Javascript Mock Interview | online video interview Questions & Answers

Online Technical Mock Interview of one of my viewers. Where I ask JavaScript Fundamental Questions , algorithm questions aand give their answers

#javaScript #interview #questions

*My Udemy Courses

Follow me for technology updates

Help me translate this video.

[assamaafzal]

one basic solution to palindrome problem can be :

const isPalindrome = n => {
let final = 0;
let givenNumber = n;
while (givenNumber > 0) {
let remainder = givenNumber % 10;
givenNumber = parseInt(givenNumber / 10);
final = final * 10 + remainder;
}
return final === n;
};

[desolatedj]

Keep zoom while candidate is writing code, its difficult to see

[matthewbarnden]

Number.prototype.isPalindrome = function(number = -1, result = []) {
if(number === -1) number = this.valueOf()
const evaluateResult = result => parseInt(result.join('')) === number
if(number <= 9) {
result.push(number)
return evaluateResult(result)
}
result.push(number % 10)
this.isPalindrome(number / 10 >> 0, result)
return evaluateResult(result)
}


const number = 101;



using a couple of techniques based on these interview videos of yours i have been watching. Many thanks learning allot!

[bawaparampreet]

My approach for Palindrome:


3) function numDigits(num) {
return Math.floor(Math.log10(num));
}


function palinCheck(number) {
if(Math.floor(number / 10) === 0) {
return true;
}


var num_digits = numDigits(number);


var right = Math.floor(number / Math.pow(10, num_digits));
var left = number % 10;








if(right !== left) {
return false;
}


var striped = number / 10; // strip the right
striped = Math.floor(striped % Math.pow(10, numDigits(striped)));


return palinCheck(striped);


}

[MohdAfeef]

Please make Angular Interview Question Series thank you techsit

[saurabhchauhan]

I am backend developer. wanted to be fullstack developer. I see interview series to get idea and check my knowledge as well. This is really a good series for people like me. Thanks a lot sir. 🙃

[cvvkshcv]

Thanks for the amazing video!


const isPalindrome = (num) => {
let numTemp = num;
let result = 0;
while (numTemp > 0) {
result = (result * 10) + Math.floor(numTemp % 10);
}
return (result === num) ? 'Palindrome' : 'Not Palindrome';
}

[Ianstudent]

The correct answer to the first question is "I don't know. Let's google it and find out in 10 seconds." Like, make sure you rephrase the question back to show understanding about what is being asked, something like "I want to prevent new property definitions, but ensure the obj is still mutable" but why lose sleep over not knowing off the top of your head the exact api call.

[viharikolla]

I have tried a lengthy way:
var a=12521;
var b=c=a;
var lengthOfA=0
var s=0;
for(i=0;i<c;i++){
lengthOfA+=1;
c/=10
}
for(j=0;j<=lengthOfA;j++){
s+=a%10;
a/=10
a=Math.floor(a);
if(a!=0){
s*=10
}
}
if(b===s){
console.log("a is palindrom")
}
what do you say?

[adityashinde]

Please make data structures and algorithms video series.

[joemadhan]

When I heard the questions in this interview my thoughts automatically route me to techsith videos where he would thought all these stuffs.

[enricomariatenca]

I did it this way
(you can just copy and paste in jsfiddle or similar)


function reverseInt(n) {
let pal = "";


while (n > 1) {
n /= 10;
pal += Math.floor((n - Math.floor(n)) * 10);
}

return parseInt(pal);
}


let int = 123456;
console.log(reverseInt(int));

[nagarjunakanamarlapudi]

hi sir i loved each and every video of yours :). Can you please explain about JavaScript design patterns also

[Ath]

Prabhu Shriram bless you Hemil 🙏 keep going bro

[abhilashatiwari]

This is really helpful, please do Angular 6+ interview also

[suryaprakash]

For last one, I will find last digit by moding the given number, and store it in a variable then cut down last digit of the number by dividing with 10. Now I will generate a reverse number(initial value is zero) by multiplying reverse number with 10 and add the extracted last digit to it. I will repeat this till number is greater than zero after cutting down the last digit.. finally I will get reversed number. I will compare with actual number..this is my approch.. pls suggest is this correct approch or not..

[mayankchandravanshi]

--- Here is my take ---

function isPalindrome(number) {
// number -> 121
const stack = [];
const originalNumber = number;
while (number > 0) {
stack.push(number % 10);
number = Math.floor(number / 10);
}
const reversedNum = Number(stack.join(""));
return reversedNum === originalNumber;
}

[wongwanchap]

How about floating number palindrome, e.g. 1234.4321, which I don't think it will work in base 10, it only work for base 2 or base 2^n.

[MuhammadIshmaell]

Here's my solution to the palindrome question.

Number.prototype.isPalindrome = function() {
let arr = [], quotient = 0, remainder = 0, init = this;
while(init > 0) {
quotient = Math.floor(init/10);
remainder = init%10;
arr.push(remainder);
init = quotient;
}
return this === Number(arr.join(""));
}


const num = 12521;
// returns true

[hovhannessardaryan]

Please, make a video about Polymorphism in JavaScrpt.