🔴 Get my Complete Frontend Interview Prep course - roadsidecoder.com/course-details ➡ Download my DSA with JS E-Book - topmate.io/roadsidecoder/491565 ✅ Follow me on Instagram to be updated about next videos - instagram.com/roadsidecoder/ If this video gets good response, I will make more DSA videos, so do share it with others 🔥
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.
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.
31:06 second largest element of the array [ 12,36,1,10,34,1,35] -> make two variable last second last , compare the last if it is greater then compare it with second last if greater smaller update ... for me
// 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));
My solution was similar to yours: function rotateArray(nums,k) { for(i = 0; i < k;i++) { nums.unshift(nums.pop()) } return nums } console.log(rotateArray([1,2,3,4,5,6,7], 3))
How did you calculate the time complexity of splice method to be O(1) at 1:03:15? In my opinion it should be O(n) since it has to modify the whole array.
let arr=[2,4,5,2,6,7,9,2,1] const duplicate=()=>{ let uni= Array.from(new Set(arr)) console.log(uni); return uni.length; } console.log(duplicate(arr)); sir we can use this method also ? for remove duplicate 1:00:00
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.
// 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));
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
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.
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!
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 , 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
//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)
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
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.
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.
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?
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.
🔴 Get my Complete Frontend Interview Prep course - roadsidecoder.com/course-details
➡ Download my DSA with JS E-Book - topmate.io/roadsidecoder/491565
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what is the time complexities of these functions 🫨🫨 🫨🫨 🫨🫨 🫨🫨
please please please dont quit this course....please keep uploading
Excellent session!
Looking forward to the whole series.
Also,
#FreeAlgoAgarwal 🚩
The best mentor of Frontend development....💞💓💗💕❤💝
❣️❣️
Wow,ab to video aur acchi hoti ja rahi hai day by day
size > k || size < k should also be there in case, size < k 49:50
yes it should be if (k>size) not (size>k)
I have purchased DSA JS e-Book, really helpful...direct links to leetcode....thank you for preparing eBook.....
Awesome, thank you!
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.
My favourite mentor Roadside coder, i always suggest his videos to everybody
❤️
Hats off to you man ❤ , the way you're explaining is so simple to understand any methods in this array video
Thanks a ton
thanks bro . we really needed quality dsa guidance in Javascript
best video..please continue this great work
Thanks ❤️🙏
Thumbnail is very good 🔥 and also the concepts are well explained in organised manner. Problems on those concepts make more understanding.
🙏❤️
Bro, thanks a lot for making this video. I've been waiting to watch a tutorial that covers most of the functions with arrays.
Thank you so much brother your all videos are very needful...!!!
Most welcome!
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
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.
✨Nice Explaintion of the and logic its very helpful plz do such videos thank you fpr knowledge✨
Most welcome 😊
eagerly waiting for next videos
31:06 second largest element of the array [ 12,36,1,10,34,1,35] -> make two variable last second last , compare the last if it is greater then compare it with second last if greater smaller update ...
for me
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));
My solution was similar to yours:
function rotateArray(nums,k) {
for(i = 0; i < k;i++) {
nums.unshift(nums.pop())
}
return nums
}
console.log(rotateArray([1,2,3,4,5,6,7], 3))
Very knowledgable content , almost all my approaches are similar to yours !! ❤😎
How did you calculate the time complexity of splice method to be O(1) at 1:03:15? In my opinion it should be O(n) since it has to modify the whole array.
1:05:44
why we are not using sets here for unique
if we remove duplicate value from array
same question
Sets would create a new way to store them , but modifying inside itself will make the code more optimized
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...
editing is really entertaining😂😂😂😂😂
Why are we giving size-k,size in splice ? 49:47 we can give size-k,k right?
let arr=[2,4,5,2,6,7,9,2,1]
const duplicate=()=>{
let uni= Array.from(new Set(arr))
console.log(uni);
return uni.length;
}
console.log(duplicate(arr));
sir we can use this method also ? for remove duplicate 1:00:00
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.
Thanks buddy appreciate it.. Keep doing the good work!!! ❤✨
// 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));
More same like this video please.,.. 🔥🔥🔥🔥🔥🔥
you should have covered the unsorted array situation in this video.but still i liked the theory and demo which you taught us thank you.
Please make strings problems they really help in interview as well as general coding
Yes, I will!
Please please make more such videos.....🙏🙏🙏🙏
Array in JS is different from core Data structures Array, in js array is a special object.
love your videos thanks dude
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
In your last question 's first method how array.splice() can have constant time complexity as splice() itself is of linear time complexity ?
Can you make more of this kind of videos? Array , object exercises
already have, you can check on my channel
Best video❤
👇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.
Thankyou very much ❣
function Rotation( arrayRot, k){
return arrayRot= [...arrayRot.slice(arrayRot.length-k,arrayRot.length),...arrayRot.slice(0,arrayRot.length-k)]
}
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 😄
Plzz Bhaiya add more videos and String question video 🙏🙏
Great explained thank you
Can we have problem solving based on dsa
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!
Can you please also add multidimensional array compare, iterate
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.
ok bro but please upload the full course we are waiting
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
Concat va spread ??
Whats the difference?
i think both work similarly, so the choice is often a matter of preference, readability, or specific use cases
//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);
not getting how the length of nums array in rotate array by k problem is 8, i mean arr=[1,2,3,4,5,6,7], here the length of array is 7 right ?
not 8
Bring Strong questions as well
Next video on objects please 🥺🙏
Already made - ruclips.net/video/XnFIX3c7xoI/видео.html
@@RoadsideCoder oh great.. Is it covering all the topics? And thank you so much for all these videos
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
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 please do this DSA series for string and object as well plz bro 😀
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!
Bro love you ❤❤.
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.
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
Bro plz do this DSA series for objects and strings plzz😢
bro if possible ,please increase your upload frequency, :)
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?
because it popped its back?
29:00
Waiting
Bhai please course ko complete karna
bro next video on strings
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.
❣
45:34
🙏
Good day greetings
ya javascript got tired of doing Push'up'
This playlist is not for beginners
Bro I have an issue how u remember all these things☠️☠️
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!
Hi guys
Heyy!
Lame joke but great video.
Bhai please course ko complete karna