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JavaScript: Array Manipulation

JavaScript: Array Manipulation

JavaScript offers different ways to manipulate arrays. However, there are methods that changes the original array and other methods that don’t. The methods that changes original array is called MUTATION. Let’s take a look at most common methods to manipulate arrays in JavaScript Now, that we know that there are methods that mutates the original array, it’s considered best practice to assign the array to const, if you know that the array that you use is not going to get mutated. It’s wise to use let otherwise. It is not a mandate that you need to use const to declare an array that will not be mutated, however, it’s easy for your colleague to understand that the array which is declared as const is not going to get changed anywhere in the code; which is why it’s considered best practice As usual, if you wanna try the examples in this post, I recommend codepen.io, but you are free to make your own choice. The most common interactions that we usually have with arrays areAdd item to array Remove item from array Update and item in the arrayAll the above mentioned actions can be done by both mutating and non-mutating methods. Instead of explaining each of the methods subjectively, I felt it would be better understood with a table. Here we go! testArray: [a, b, c, d, e, f, g, h, i];Code Original Array Processed Array Returned Val Is Mutated Action MethodtestArray.push('j') [a,b,c,d,e,f,g,h,i] [a,b,c,d,e,f,g,h,i,j] 10 true add push()testArray.unshift('z') [a,b,c,d,e,f,g,h,i,j] [z,a,b,c,d,e,f,g,h,i,j] 11 true add unshift()testArray.concat('k') [z,a,b,c,d,e,f,g,h,i,j] [z,a,b,c,d,e,f,g,h,i,j] [z,a,b,c,d,e,f,g,h,i,j,k] false add concat()['y', ...testArray, 'l'] [z,a,b,c,d,e,f,g,h,i,j] [z,a,b,c,d,e,f,g,h,i,j] [y,z,a,b,c,d,e,f,g,h,i,j,l] false add ...testArray.pop() [z,a,b,c,d,e,f,g,h,i,j] [z,a,b,c,d,e,f,g,h,i] j true remove pop()testArray.shift() [z,a,b,c,d,e,f,g,h,i] [a,b,c,d,e,f,g,h,i] z true remove shift()testArray.splice(0, 2) [a,b,c,d,e,f,g,h,i] [c,d,e,f,g,h,i] [a,b] true remove splice()testArray.filter(a => a!== 'c') [c,d,e,f,g,h,i] [c,d,e,f,g,h,i] [d,e,f,g,h,i] false remove filter()testArray.slice(1, 6) [c,d,e,f,g,h,i] [c,d,e,f,g,h,i] [d,e,f,g,h] false remove slice()testArray.slice(2) [c,d,e,f,g,h,i] [c,d,e,f,g,h,i] [e,f,g,h,i] false remove slice()testArray.splice(2, 1,30, 31) [c,d,e,f,g,h,i] [c,d,30,31,f,g,h,i] [e] true update splice()testArray.map(x => x ==='d' ? 29 : x) [c,d,30,31,f,g,h,i] [c,d,30,31,f,g,h,i] [c,29,30,31,f,g,h,i] false update map()I hope the above table is self-explanatory, however if you need to check the values of the testArray with real code, feel free to visit programmatic output where the exact same table is derived programmatically. Codepen pin link

Cookie Policy

Cookie Policy

Ever wondered why do websites have no decline option in their cookie policy popup? Well, lets’ see. Today, almost all sites in the internet uses cookies that is stored temporarily in the system cache to ease authentication of the same user in the same system. I’m talking about well-established sites that makes some use of the information in your local machine (THE COOKIE!). Why do I need it? As per GDPR, every website that tracks their visitors should have the visitors consent to use their data to better manage what they need. So if you are tracking your website users and making decisions based on their user data, then ‘YES’, you have to force the user to accept your cookie policy. Do I need it? Before we get into the fun part, we gotta understand what this cookie policy can do. Let’s say you own a blogging portal with ‘Disqus’ comment system enabled. Obviously, any visitor need to have a ‘Disqus’ account in order to comment. This means, like it or not ‘Disqus’ can track your visitors on which sites they visit and which site they comment and maybe manipulate your data based on their need. Now, let’s get into the fun part. As long as you plan to use your visitor’s data and manipulate something from it, you gotta have the cookie policy in your site. However, what do you think you can do with your visitor’s data? As soon as that question is asked, we start to think big! Maybe we can find users locations. “Ok, then what?” If you are just planning or started a blog for yourself like me, all we have is our petty subscription list, which is a list of email ids’ that will be almost empty for atleast couple of months! What can possibly be done with that except sending emails about new posts, which again seldom occurs? Therefore, my view of adding a cookie policy is, if you are just starting a blog; just make sure that you are not adding any tracking 3rd party code that tracks your blogs visitor. That means, in case of WordPress websites, the plug-ins that you use in your site may need you to register in order to use it, but not for your visitors. That is it. You have a long way to go before you even think about adding the cookie policy. Just focus on adding new and useful posts. After all, who wants to accept a policy that has no choices?

Binary Search

Binary Search

Before we get into the syntax and logic of binary search algorithm, let’s ask ourselves this question. Why do we need binary search at all in the first place? To understand, let’s work with an example here. Let’s say you were given a name in a letter pad (say ‘Michael’) and you were asked to find that person from a group of 50 people who are standing by name order i.e., Adam stands first, then Alex, then Bob and goes on. You get the idea. Now, what you do is, pick the 25th person and ask his/her name. Let’s say the person’s name is ‘Monica’. Now you know that ‘Michael’ is somewhere before ‘Monica’ as everyone is standing in ascending order. The reason I picked the 25th person, is coz’ picking that person would give me the max reduction. In the sense, I can ignore the other 25 people after ‘Monica’, since; I know that ‘Michael’ cannot come after ‘Monica’. The idea is filtering 50% of the wrong choices in every question you ask, will give you the fastest route to the destination. To make it clearer, let’s say I pick a random number (say 35) instead of picking the 25th person. I ask the 35th person, what his/her name is? Now there are two possible outcomes. Either his/her name can be ‘Stuart’ or ‘Henry’. Now, you have a 50 percent chance that you might get either one of these names. If the person says ‘Henry’ then you are lucky, coz’ you only need to filter remaining 15 people to find out your match, but guess what, what if the person’s name is ‘Stuart’, you just got burned here! In other words, instead of filtering from your best bet number, which is 25, now you need to filter from 35. Hope you get the reason why you need to filter by half every time to find the quickest way to find your match. If all the names in the world are written down together in order and you want to search for the position of a specific name, binary search will accomplish this in a maximum of 35 iterations, which means with asking 35 questions, you will be able to find the person that you are looking for. Isn’t that cool! Hence, always try to order your array or collection in some way and make sure to index it i.e., if you are adding a row in your excel spreadsheet which you know that the records are gonna increase in the course of time, make sure to add a serial number to each row, so that you can order by ascending/descending anytime to sort the list and, once the list is sorted, perform the binary search to find the record you need. So, to answer the question ‘why’, binary search algorithm is one of the fastest way to find a random number in a set of indexed numbers, instead of searching the numbers sequentially. Now that’s why! Alright, funz over. Let’s get to work now. Below is the program to implement binary search in VB Script Dim testarr(100000) Dim n, i, resn = 5643For i = 1 To 100000 testarr(i) = i Nextres = CStr(binser(testarr, n)) Wscript.echo resFunction binser(arr, tar) Dim low, high low = LBound(arr) high = UBound(arr) iter = 0 Do While low <= high i = Round((low + high) / 2) If tar = arr(i) Then binser = True Wscript.echo iter Exit Do ElseIf tar < arr(i) Then high = i - 1 Else low = i + 1 End If iter = iter + 1 Loop If Not binser Then binser = False End IfEnd FunctionNote: Binary search works only with sorted arrays, it just won’t work in case of unsorted arrays. Imagine the people are standing in unsorted order in the above example, in which case, you just got double burned! And the odds of finding a person (out of 50) is you gotta ask atmost 49 questions to find the person who you are looking for. You gotta be really lucky otherwise!