JavaScript LIVE Coding Interview Round (Mock)

In this video, we have a candidate, a fresher, who tries to solve a live coding problem in a javascript interview

#reactjs #javascript #interview #leetcode

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[varandeepsahota]

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.

[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);

[karthikhs]

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<1) // last filter for only first occurance

[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);

[YourSecularBro]

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);

[sannisinha]

It's very helpful... please continue😊

[itechinnovations]

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

[AbdurRahimKhan-gzjo]

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

[Rohit-zskn]

const arr = [5, 7, 9, 11, 13, 15, 17, 21];

let index = 0;
for (let oddNum = arr[0]; oddNum <= arr[arr.length - 1]; oddNum += 2) {
if (arr[index] !== oddNum) {
console.log(oddNum);
break;
}
index++;
}

[khanapeena]

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

[ApurvaKashyap-kjqz]

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

[sanvijha]

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

[yesuraju-tfyp]

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<=last){
mid = Math.floor((first+last)/2);
first = mid+1;
else last = mid-1;
}

return first_element + 2*first;

}

console.log(missOdd([5, 7, 11, 13, 15, 17]));

😀

[GurpreetSingh-gogsy]

const input = [5, 7, 9, 11, 15, 19];
for(let i=1;i<input.length;i++){
if(input[i-1]+2 != input[i]){
console.log(input[i-1]+2)
break
}
}

[MehulJadavIndia]

const input = [5, 7, 9, 11, 13, 15, 17, 21];

for (i = 0; i <= input.length; i++) {
let number = input[i];
let oddNumber = input[i + 1];
let margin = oddNumber - number;
if (!(margin === 2))
return console.log(
"This number is missing:",
oddNumber - 2
);
}

[happylaughy]

const arr= [3, 5, 7, 9, 13]
let odd=arr[0];
for(let i=0;i<arr.length;i++){
if(arr[i]!=odd){
console.log (odd);
}
else{
odd+=2
}
}

[Rubyd]

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

[MrHimanshuji]

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<=inputEnd; i++) {
if( i%2 !==0 && !input.includes(i)) {
missDigit.push(i);
}
}

console.log(missDigit);

[AkashSingh-xfbd]

let input =[5, 7, 11, 13, 15, 17];
let out = 13
function s(input){
for(let i= 0; i<input.length; i++){
let temp = input[i]
if(temp+2 !== input[i+1])
{
return input[i]+2
}
}
}
// s(input)
console.log(s(input))

[yournanban]

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