Arrays - Data Structures and Algorithms in Javascript | Frontend DSA Interview Questions
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- เผยแพร่เมื่อ 21 ก.ค. 2024
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In this video, we will dive into Array methods in JavaScript and explore their various use cases. We will also discuss commonly DSA interview questions related to Arrays and how to solve them using JavaScript.
This is the most Entertaining and Knowledge Packed Data Structures and Algorithms in Javascript Interview Course on TH-cam !
#javascriptinterviewquestions #datastructures #datastructuresandalgorithms
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-------------------------------------------------------------------------
00:00:00 Intro
00:00:54 Creating an Array
00:01:34 Initializing an Array
00:02:24 What all can array store?
00:03:24 Array length Property
00:03:54 Add / Remove items in Array
00:03:58 Push and Pop method
00:05:32 shift and unshift method
00:06:29 Looping the Arrays
00:06:49 "for" loop
00:07:37 "while" loop
00:08:09 Inbuilt Traversal Methods
00:08:25 map() in Arrays
00:10:14 filter() in Arrays
00:10:59 reduce() in Arrays
00:12:55 some() in Arrays
00:13:25 every() in Arrays
00:14:01 find() in Arrays
00:14:42 Spread and Rest Operators
00:17:51 More Array Methods
00:17:56 concat() in Arrays
00:19:19 slice() in Arrays
00:21:13 splice() in Arrays
00:22:47 fill() in Arrays
00:23:58 findIndex() in Arrays
00:25:09 flat() in Arrays
00:26:19 Technical Glitch
00:26:48 reverse() in Arrays
00:27:18 sort() in Arrays
00:29:44 DSA Interview Questions on Array in Javascript
00:29:50 Ques 1 - Second Largest Number
00:30:28 Solution #1 - Brute Force Approach
00:34:46 Time Complexity of this Algorithm
00:36:33 Solution #2 - Optimised Approach
00:39:30 Explanation of Solution #2
00:41:29 Time Complexity of this Algorithm
00:42:26 Space Complexity of this Algorithm
00:42:49 Ques 2 - Rotate Array by K
00:43:51 Solution #1 - Using Inbuilt JS Functions
00:48:40 Time Complexity of this Algorithm
00:49:24 Solution #2 - Without Inbuilt Functions
00:56:26 Time Complexity of this Algorithm
00:57:29 Space Complexity of this Algorithm
00:58:16 Ques 3 - Remove Duplicates from Sorted Array
00:59:20 Solution #1 - With Inbuilt JS Methods
01:03:08 Time Complexity of this Algorithm
01:03:32 Space Complexity of this Algorithm
01:03:41 Solution #2 - Two Pointer Approach
01:07:28 Explanation of above algorithm
01:10:22 Time / Space Complexity of this Algorithm
-------------------------------------------------------------------------
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If this video gets good response, I will make more DSA videos, so do share it with others 🔥
please please please dont quit this course....please keep uploading
thanks bro . we really needed quality dsa guidance in Javascript
Wow,ab to video aur acchi hoti ja rahi hai day by day
Excellent session!
Looking forward to the whole series.
Also,
#FreeAlgoAgarwal 🚩
Very knowledgable content , almost all my approaches are similar to yours !! ❤😎
Bro, thanks a lot for making this video. I've been waiting to watch a tutorial that covers most of the functions with arrays.
Thanks buddy appreciate it.. Keep doing the good work!!! ❤✨
eagerly waiting for next videos
Hats off to you man ❤ , the way you're explaining is so simple to understand any methods in this array video
Thanks a ton
Great explained thank you
Thumbnail is very good 🔥 and also the concepts are well explained in organised manner. Problems on those concepts make more understanding.
🙏❤️
Khatrnak bhaii.. All array clear in one video my frnd also thx me for your video🎉❤
can anyone explain if secondLargest ([2, 3, 44, 5, 44, 3, 7, 8, 11]) in the arr the optimised code will give 44 as second largest
During the fifth iteration, the value of arr[i] is 44, which is equal to the current value of largest. Therefore, the else if condition is true, and the value of secLar is updated to the current value of largest, which is 44. The value of largest remains 44.
During the sixth iteration, the value of arr[i] is 3, which is smaller than largest. The else if condition is also false, so nothing happens.
During the seventh iteration, the value of arr[i] is 7, which is smaller than largest. The else if condition is also false, so nothing happens.
During the eighth iteration, the value of arr[i] is 8, which is smaller than largest. The else if condition is also false, so nothing happens.
In the ninth iteration, the largest value is 44 (which was updated during the eighth iteration), and the current value of arr[i] is 11, which is smaller than both the current largest and second-largest values. Therefore, there are no updates made to the values of largest or secLar during this iteration.
✨Nice Explaintion of the and logic its very helpful plz do such videos thank you fpr knowledge✨
Most welcome 😊
Great content, would be awesome if you could upload more and continue with DSA's (going into string manipulation and potentially some common toy problems in JS!) 🙏👑
#FreeAlgoAgarwal
can anyone explain if secondLargest ([2, 3, 44, 5, 44, 3, 7, 8, 11]) in the arr the optimised code will give 44 as second largest
During the fifth iteration, the value of arr[i] is 44, which is equal to the current value of largest. Therefore, the else if condition is true, and the value of secLar is updated to the current value of largest, which is 44. The value of largest remains 44.
During the sixth iteration, the value of arr[i] is 3, which is smaller than largest. The else if condition is also false, so nothing happens.
During the seventh iteration, the value of arr[i] is 7, which is smaller than largest. The else if condition is also false, so nothing happens.
During the eighth iteration, the value of arr[i] is 8, which is smaller than largest. The else if condition is also false, so nothing happens.
In the ninth iteration, the largest value is 44 (which was updated during the eighth iteration), and the current value of arr[i] is 11, which is smaller than both the current largest and second-largest values. Therefore, there are no updates made to the values of largest or secLar during this iteration.
Thank you so much brother your all videos are very needful...!!!
Most welcome!
More same like this video please.,.. 🔥🔥🔥🔥🔥🔥
best video..please continue this great work
Thanks ❤️🙏
I have a MERN app with Frontend and Backend, I am having trouble with Axios when I deployed the app to Amazon Web Service (EC2). Have you encountered this with Axios?
Brother please upload further, your videos are very helpful and we'll explained.
can anyone explain if secondLargest ([2, 3, 44, 5, 44, 3, 7, 8, 11]) in the arr the optimised code will give 44 as second largest
During the fifth iteration, the value of arr[i] is 44, which is equal to the current value of largest. Therefore, the else if condition is true, and the value of secLar is updated to the current value of largest, which is 44. The value of largest remains 44.
During the sixth iteration, the value of arr[i] is 3, which is smaller than largest. The else if condition is also false, so nothing happens.
During the seventh iteration, the value of arr[i] is 7, which is smaller than largest. The else if condition is also false, so nothing happens.
During the eighth iteration, the value of arr[i] is 8, which is smaller than largest. The else if condition is also false, so nothing happens.
In the ninth iteration, the largest value is 44 (which was updated during the eighth iteration), and the current value of arr[i] is 11, which is smaller than both the current largest and second-largest values. Therefore, there are no updates made to the values of largest or secLar during this iteration.
In your last question 's first method how array.splice() can have constant time complexity as splice() itself is of linear time complexity ?
Please please make more such videos.....🙏🙏🙏🙏
Moving forward from that joke was a better decision than anything else
// Ques 2 - Rotate Array by K
// Given an integer array nums, rotate the array to the right by k steps,
// where k is non - negative.
// Input: nums = [1,2,3,4,5,6,71, k = 3
// Input: nums = (-1, -100,3,99], k = 2Output: [5,6,7,1,2,3,4]Output: [3,99,-1, -100]
solution :
function rotateIT(nums, a) {
for (let i = 0; i < a; i++) {
nums.unshift(nums.pop());
}
return nums;
}
let num1 = [1, 2, 3, 4, 5, 6, 7]
let num2 = [-1, -22, 33, 4 - 2]
console.log(rotateIT(num1, 3));
Bro , last vedio which is Remove duplicate form array ,last program output for length giving crt answer...if u print sorted array it is giving a wrong output
Why are we giving size-k,size in splice ? 49:47 we can give size-k,k right?
Can we have problem solving based on dsa
I have purchased DSA JS e-Book, really helpful...direct links to leetcode....thank you for preparing eBook.....
Awesome, thank you!
Best video❤
Thankyou very much ❣
ok bro but please upload the full course we are waiting
Plzz Bhaiya add more videos and String question video 🙏🙏
So indepth and crystal clear explanation....🎉
You can check my complete course for more - roadsidecoder.com/course-details
@@RoadsideCoder purchased complete course
@@RoadsideCoder really helpfull interview guide to crack interviews
Array in JS is different from core Data structures Array, in js array is a special object.
Would like to see algo Agarwal conduct one session -
In your 2'nd problem your approach is not working if the k is >= 2*array.length please check if it's okay or not
Can you please also add multidimensional array compare, iterate
Bro please do this DSA series for string and object as well plz bro 😀
Bring Strong questions as well
because it popped its back?
Please make strings problems they really help in interview as well as general coding
Yes, I will!
bro if possible ,please increase your upload frequency, :)
Hi Sir I am building your Chat application but I have seen that people are having issues with deployment. Can you create a video on just the deployment part or suggest me an alternative
can anyone explain if secondLargest ([2, 3, 44, 5, 44, 3, 7, 8, 11]) in the arr the optimised code will give 44 as second largest
During the fifth iteration, the value of arr[i] is 44, which is equal to the current value of largest. Therefore, the else if condition is true, and the value of secLar is updated to the current value of largest, which is 44. The value of largest remains 44.
During the sixth iteration, the value of arr[i] is 3, which is smaller than largest. The else if condition is also false, so nothing happens.
During the seventh iteration, the value of arr[i] is 7, which is smaller than largest. The else if condition is also false, so nothing happens.
During the eighth iteration, the value of arr[i] is 8, which is smaller than largest. The else if condition is also false, so nothing happens.
In the ninth iteration, the largest value is 44 (which was updated during the eighth iteration), and the current value of arr[i] is 11, which is smaller than both the current largest and second-largest values. Therefore, there are no updates made to the values of largest or secLar during this iteration.
Bro plz do this DSA series for objects and strings plzz😢
Can you make more of this kind of videos? Array , object exercises
already have, you can check on my channel
Bhaiya, please late dal rhe ho chalega but finish kar dena bhot courses pade hai Jo kabhi complete he nhi hue 🫠hats off to work you did till now🤩
I promise to bring you all the most high quality DSA with JS content on youtube. Plus this video will be 1 hour+ so even if its late, I will cover most things in one video, so u all get maximum value!
Bhai please course ko complete karna
Waiting
Hey Piyush, can you please make a video on how we can send file to that chat application like image, pdf and zip file and the message seen this will add more value to that application.
can anyone explain if secondLargest ([2, 3, 44, 5, 44, 3, 7, 8, 11]) in the arr the optimised code will give 44 as second largest
During the fifth iteration, the value of arr[i] is 44, which is equal to the current value of largest. Therefore, the else if condition is true, and the value of secLar is updated to the current value of largest, which is 44. The value of largest remains 44.
During the sixth iteration, the value of arr[i] is 3, which is smaller than largest. The else if condition is also false, so nothing happens.
During the seventh iteration, the value of arr[i] is 7, which is smaller than largest. The else if condition is also false, so nothing happens.
During the eighth iteration, the value of arr[i] is 8, which is smaller than largest. The else if condition is also false, so nothing happens.
In the ninth iteration, the largest value is 44 (which was updated during the eighth iteration), and the current value of arr[i] is 11, which is smaller than both the current largest and second-largest values. Therefore, there are no updates made to the values of largest or secLar during this iteration.
bro next video on strings
is JS will not work for Faang Companies as small community and and not so much resources available or i continue JS with DS.
For FAANG companies, comtinue JS with DSA
@@RoadsideCoder thanks keep posting video on DSA more and more 😄
brother can we use this instead of that whole big code
function check (arr){
return arr.sort((a,b)=> b-a)[1]
}
console.log(check([1,2,3,1]))
not allowed in interviews
@@RoadsideCoder but if i apply in a company where they want javascript and they are paying 25 k a month then it’s applicable?
Piyush brother, please make videos on linked list also
can anyone explain if secondLargest ([2, 3, 44, 5, 44, 3, 7, 8, 11]) in the arr the optimised code will give 44 as second largest
During the fifth iteration, the value of arr[i] is 44, which is equal to the current value of largest. Therefore, the else if condition is true, and the value of secLar is updated to the current value of largest, which is 44. The value of largest remains 44.
During the sixth iteration, the value of arr[i] is 3, which is smaller than largest. The else if condition is also false, so nothing happens.
During the seventh iteration, the value of arr[i] is 7, which is smaller than largest. The else if condition is also false, so nothing happens.
During the eighth iteration, the value of arr[i] is 8, which is smaller than largest. The else if condition is also false, so nothing happens.
In the ninth iteration, the largest value is 44 (which was updated during the eighth iteration), and the current value of arr[i] is 11, which is smaller than both the current largest and second-largest values. Therefore, there are no updates made to the values of largest or secLar during this iteration.
Hi brother I'm going to start your Chat application playlist. The Application is just God level. But if I download the versions which you have installed and start developing will everything work fine. Tell me if I'm good to go
Yes it will work fine!
@@RoadsideCoder OK word!! Starting it today itself
46:22 how the length of array become 8 ,as you mentioned in video arr = [1,2,3,4,5,6,7] . it is 7 ryt ?
Also the index count should start from zero instead of one
@@krishnat9767the it should be 6 right?
Length of an array is the no. of elements and indexes start from 0. so length should be 7 and no of indexes should be 6 - [0,1,2,3,4,5,6]
Yehi toh...Galat bataya usne solution 1 and optimised raw wala bhi...Yaar isko zyada knowledge nahi he...
Good day greetings
🙏
❣
can anyone explain if secondLargest ([2, 3, 44, 5, 44, 3, 7, 8, 11]) in the arr the optimised code will give 44 as second largest
During the fifth iteration, the value of arr[i] is 44, which is equal to the current value of largest. Therefore, the else if condition is true, and the value of secLar is updated to the current value of largest, which is 44. The value of largest remains 44.
During the sixth iteration, the value of arr[i] is 3, which is smaller than largest. The else if condition is also false, so nothing happens.
During the seventh iteration, the value of arr[i] is 7, which is smaller than largest. The else if condition is also false, so nothing happens.
During the eighth iteration, the value of arr[i] is 8, which is smaller than largest. The else if condition is also false, so nothing happens.
In the ninth iteration, the largest value is 44 (which was updated during the eighth iteration), and the current value of arr[i] is 11, which is smaller than both the current largest and second-largest values. Therefore, there are no updates made to the values of largest or secLar during this iteration.
you have wrongly understood the problem in the if else(arr[i] !== firstlargest ) which means in the third index the 44 is overwritten by the immediate value of an array and those the immediate value we will compare with the arr[i] thats the reason we provided else if(arr[i] !==firstlargest ) i think you forget to enter "!" operator in the edge case statement
Next video on objects please 🥺🙏
Already made - th-cam.com/video/XnFIX3c7xoI/w-d-xo.html
@@RoadsideCoder oh great.. Is it covering all the topics? And thank you so much for all these videos
Many ytubers started DSA playlists but no one complete all topics
Because they got subscribers so they will start collaboration vedio😢
Don't worry, I will complete it fully!
29:00
Bro love you ❤❤.
//Ques 1 - Second Largest Number
// Given an array Arr of size N, print second largest
// distinct element from an array.
// Input: [12, 35, 1, 10, 34, 1]
// Input: [10, 5, 10] ->>›››Output: 34 Output: 5
let aa = [12, 35, 11, 10, 34, 1]
let q = [10, 5, 10]
let UNiq = q.filter((value, index, self) => self.indexOf(value) === index)
let bb = aa.sort((a, b) => a - b)
let c = bb[bb.length - 2]
let DD = UNiq.sort((a, b) => a - b)
let d = DD[DD.length - 2]
console.log(c)
console.log(d)
//with function
let aa = [12, 35, 11, 10, 34, 35, 1]
let q = [10, 5, 10]
function sortIng(arr) {
let uniquearr = arr.filter((value, index, self) => self.indexOf(value) === index)
uniquearr.sort((a, b) => a - b);
if (uniquearr.length < 2) {
return null;
}
return uniquearr[uniquearr.length - 2];
}
let bbb = sortIng(aa);
let bbb2 = sortIng(q);
console.log(bbb);
console.log(bbb2);
// More efficiennt solution for second max.
let arr=[10,0,90];
function secondMax(arr){
let max= arr[0];
let secondMax = -Infinity;
for(let i=1; i max){
secondMax = max;
max = arr[i];
}else if(arr[i] >secondMax && arr[i] < max) {
secondMax = arr[i];
}
}
return secondMax;
}
console.log(secondMax(arr));
45:34
👇More optimised code 👇
function secondLar(arr) {
let max= Number.NEGATIVE_INFINITY;
let secondMax=Number.NEGATIVE_INFINITY;
for(let i=0; i max){
secondMax=max;
max= arr[i];
}
}
return secondMax;
}
No need of else if condition
This isn't correct. What if you had array of the following: [4,7,6,5,2 1]. The second largest would be 6, but this code would return 4. That's why the second if statement is needed. In his first if statement however, he should've set secondLargest before updating the largest though because doing it the way he did it will cause both variables to store the dame value.
@@raiga98 yes, that's for an edge case.
ya javascript got tired of doing Push'up'
un-optimized solutions are the easiest to understand
can anyone explain if secondLargest ([2, 3, 44, 5, 44, 3, 7, 8, 11]) in the arr the optimised code will give 44 as second largest
During the fifth iteration, the value of arr[i] is 44, which is equal to the current value of largest. Therefore, the else if condition is true, and the value of secLar is updated to the current value of largest, which is 44. The value of largest remains 44.
During the sixth iteration, the value of arr[i] is 3, which is smaller than largest. The else if condition is also false, so nothing happens.
During the seventh iteration, the value of arr[i] is 7, which is smaller than largest. The else if condition is also false, so nothing happens.
During the eighth iteration, the value of arr[i] is 8, which is smaller than largest. The else if condition is also false, so nothing happens.
In the ninth iteration, the largest value is 44 (which was updated during the eighth iteration), and the current value of arr[i] is 11, which is smaller than both the current largest and second-largest values. Therefore, there are no updates made to the values of largest or secLar during this iteration.
Lame joke but great video.
Hi guys
Heyy!
This playlist is not for beginners
Why you gyus just make a video without board and pen 🖊️ with I think I need to creat a video beacuse of awo kuch bhi bna do Bina ache se samghye
Ok from next video I will use it!
function Rotation( arrayRot, k){
return arrayRot= [...arrayRot.slice(arrayRot.length-k,arrayRot.length),...arrayRot.slice(0,arrayRot.length-k)]
}
Bhai please course ko complete karna