussed above will revolve around similar concepts.

Common questions for IT interviews

  • What is a message ?
  • What is a feed ?
  • How many types of access modifiers are there ? Name and define each of them.
  • What is the application layer?
  • Name the main types of constructors.
  • Explains some main types of access modifiers.
  • What is the transport layer?
  • What is the difference between an interface and an abstract class?
  • Describe a singleton class.
  • What is an abstract class ?
  • What is an abstract keyword ?
  • Provide C source code
  • What is a byte stream ?
  • Name a type of balanced binary tree. It explains how it is implemented.
  • What is a view?
  • How does the central processing unit (CPU) work in a computer system?
  • What is GitHub, and how do you use it ?
  • Explain how a wrapper class works.
  • What is an array ?
  • What does a system work ? Give some examples of common operating systems in use today.
  • Comparison of method and constructor.
  • What is a data structure ?
  • List the steps for creating an object.
  • Explains the purpose of the data link layer.
  • What is recursion?
  • Explain the difference between overload and overwrite.
  • What JavaScript frameworks do you know?
  • What do you know about the software development cycle?
  • What is polymorphism?

Here are some examples of technical interview questions you can expect at your next interview. The best way to prepare would be to review basic CS concepts first. Then, start to compare concepts, identify similarities and differences, and understand the structure and components involved in each concept. You should also check websites like Glassdoor for the types of questions asked of previous applicants for a similar position.

If you are in an interviewee, the interviewer may ask you to answer questions on a whiteboard. . These whiteboard interview questions often ask you to troubleshoot programming issues in real time. In a telephone interview, however, the goal will be to provide your understanding of the concepts and their application.

Behavioral interview questions

When preparing for an IT interview, do not assume that the hiring manager will only ask questions. technical issues. The behavioral interview questions have become an important part of the recruiting process in most companies. This applies to both tech giants and startups.

Behavioral questions are typically worded like this:

  • Tell me about a time when ...
  • Describe a situation where ...
  • Give me an example of a time ...

The purpose of the behavioral questions is to allow the employer to understand the results you have obtained in your past work experience. For example, how you applied your knowledge and skills from your previous job to take concrete " action ‚". This action could be a task that you completed successfully, a team that you managed, or an actual issue that you resolved.

The interviewer can also ask general questions. For example:

Make sure to review your resume several times before the interview so that you can answer any resume-related questions as well.

Ask your questions

The hard part is over ; however, the last part that a lot of people also tend not to pay a lot of attention is very important. In this final step, you can show the interviewer that you are interested in the job and that you have been attentive throughout the interview process.

So at the end when an interviewer asks "Do you have questions for me?" You should always have a list of 2-3 questions. You can ask a follow-up question based on the job description or ask questions about the team.

Conclusion

Preparing for the interview is not an easy process. It takes time and dedication. And this is especially true when it comes to preparing for IT interviews. Each company has a different way of interviewing. You can see sample interview questions for the job on sites like Glassdoor. That way, you can get a feel for the type of questions they might be asking you.

Also be sure to go through the hiring process after the interview. If you don’t get a response from the hiring manager within days, you should contact them and request updates. This shows them that you are still interested in the job and that you know how to take initiatives.

Most Difficult Javascript Interview Questions around: Questions

Removing white space around a saved image in matplotlib

2 answers

I need to take an image and save it after some process. The figure looks fine when I display it, but after saving the figure, I got some white space around the saved image. I have tried the "tight" option for savefig method, did not work either. The code:

  import matplotlib.image as mpimg
  import matplotlib.pyplot as plt

  fig = plt.figure(1)
  img = mpimg.imread(path)
  plt.imshow(img)
  ax=fig.add_subplot(1,1,1)

  extent = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
  plt.savefig("1.png", bbox_inches=extent)

  plt.axis("off") 
  plt.show()

I am trying to draw a basic graph by using NetworkX on a figure and save it. I realized that without a graph it works, but when added a graph I get white space around the saved image;

import matplotlib.image as mpimg
import matplotlib.pyplot as plt
import networkx as nx

G = nx.Graph()
G.add_node(1)
G.add_node(2)
G.add_node(3)
G.add_edge(1,3)
G.add_edge(1,2)
pos = {1:[100,120], 2:[200,300], 3:[50,75]}

fig = plt.figure(1)
img = mpimg.imread("image.jpg")
plt.imshow(img)
ax=fig.add_subplot(1,1,1)

nx.draw(G, pos=pos)

extent = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
plt.savefig("1.png", bbox_inches = extent)

plt.axis("off") 
plt.show()
228

Answer #1

You can remove the white space padding by setting bbox_inches="tight" in savefig:

plt.savefig("test.png",bbox_inches="tight")

You"ll have to put the argument to bbox_inches as a string, perhaps this is why it didn"t work earlier for you.


Possible duplicates:

Matplotlib plots: removing axis, legends and white spaces

How to set the margins for a matplotlib figure?

Reduce left and right margins in matplotlib plot

228

Answer #2

I cannot claim I know exactly why or how my “solution” works, but this is what I had to do when I wanted to plot the outline of a couple of aerofoil sections — without white margins — to a PDF file. (Note that I used matplotlib inside an IPython notebook, with the -pylab flag.)

plt.gca().set_axis_off()
plt.subplots_adjust(top = 1, bottom = 0, right = 1, left = 0, 
            hspace = 0, wspace = 0)
plt.margins(0,0)
plt.gca().xaxis.set_major_locator(plt.NullLocator())
plt.gca().yaxis.set_major_locator(plt.NullLocator())
plt.savefig("filename.pdf", bbox_inches = "tight",
    pad_inches = 0)

I have tried to deactivate different parts of this, but this always lead to a white margin somewhere. You may even have modify this to keep fat lines near the limits of the figure from being shaved by the lack of margins.

Most Difficult Javascript Interview Questions exp: Questions

exp

How do I merge two dictionaries in a single expression (taking union of dictionaries)?

5 answers

Carl Meyer By Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place.

>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{"a": 1, "b": 10, "c": 11}

How can I get that final merged dictionary in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.)

5839

Answer #1

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x.

  • In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

    z = x | y          # NOTE: 3.9+ ONLY
    
  • In Python 3.5 or greater:

    z = {**x, **y}
    
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with keys and values of x
        z.update(y)    # modifies z with keys and values of y
        return z
    

    and now:

    z = merge_two_dicts(x, y)
    

Explanation

Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first.

>>> z
{"a": 1, "b": 3, "c": 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, "foo": 1, "bar": 2, **y}

and now:

>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    z.update(y)
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

def merge_dicts(*dict_args):
    """
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    """
    result = {}
    for dictionary in dict_args:
        result.update(dictionary)
    return result

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key-value pairs in g will take precedence over dictionaries a to f, and so on.

Critiques of Other Answers

Don"t use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"

Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.

and

Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{"a": 1, "b": 10, "c": 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z

Usage:

>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

itertools.chain will chain the iterators over the key-value pairs in the correct order:

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
    z = x.copy()
    z.update(y)
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954
$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries

5839

Answer #2

In your case, what you can do is:

z = dict(list(x.items()) + list(y.items()))

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value:

>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python 2, you can even remove the list() calls. To create z:

>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python version 3.9.0a4 or greater, then you can directly use:

x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
print(z)
{"a": 1, "c": 11, "b": 10}

5839

Answer #3

An alternative:

z = x.copy()
z.update(y)

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