JavaScript LIVE Coding Interview Round (Mock)

[@varandeepsahota1314]

Binary search is POSSIBLE here. We can determine which element should be present at ith index using simpe A.P formula and if correct element present, move right else move left side. Thanks for this content.

[@Its_Abhi_Thakur]

let arr = [5, 7, 9, 11, 15, 17]; let missing = 0; for (let i = 0; i < arr.length; i++) { let el = arr[i]; let next_el = arr[i + 1]; if (next_el - el !== 2) { missing = el + 2; break; } } console.log(missing);

[@coderdost]

good and simple

[@Its_Abhi_Thakur]

@@coderdost thank you 😇

[@ahmadraiyan165]

let arr=[5,7,9,11,15,17] let arr2=[] for (let i = 0; i < arr.length; i++) { if ((arr[i]-arr[i-1])!==2&&(arr[i]-arr[i-1])==4) { arr2.push(arr[i-1]+2) } } console.log(arr2);

[@karthikhs2530]

const input= [5,7,9,11,15,17] input.filter((e,i)=>{return(input[i+1]-input[i])>2}) .map((e)=>e+2) .filter((e,i)=>i

[@thesujalogy]

my code works with any case: const input = [1,2,3,4,5,6,7,9]; function checkMate(arr) { let diff = arr[1] - arr[0]; let i = 0; let j = arr.length-1; let ans; while (i < j) { if(arr[i] + diff === arr[i+1]) { i++; } else { ans = arr[i] + diff; console.log(ans); break; } } } checkMate(input);

[@Rohit-zs7kn]

const arr = [5, 7, 9, 11, 13, 15, 17, 21]; let index = 0; for (let oddNum = arr[0]; oddNum

[@MehulJadavIndia]

const input = [5, 7, 9, 11, 13, 15, 17, 21]; for (i = 0; i

[@yesuraju-tf1yp]

function missOdd(arr){ let first=0, last=arr.length-1, mid = 0 ,first_element = arr[0]; if(first_element+2*last == arr[last] ) return "no element is missed"; while(first

[@itechinnovations2200]

Map all the values inside array then go for a loop that iterates within the req range then for each odd no check it is present in map or not if not then print and break

[@vin2368]

const Arr3 = [6,9,12,18,24,30] const n = 3 function getMissingOddNumbers(arr){ let output=[]; for(let i = 0;i

[@AashMusic1]

const lst = [5, 7, 9, 11, 15, 17]; let res = [] for (let i =0 ; i < lst.length; i++){ let current = lst[i] let nextOdd = current + 2 if(lst[i+1]===nextOdd){ continue; }else{ res.push(nextOdd); break; } } console.log(res)

[@werecoders]

Very simple for (let index = 0; index < num.length; index++) { let current = num[index]; let next = current + 2; if (num.includes(next)) { continue; } else { console.log(next); break; } }

[@abnow1998]

It is in AP(Arithmetic Progression) Just check Common Difference. Formula: a+(n-1)*d a is first number, d is common diff

[@madhursangeet29]

const input = [1,5,7,9,11,15,17,21,25]; let inputFirst = input[0]; let inputEnd = input.at(-1); let missDigit = []; for(let i = inputFirst; i

[@GurpreetSingh-gogsy]

const input = [5,7,9,11,15,19]; for(let i=1;i

[@ApurvaKashyap-kj6qz]

Because binary search is applicable where we r given a target number to search in an array .. which is not in this case .

[@PunkSage]

Shortest I could think of: const i = [5,7,9,11,15,17]; const find = (a)=> { const f = (a[0] + a[a.length-1])/2*(a.length+1); const s = a.reduce((t, c)=> t+c, 0); return f-s; } console.log(find(i));

[@AkashSingh-xf6bd]

let input =[5,7,11,13,15,17]; let out = 13 function s(input){ for(let i= 0; i

[@khanapeena2191]

in your code for loop should iterate till array.length-1

[@tonmaysardar3331]

const input=[1,3,5,7,11,13,15,17] let b; for (let i=1;i

[@syedwaseem3793]

function findMissingOddNumber(numbers) { // Loop through the array of numbers for (let i = 0; i < numbers.length - 1; i++) { // Check if the next number is not exactly 2 more than the current number if (numbers[i + 1] - numbers[i] !== 2) { // If not, the missing number is between these two numbers return numbers[i] + 2; } } // Return null if no number is missing return null; } const input = [5, 7, 9, 11, 15, 17]; const missingNumber = findMissingOddNumber(input); console.log(missingNumber); // Output: 13

[@happylaughy2777]

const arr= [3,5,7,9,13] let odd=arr[0]; for(let i=0;i

[@bathininaveen5198]

function findMissingOdd(arr){ for(let i =0;i 2) return arr[i] + 2; } return "No element is Missing!" }

[@WalkingtotheTruth]

Binary Search is possible here is the answer: function findMissingOddNumber (arr) { let left = 0 let right = arr.length - 1 let startingNumber = arr[0] let ans = -1 while (left

[@Rubyd777]

I guess using some mathematics can be helpful. But i would use it only if array was large and we need to solve it in minimum time complexity else i would use sorting approach 1) Find max and min at once from array 2)find sum of odd numbers from 1 to nth term(let x) 3)find total sum of elemnts given in array (let y) 4)find total sum of odd number from min to max that we found above(let z) 5)lastly: (x-y)-(x-z) This can be simplified but you compolsary need x to find y and z Please correct if i am wrong. I just ried my approach

[@PunkSage]

There is even a much simpler approach. Find my comment above.

[@enthusiatic_coder6801]

const input = [5, 7, 9, 11, 13, 17, 19, 21]; const compare = (a, b) => { return a - b; }; for (let i = 0; i < input.length; i++) { // console.log(input[i]); if (input[i] - input[i + 1] !== -2) { console.log('The missing number is', input[i] + 2); input.push(input[i] + 2); input.sort(compare); break; } } console.log(input);

[@yournanban3812]

const input = [5,7,9,11, 15,17]; let missingNum = ""; for(let i= 0 ; i < input.length; i++){ if (input[i] + 2 != input[i+1]){ missingNum = input[i]+2; break; } } console.log(missingNum) //output missing odd number is 13

[@akkburf5450]

let arr = [5,7,9,11,15,17] function missNumber(arr){ for(let i =0 ; i < arr.length ; i++ ){ if(!(arr[i] + 2 === arr[i + 1])) return arr[i] + 2 } } console.log(missNumber(arr))

[@prathamlashkari4230]

let input = [5, 7, 9, 11,15, 17]; for (let i = 0; i < input.length; i++) { let diff = input[i + 1]-input[i]; if (diff !== 2){ console.log(input[i]+2); break; } }

[@niju7489]

//AP addition formula s= n/2(frst +last) let len = input.length ; return (input[0] +input[len-1])(len+1)/2 - input.reduce((a,c)=>a+c);

[@mohammedrayyan3029]

const input7 = [5, 7, 9, 11, 15, 17, 21, 25]; const output7 = 13; function missingNumber(input7) { let index; for (index = 0; index < input7.length; index++) { if (input7[index] > output7) { break; } } input7.splice(index, 0, output7); return input7; } console.log(missingNumber(input7), "missing");

[@abdulbasith650]

let arr = [1,3,5,7,9,13] function misodd(arr){ let misnum; for(let i=0;i

[@addep-kardy]

This does work fine! const inp = [5,7,9,11,15,17]; const outp = []; for(let i=0; i < inp.length; i++){ if(inp[i]+2 != inp[i+1]){ outp.push(inp[i]+2); } } console.log(outp);

[@chiraglegend2670]

function nextOdd(input){ for(let i=0 ; i < input.length ; i++){ let val = input[i]+2 if(val != input[i+1]){ return val } } }

[@nitishgupta8393]

Q1: const missingOdd =(arr)=> { for(let i=0; i

[@shazaibahmad7807]

Bro simple is that one if condition if(arr[I+1]-arr[I]===4) { return are[I+1]+2; } else{ return; }

[@suryateja9881]

My solution for Sum 1: let input = [21,23,25,27,31,33,35,37,39,41,45,47,49,53]; for (let i=input[0];i

[@YouTubeflixyt]

Great quality content keep it up❤️

[@nevilmistry740]

const arr = [3,5,7,9,11,13,17,19] //op -> 13 const arrLength = arr.length; let checker = 3 console.log(check(arr,arrLength,checker)) function check(arr,arrLength,checker){ for(i=0;i

[@tejasshinde2011]

let temp =input[0]; for (let i = 0; i < input.length && input.includes(temp + 2); i++, temp += 2); console.log(temp + 2);

[@maazhasan-mi8le]

input.forEach((element,index)=>{ if (element + 2 !== input[index + 1] && input.length > index + 1) { console.log('missing number is ', element + 2) } } )

[@akash_gupta_2090]

Missing odd numbers function missingFirstOddNumbers(arr) { const len = arr.length; let missingNumber = null; for (let i = 0; i < len; i++) { let _curr = arr[i]; let next = arr[i + 1]; if (next - _curr != 2) { missingNumber = _curr + 2; break; } } return missingNumber; }

[@Doglapan64]

for (let i=0; i

[@ankursharma1822]

let arr = [5,7,9,11,15,17]; for(value of arr){ let a = value +2; if(!arr.includes(a)){ console.log(a); break } }

[@MilindP]

const input = [5,7,9,11,15,17] let count = input[0]; input.map(e=>{e===count?count +=2:count}) console.log(count)

[@aparnas1613]

Best simple code

[@MilindP]

@@aparnas1613 🙂

[@coderdost]

Good logic. only thing use forEach in place of map. Map should only be used when array output is expected. Since map builds a new array, calling it without using the returned array is an anti-pattern. Reference : [developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map]

[@rajatgour9686]

ok so let assume given arry is sorted const input = [5,7,9,11,15,17] let output = null for(let start = input[0]+2;start

[@nevilmistry740]

2 nd answer const arr = [5,7,9,11,13,17,19] const arrLength = arr.length; let checker = 3 console.log(check(arr,arrLength,checker)) function check(arr,arrLength,checker){ for(i=1;i

[@nevilmistry740]

just remove console😅😅😅

[@40fps143]

for last question i thought of this -- console.log(input.map((el) => el *n))

[@FreeTrial0315]

const arr =[5,7,9,11,15,17,19,23]; const data = new Set(arr) for(i=5; i

[@ankitarya7492]

let arr = [1, 5, 7, 9, 11,13]; let j = arr[0]; for (let i = 0; i < arr.length; i++) { if (arr[i] !== j) { console.log(j) break } j+= 2 } KeepItSimple

[@kiranm5419]

const input = [3,5,7,9,11,15,17]; for(let i=0;i

[@unboxtheboxed-ox3cp]

function oddcalc(arr){ for(let i=0;i2){ missing = (arr[i]+arr[i+1])/2; return missing; } } } let val =oddcalc([5, 7, 9, 11, 13, 15, 19]); console.log(val)

[@amazondeals7284]

Solution 1: const input = [5,7,9,11,15,17] const findMissing =(arr)=>{ let temp = arr[0] for(let i= 0; i< arr.length; i++){ if(temp !== arr[i]){ return temp; } temp += 2 } } console.log(findMissing(input))

[@amazondeals7284]

To change my algorithm to multiple of three, I have to increase the temp to 3.

[@mohaimin95]

function missingOdd(arr) { for(let i = 1; i < arr.length; i++) { const last = arr[i - 1]; const current = arr[i]; if(current - last > 2) { return last + 2; } } }

[@rahulkr7349]

the question was simple. not that hard but the interview makes your heartbeat sound like a banging drum.

[@coderdost]

Yeah its the pressure which acts..

[@shivagyawali2590]

const input = [5, 7, 9, 11, 15, 17]; const missingOdd = input.reduce((prev, curr, index) => { if (index === 0) return curr; const nextVal = prev + 2; if (nextVal !== curr) return nextVal; return prev; }); console.log(missingOdd);

[@coderdost]

try with this : [5, 7, 11, 13, 15, 17];

[@adnaan28]

const data = [5, 7, 11, 15, 17] for(let i =data[0]; i

[@Snagesh3770]

const input = [5, 7, 9, 11, 15, 17]; function missingOddNumbers(arr) { let min = Math.min(...arr); let max = Math.max(...arr); let oddNumbers = []; for (let i = min; i !arr.includes(x)); return res } console.log(missingOddNumbers(input));

[@NavneetTaneja-d1y]

easy first sort then extract min and max after than loop range from min to max than extrat odd number list that number not in arr that is missing after first missing get break the array

[@FarhanKhan-ox4iy]

I see in the comments no one has used two pointers for this. With that we can achieve linear time complexity. Use start and end. Run while loop until end >= start. start++ and end--

[@PunkSage]

No reason to use it here. All you need to do is to calculate the sum of numbers.

[@darklord9500]

function findMissingNumber(arr) { let missingNumber = null; for (let i = 0; i < arr.length; i++) { if (arr[i] !== arr[0] + i * 2) { missingNumber = arr[0] + i * 2; break; } } return missingNumber; }

[@teentexh]

let array=[5,7,9,11,15,17]; let value=array[0]; for(let a of array){ if(a===value){ value+=2 ; }else{ console.log(value); break; } }

[@stefumies]

const missingNumber = input.reduce((ac,c,i,a) => i == 0 ? a : ((c - a[i-1])> 2) ? a[i-1] + 2 : ac ,null)

[@sannisinha58]

It's very helpful... please continue😊

[@AbdurRahimKhan-gz1jo]

Hey, @coderdost please do more videos like this...

[@coffee0919]

function findOdd(arr) { for (let i = 0; i < arr.length; i++) { if (arr[i] + 2 !== arr[i + 1]) { return arr[i] + 2; } } return -1; } Is that work?

[@brainnbeyond]

what I had done for(let i=0; i< input.length; i++){ if(input[i+1] - input [i] > 2){ console.log(input[i] + 2) } } is it right ?

[@coderdost]

yes. replacing 2 by "n" can make it more dynamic to and useful fo any kind Arithmetic progression

[@laylahashmi4234]

function missingOddNum(arr) { let completeArr = [arr[0]]; for (let i = 0; i < arr.length; i++) { let value = completeArr[i] + 2; completeArr.push(value) } return completeArr.filter((num) => !arr.includes(num)) }

[@584_tyit_rohitghadge9]

let arr=[1,3,9,11,15,17]; let emp=[]; let missingvalue=(arr)=>{ let min=Math.min(...arr); let max=Math.max(...arr); for(let i=min;i

[@zimbasardhara3450]

let data =[1,3,5,7,9,11,15,17,19,21] let data2=[]; for(let i=0;i

[@harshjoshi6496]

@coderdost Binary search is possible here, Solution: (Explanation at last) const input = [5, 7, 9, 11, 15, 17,19, 21]; first = 0 last = input.length-1 mid = (first+last)/2 while (first

[@coderdost]

nice exaplanation

[@sanvijha]

We can not apply binary search for this problem as we have to evaluate key from the array itself.

[@arbazhussain6751]

Sir, I have tried to solve this question. Is it correct? // find the first missing ODD number let input = [5,7,9,11,15,17]; for(let i = input[0]; i < input[input.length-1]; i++){ if(i % 2 !== 0){ if(!input.includes(i)){ console.log(i); break; } } }

[@coderdost]

Even though it works. But there are some unnecessary iterations like "i" is 5,6,7,8,9 etc.. these values are not all needed like 6, 8 , 10 etc. You can optimize this solution to a more simpler method.

[@rajatraj3160]

sort the array use map and return the element which which has remainder 1. Or u can use sort along with map

[@coderdost]

@@rajatraj3160 array is already sorted.. so its even simpler now. and we can't evaluate using remainder, as here all numbers are ODD only. problem is to find out 1 missing number from sequence.

[@AbhishekKumar-xw1cm]

see this let input = [5,7,9,11,15,17] let n = 2 const findMissing = (input,n) => { for(let i =0 ; i < input.length ;i++){ if(input[i]+n !== input[i+1]){ return input[i]+n } } } console.log(findMissing(input,n))

[@uditkhandelwal6330]

USING BINARY SEARCH -> const input1 = [5, 7, 11, 13, 15, 17, 19]; const input2 = [19, 17, 13, 11, 9, 7, 5]; const findMissingOddNumber = (arr) => { let l = 0; let r = arr.length - 1; let d; if(arr[l]

[@sumitkumar-in2si]

Is this correct? let input=[5,7,9,11,15,17]; for(let i=0;i

[@coderdost]

Yes

[@rahil516]

First view ; Sir your videos are very helpful appreciated

[@nikhilnikki9627]

IS THIS CORRECT WAY: const input=[5,7,9,11,15,17] let first =input[0] for (let i of input){ if (i ===first){ let next = first+2 first=next } else{ console.log(first) } }

[@coderdost]

yes, but I think you can remove "first" and use only 1 variable "next" let next =input[0] for (let i of input){ if (i ===next){ let next = next+2 } else{ console.log(next) } }

[@Doglapan64]

@coderDost Please let us know how to break forEach loop ? without using try-catch

[@a.spragadeeshbalaji2917]

//FIND THE MISSING ODD NUMBER WITHIN THE RANGE SPECIFIED IN THE ARRAY const input = [5,7,9,11,15,17]; for(let i=input[0]; i

[@namangaur5329]

Binary search is not helpful in this case bcz in binary search we find a given number within array. But here the number itself not determined. So first we need to find the number and then we can use the bs to find the appropriate position for the number. I'm hoping that the answer is correct..😅 But this type of content is really helpful for us ... Thank u sir for making such type of content❣️

[@coderdost]

True story

[@WalkingtotheTruth]

Binary Search is possible here is the answer: function findMissingOddNumber (arr) { let left = 0 let right = arr.length - 1 let startingNumber = arr[0] let ans = -1 while (left

[@WalkingtotheTruth]

@@pawansinghrawat7886 const mid = Math.floor((left + right) / 2) this is the mid,we are traversing logn times not whole array,you can run this code and anlysis yourself, thank you.

[@notyournormaldev1419]

yes it is possible, i am the one in this video@@WalkingtotheTruth

[@payalkothari4862]

let missingNumber; for (let i = input[0]; i

[@lalitsinghnegi1562]

aaj meri joining thi or m MERN stack profile ke liye gya tha or saalo ne python ( flask ) ka project de dia ... pehle se bna hua tha ab usme new functionalities dd krni h or mujhe python aati h nhi .. ab m kya kru

[@coderdost]

I think learn python because JS and python are almost same. and flask is much easier like express kzbin.info/www/bejne/Z3yWq62lpK98gK8

[@lalitsinghnegi1562]

@coderdost aap shi keh rhe h but agr wo pehle ye baat bta dete ya post m mentioned kr dete to m join hi nhi krta .. but ab bna hua project h to mere kch samjh m aa nhi rha h .. smjh nhi aara kya kru

[@coderdost]

There are some hard realities of tech sector...they hire for some profile but give work in some other...specially to freshers.. Its hard to judge many companies..

[@lalitsinghnegi1562]

@coderdost yes bro .. i think i should quit ...

[@coderdost]

In sab facts ko batane k liye ek alag channel banane ka socha h ... soon will plan.. code wale channel pe to mix nhi krenge

[@jsview5315]

Hi sir I am fresher Frontend dev i have completed projects and portfolio too and have little bit DSA knowledge but Machine coding and this type question se daar lgta hai so job ki tention aa rahi hai . Please suggest anything

[@coderdost]

Everyone fears such questions as they are matter of practice. So just keep doing such questions more and more. And in real round they will feel simpler. Without practice it will be hard

[@jsview5315]

@@coderdost Thanks for the reply sir your content is awesome.

[@AvadheshVerma24]

for (let i = 0; i < arr.length; i++) { let c = arr[i]; let next = c + 2; if (!arr.includes(next)) { console.log(next); break; } }

[@ankurpandey3152]

Wow nice interview

[@nileshnilu7902]

Make more video like this very her.

[@immdipu]

function missingOdd(array){ let firstDigit = array[0]; let missingDigit = null; while(missingDigit === null){ if(firstDigit % 2 !== 0){ if(!array.includes(firstDigit)){ missingDigit = firstDigit; return missingDigit; } } firstDigit++; }

[@coderdost]

Why we will need to check -> firstDigit %2, as all are odd only. and will have mod as 1.

[@immdipu]

@@coderdost all are not odd. firstDigit will increase by one each time the while loop runs. so it means the first time while loop runs firstDigit will be 5 and second time the while loop run the value of firstDigit will increase by one i.e 6 . 6 is not a odd number so that why we need firstDigit%2

[@coderdost]

@@immdipu OK, I think you can optimize by reducing the iteration and skipping the even increments totally

[@AbhishekSharma-fg1ho]

its front end dev interview?? and which type of qus asking in front end dev

[@coderdost]

its only coding round for javascript (front-end). which is not compulsory in all companies Some other interview playlists: REACT Interview Shorts : bit.ly/3VfIrMi JAVASCRIPT Interview Shorts: bit.ly/3XhHRQ1 More REACT video Interviews - bit.ly/3QAjAln

[@DhirajkaReview26]

Sir one question currently i am cdac student they teaches so many concepts rapidly and some topics I don't understand so can u tell me about is that our interview is which base and how should I have to prepare for interview the level Interview they can u tell me about that ??

[@coderdost]

for long term only your knowledge matter, how many projects you have worked on. What you can make. preparing for interview is only a short term things for placements. Interviews are mostly based on popular tricky questions. Just go on internet and check the question related to JS or React. Same 50 questions are asked mostly We made a playlist of popular question in JS and React here : REACT Interview Shorts : bit.ly/3VfIrMi JAVASCRIPT Interview Shorts: bit.ly/3XhHRQ1

[@DhirajkaReview26]

@@coderdost thanks sir , i saved those playlist and sir one more thing as knowledge perspective my explanation and understanding is good but sometimes I can't write because of syntax is that ok ? As i herad from student interview just how much i understand and even if writte normal code they say is ok for them to hire , like now they start adv java as seen jdbc code to connect i don't they will write that whole code in 1 hour interview can tell me about those things , i have your adv java one video also for reference 🤓 can explain this point how much interview can ask like 5 min code on notepad ?

[@coderdost]

@@DhirajkaReview26 Most of them look for approach rather than exact code. So your code or Pseudocode should explain the steps. They look if you can think of logic, why this logic, what are limitation of that logic, are their any bounds in which that logic will work, are there performance issue with that logic,... They try to change some part of that question also - to check if you can now think of different logic also, to check if you have not mugged up from somewhere. If they have to check syntax and code prefection they will always ask to code on some machine.

[@DhirajkaReview26]

@@coderdost thank u so much to clear my questions, getting proper suggestions and knowledge From u it's so much for me , thanks lot 🙏

[@pritamkeshri8389]

// ===========Missing odd Number const arr5 = [5, 7, 9, 11, 15, 17]; const arr5output = 13; function findOddNumber(input) { let output; for (let i = 0; i < input.length; i++) { let current = input[i]; let future = input[i + 1]; if (future && future - current !== 2) { output = current + 2; } } return output; } console.log(findOddNumber(arr5));

[@mitesh5189]

const b = [5,7,11,13,17] function missOdd(arr) { let i = arr[0]; let a =0 while(true){ if(arr[a]===i){ i +=2; a++; } else{ console.log(i); break; } } } missOdd(b);

[@coderdost]

simple and effective. using 'a' as index and 'i' as element is only distracting part.

[@DeepakGupta-yk2ep]

Sir please make more Javascript interview videos

[@coderdost]

Yes 👍🏻

[@gojo55x]

Can you please tell me sir how much experience he has.....

[@coderdost]

Fresher

[@nazirrizwan]

Sir how apply and what procedure to all about this interview

[@coderdost]

There is a form which was put on community tab. Or directly book slot via emailing us. Email is in about us section. Business email section

[@PIYUSH-lz1zq]

Bro , coding rounds ks ek alag se playlist bana do !!!

[@coderdost]

Will do

[@PIYUSH-lz1zq]

@@coderdost u uploaded only 2 video on js coding round ??

[@coderdost]

@@PIYUSH-lz1zq 3 videos. all in react interview playlist right now.. will move to separate one

[@AkashSingh-xf6bd]

Coder Dost, Hi I did not get you "n" concept in last can you explain me

[@coderdost]

like here N was 2. as each next number was +2. So just asked if we can code for any AP series like 3,6,9 - where N=3 and so on.

[@AkashSingh-xf6bd]

@@coderdost okay, now I get this

[@montugohain48]

Great video ,really helpful, I also tried is to solve and came up with this one.. class ArithmeticProgression { arr = []; constructor(a, n, num) { for (let i = 0; i < num; i++) { this.arr.push(a + n * i); } } AP_generator() { return this.arr; } AP_sum() { return this.arr.reduce((acc, cur) => acc + cur); } } const input = [5, 7, 9, 11, 15, 17]; const CompleteAP = new ArithmeticProgression( input[0], input[1] - input[0], ++input.length ); const givenTotal = input.reduce((acc, cur) => acc + cur); const ans = CompleteAP.AP_sum() - givenTotal; console.log(ans);

[@coderdost]

Answer might be right. But its a complicated approach also uses a lot of operations of summing whole array twice etc. both space and time complexity becomes high

[@CryptoAirdropnow]

is there any way to give a mock interview with you

[@coderdost]

you can email. at email given on about page.

[@kiranm5419]

const input = [5,7,9,11,15,17]; const a = []; const b = []; for(let i=0;i

[@rajulorencemurmu3838]

My approach is:- Since they all are odd then what we can do is create an array and store all odd elements in that array like 5 to 17 then we can compare them both using loops if the first element of new array does not match to the first element of previous array then we print that.

[@coderdost]

but creating another array is extra effort. it will increase the space complexity. you can use same logic on original array and check if next element is not the next Odd number ( previous + 2)

[@vision_lc]

I make this question sir with not cheating sir