React 2023 MasterClass : bit.ly/3I7MYN7 React Interviews : bit.ly/3QAjAln JavaScript Interview Shorts: bit.ly/3XhHRQ1 React Interview Shorts : bit.ly/3VfIrMi Candidate : Raghbir Singh : www.linkedin.com/in/raghbir-singh-a1183a219/ Interviewer: Abhishek Rathore : www.linkedin.com/in/abhishek87/ More REACT Interviews - bit.ly/3QAjAln REACT Interview Shorts : bit.ly/3VfIrMi JAVASCRIPT Interview Shorts: bit.ly/3XhHRQ1 REACT JS Course (Redux & HOOKS) : bit.ly/3iMethN Course of JAVASCRIPT : bit.ly/3u049tf Course of Modern JS.(ES6) : bit.ly/3DvYCh6

@krishnodas6678

Жыл бұрын

In question 1 it is just matching the same indices in both object How the program is figuring out e:12

const input1 = { a: 1, b: 2, c: 3, d: 10, e: 12 }; const input2 = { a: 2, e: 12, f: 6, d: 10 }; let output = {}; Object.keys(input1).forEach((item) => { if (input2[item] === input1[item]) { output[item] = input1[item]; } }); console.log(output);

Question 3. const input = [2,7, 11,4, -2]; const output = [11,4, -2, 2, 7]; Solution: const input = [2,7, 11,4, -2]; input.push(input.shift()); input.push(input.shift()); console.log(input);

@codewithadii

Жыл бұрын

its better approach

@nitishgupta8393

Жыл бұрын

smart, but interviwer always wants you to solve without any inbuilt method. to see you logic.

Abhishek is a very patient and makes the interviewee comfortable. Loved it!

I am not a js user but I have a logic for Question 3 which is in cpp int rotate(vector& nums, int k) { vector temp(nums.size()); for(int i = 0 ; i

I have discovered your channel just a month ago , and seriously I gain much more confidence on JavaScript then before. Sir You are a great teacher❤❤Your every video really a masterpiece . Thank you so much sir. Please upload content on live small React project. Currently I am learning React. Paid courses also have no compare to your content,100% True solid knowledge. "Agar koi aeisa banda bhi aapka video dekh liya ekbar,jisko kuch knowledge hi nhi hai, vo bhi 3 months mei ek front end developer ban sakta hai" great teacher ! great content ! this channel really deserve Millions of views.

fabulous content it really very helpful

This video was greatefull who are learing and wants to become as programmer for fresh beginning

in solution 3 : you can use a recursive reverse array approach to reduce time complexity to O(2n)

Very useful Video 👍

Awesome keep making such video

let input1 = {a:1, b:2 , c:3 ,d:4} let input2 = {a:3, b:2, e:4 , d:4} for (const key1 in input1) { if (key1 in input2 && input1[key1] === input2[key1]) { console.log(key1, input1[key1]); } }

Sir I am requesting you please make a series on DSA in JavaScript for beginners 🥲 🙏🏼

@coderdost

Жыл бұрын

Noted. This already in listed but will require some time. As beginners DSA and then advanced DSA lot of topics are there

@shahhussain56

Жыл бұрын

@@coderdost Thank you so much sir. I will finish first your MERN stack videos and wait for the DSA.

@karanbhoite9552

Жыл бұрын

@@coderdost Is dsa really required , if we are working as frontend dev, and if needed then how much .?

@coderdost

Жыл бұрын

@@karanbhoite9552 DSA is required for cracking some good jobs, I haven't find any use of DSA for junior developers (although some dsa basics can help in writing good code, but those are very easy ones) . When you become senior there are chances that you will build some systems and they might need to have better data structure. But in front-end that kind of complete architecture is designed only by really expreience developer (5-10 yrs). Majorly people learn DSA for job interviews. Hope that trend ends

@AnkitKumar-xy1xo

10 ай бұрын

Don't learn dsa in JavaScript. It would be better to pick some other language like c++.

to find second largest number : //asumming there atleast 2 numbers in array nums=[2,45,3,4,2,9,45] // initializing max which will keep track of max int and res which will be second largest let max=nums[0] let res=nums[1] if(nums[1]>nums[0]){ max=nums[1] res=nums[0] } //starting loop from the third number coz first two are already assigned; for(let i=2; imax){ res=max max=nums[i] } else if(nums[i]>res && nums[i] !=max){ res=nums[i] } } console.log(res) TIME COMPLEXITY O(N) only traversing through entire array once SPACE O(1)

Thanks for uploading!

1st question let input1 = {a:1,b:2,c:4,f:11,t:10} let input2 = {a:2,b:2,c:3,f:11} // take out common key with same value {b:2,f:11} console.log(TakeOutCommon(input1,input2)) ***CODE*** function TakeOutCommon(ob1,ob2){ if(Object.keys(ob1).length===0 || Object.keys(ob2).length===0)return {} // if any input object is empty let ansObject = {} for(let k in ob1){ if( ( ob1[k] && ob2[k] ) && ob1[k]===ob2[k]){ // checking if both object have same key..... if they have same key then checking for its value ansObject[k]=ob1[k] } } return ansObject }

Thanks sir for mock Interview

Nice, It was a helpful video

vscode text-editor are allow or not?

Really so helpful.

For the second question it is ok to sort array in descending order and then make two a,b and using shift to remove the first element .and b will have value of 7 second largest element from array1

@coderdost

Жыл бұрын

No. As we can also have repeated numbers so its not sure that sorted 2nd index will give the result

in 2nd question i think the using of set method is good instead of solving with complexity

@vaibhavjaiswal1309

9 ай бұрын

you can directly solve that question in o(N) with help of two variable in single loop

@codewithadii

9 ай бұрын

@@vaibhavjaiswal1309 yes you are right

I am familiar with some of the concepts of DSA but not all and currently i am learning web development and want to get hired off campus because i've been completed my graduation so should i have to learn dsa and solve questions or i just focus on technology and try to build projects.

Very helpful 👍

please release as many videos as possible it would be helpful for us who are learning programming from youtube

@coderdost

Жыл бұрын

yes trying that. will soon have more videos

U can do Let input=[2,7,11,4,-2] for(let I =0; I

@JAYPATEL-hn2nn

Жыл бұрын

In this way u can do this left or right using pop,push, shift and unshift

const input = [2, 7, 11, 4, -2] let spl = input.splice(0, 2) console.log(spl); //[2,7] console.log(input); //[11,4,-2] output = input.concat(spl) //1. using concat to merge 2 arrays // output = [...input, ...spl] //2. using spread operator to merge 2 arrays // output = input.push(...spl) //3. Merge using array.push() console.log(output) //output = [11,4,-2, 2, 7]

question : 1 ES6 version const obj1 = {a : 1, b : 2, c : 3, d: 4, e : 5} const obj2 = {e : 4, c : 3, r : 6, a:1, f : 9, p : 4} const keys1 = Object.keys(obj1) const keys2 = Object.keys(obj2) const resultObject = {} const commonKeyCollector = keys1 > keys2 ? keys2.filter((key2) => keys1.includes(key2)) : keys1.filter((key1) => keys2.includes(key1)) commonKeyCollector.map((key) => { if(obj1[key] === obj2[key]){ resultObject[key] = obj1[key] }}) console.log(resultObject)

Solution 2. I didn't used any inbuilt function. Is my approach right? let arr = [1,2,-2,11,7,1] let big = arr[0]; let big2; let bigI; for(let i=0; i

I solved the 1st problem in this way cause I thought we can't loop through objects directly😆😆😆 function objCompare(a, b){ const first=Object.entries(a) const second=Object.entries(b) let arr=[] for(let i=0;i

Question-3 const input=[2,7,11,4,-2,]; const output=[11,4,-2,2,7]; Solution: function rotateArr(input){ let result=[]; for(let i=2;i

Subscribed🎉

const value_1 = [1, 2, -2, 11, 7, 1]; // output 7 const value_2 = [1, 4, 7, 2, 4, 7]; // output 4 function nextBiggest(arr) { let max = 0, result = 0; for (const value of arr) { if (value > max) { result = max; max = value; } else if (value result) { result = value; } } return result; }

@rishabrajverma2883

Жыл бұрын

let arr=[5,4,8,2,3,6]; function Secondlargest(arr){ return arr.sort()[arr.length-2]; } console.log(Secondlargest(arr));

It will be easy if you used inbuilt functions

Question 3 : const input = [2, 7, 11, 4, -2]; function rot2IndLeft(array) { let output = []; for (let i = 2; i output.push(array[i]); } output.push(array[0], array[1]); console.log(array); return output; } const output = rot2IndLeft(input); console.log(output);

const input=[1,4,7,2,4,7] let a=0; let b; for (let i=0;ia){ b=a; a=input[i] } } console.log(b)

const arr=[1,5,-2,11,7] let res= arr.sort() console.log(res[arr.length-1])

Thanks Bhai.....

we can find the second item in the array in just one iteration ,just maintaining two Fmax,Smax ,Variable,Time complexity =O(N) . we do not need to sort the array and also no need of set ds

@coderdost

Жыл бұрын

Great. Can you share this solution.

@ArunKumar-gx8iv

Жыл бұрын

@@coderdost As i Am A Java Guy ,Currently Learning JS from Your videos JAVA CODE: Time Complexity =O(N) AND SC=O(1) static int getSecondLargestItem(int[] arr) { int n=arr.length; //if array Length is less than 2 then,there is no second largest item if(nS_max && arr[i]!=F_max){ S_max=arr[i]; //update second-largest item } } return S_max; } But Equivalent JAVASCRIPT CODE : function getSecondLargestItem(arr) { var n = arr.length; //if array Length is less than 2 then,there is no second largest item if (n return -1; } var F_max = Number.MIN_SAFE_INTEGER; //First-largest item in an array var S_max = Number.MIN_SAFE_INTEGER; //Second-largest item in an array for (var i = 0; i //check if current element is greater than first-largest item if (arr[i] > F_max) { S_max = F_max; //update second-largest item F_max = arr[i]; //update first-largest item } //check if current element is greater than second-largest item and not equal to first-largest item else if (arr[i] > S_max && arr[i] != F_max) { S_max = arr[i]; //update second-largest item } } return S_max; }

@coderdost

Жыл бұрын

@@ArunKumar-gx8iv Great Approach.

let input = [1,4,7,2,4,7] function secondLargest(input) { let num = [...new Set(input)].sort().reverse() console.log(num[1]); } secondLargest(input) is it right????

Hi sir , this is my solution of finding 2nd largest element; *Not use of reverse inbuilt function function find2ndlargest(arr) { let smax = 0; let stack = []; for (let i = 0; i while (stack.length !== 0 && stack[stack.length - 1] smax = stack[stack.length - 1]; stack.pop(); } stack.push(arr[i]); } return smax; } console.log(find2ndlargest(arr));

question 2 function task2(arr) { let arrCopy = arr.slice().sort((a, b) => b - a); // reverse sort let output = arrCopy.find((x) => x return output }

You said he is a non CS-student but how does he know these algorithms though?

q3: const rotate2place = (arr)=> { let res = [] for (let i=0; i if(i+2 res.push(arr[i+2]) } else { res.push(arr[(i+2) - arr.length ] ) } } return res }

I usually reject the candidate if he/she use Array.sort() for finding the second largest element. I expect this answer to be solved in O(n) instead of using Array.sort() which takes O(nlogn). People use the sort method blindly, without even realising the time complexity.

@succeedwithuttam2271

Жыл бұрын

How to find second largest number in native javascript: nums=[2,45,3,4,2,9,45] max=nums[0] secondmax = Number.NEGATIVE_INFINITY //finding largest number nums.map((num)=>{ if(num > max){ max=num } }) //removing largest number const arr = nums.filter(function(item) { return item !== max }) //finding second largest number arr.map((num)=>{ if(num > secondmax){ secondmax=num } }) console.log(arr) console.log(max) console.log(secondmax)

@AbhishekTrivedi07

Жыл бұрын

@@succeedwithuttam2271 - Why are you removing the max from the array? You can use it to find 2nd Max. In fact, you can do all this in a single loop, which will be the optimised solution.

@mysteryman2213

Жыл бұрын

​@@AbhishekTrivedi07 to find second largest number : //asumming there atleast 2 numbers in array nums=[2,45,3,4,2,9,45] // initializing max which will keep track of max int and res which will be second largest let max=nums[0] let res=nums[1] if(nums[1]>nums[0]){ max=nums[1] res=nums[0] } //starting loop from the third number coz first two are already assigned; for(let i=2; imax){ res=max max=nums[i] } else if(nums[i]>res && nums[i] !=max){ res=nums[i] } } console.log(res) TIME COMPLEXITY O(N) only traversing through entire array once SPACE O(1)

@a.spragadeeshbalaji2917

4 ай бұрын

const arr = [2,4,5,3,4,2,9,45]; //sorting the array for(let i=0,j=1;j arr[j]){ arr[j] = arr[i] + arr[j]; arr[i] = arr[j] - arr[i]; arr[j] = arr[j] - arr[i]; } } console.log("The second largest number in this array is "+arr[arr.length - 2]);

last one : const rotate = (arr) =>{ let res = arr.splice(0,2) arr.push(...res) return arr } console.log(rotate(arrVal))

// question 1 // const input_1 = { a:1 , b:2 , c:3, d:10 , e:12 } // const input_2 = { a:2 , e:12 , f:6, d:10 } // const output = {} // for(let item in input_1){ // if(input_1[item] === input_2[item]){ // output[item] = input_1[item] // } // } // console.log(output) // question 2 // const input_1 = [1,2,-2,11,7,1] // const temp = Math.max(...input_1) // const temp_2 = input_1.filter(item => item != temp) // const result = Math.max(...temp_2) // console.log(result) // question 2 // let max_1 = -Infinity; // let max_2 = -Infinity; // for(let item of input_1){ // if(item > max_1){ // max_2 = max_1 // max_1 = item // } // else{ // if(item > max_2) // { // max_2 = item // } // } // } // console.log(max_2) // question 3 // const input = [2,7,11,4,-2] // const result = [...input] // for(let index in input){ // result.splice(index - 2 , 1 , input[index]) // } // console.log(result);

@anandsen8893

Жыл бұрын

Salary kitne loge discuss kar le 😉

@lalitsinghnegi1562

Жыл бұрын

@@anandsen8893 jitne se ghar chl jaye utni dedo sir ...

solution for rotation of the array llet input1 = [2, 7, 11, 8, 19]; const rotateArray = (input) => { for (let i = 0; i if (input1[i] === input) { input1[i - 2] === undefined && input1.push(input1[i]) && input1.shif(); input1[i] = input1[i - 1]; input1[i - 1] = input1[i - 2]; input1[i - 2] = input; break; } } console.log(input1); }; rotateArray(input1[3]);

const input = [2, 5, 7, 8, 3, 4]; // output should be [7,8,3,4,2,5] const output = []; function main(reverseVal) { for (let i = reverseVal; i

1st question --> const input = [1, 2, -2, 11, 7, 1] function scndLarg(arr) { let newArr = arr.sort((a, b) => a - b); let val = [] for (let i = 0; i if (!val.includes(newArr[i])) { val.push(newArr[i]) } } return val[val.length -2] }

nice sir

pajji cha gye tusse

const input = [1,3,5,6,3,-1] const output= [5,6,3,-1,1,3] const rotateArr =(arr,numberOfRotation=0)=>{ if (numberOfRotation>0){ let temp = arr.pop() arr.unshift(temp); } return arr } console.log(rotateArr(input,2))

♥️

function secondLargest(input1){ input1.sort((a,b)=>b-a); for(let i=0;i

et input1 = [1, 2, -2, 11, 7, 1, 11, 12]; let input2 = [1, 4, 2, 4, 7]; function getSecondValue(input) { let data = input.sort((a, b) => { return a - b; }); return data; } let val = getSecondValue(input2); console.log(val); for (let i = 0; i if (i === val.length - 2) { console.log(val[i]); break; } }

ques1:Better approach const output = {}; for (const key in obj1) { if (obj2.hasOwnProperty(key) && obj1[key] === obj2[key]) { output[key] = obj1[key]; } } console.log(output);

👍🏻

const input1=[1,2,11,22,33,44,22,33,44]; let input2=Array.from(new Set(input1)); input2.sort((a,b)=>b-a); console.log(input2[1]);

/** Question Number 2 */ const input = [1,2,-2,11,7,1] input.sort((a,b)=> b-a) function secondLargest(input){ for(value of input){ let temp=input[0] if(value!==temp){ return value break } } } const result = secondLargest(input) console.log(result)

question 3 function task3(arr, elemPosition) { let arrCopy = [...arr]; let arr2 = arrCopy.splice(elemPosition); return [...arr2, ...arrCopy]; }

question 1: const input1 = {a:1, b:2, c:3, d:10, e:12} const input2 = {a:2, e:12, f:6, d:10} let output = {} for(let [key, value] of Object.entries(input1)) { if(input2[key] === value) { output[key] = value } } console.log(output)

🥰🥰🥰

second question ---- const secLar = function (arr) { const max = Math.max(...arr); const index = arr.indexOf(max); const del = arr.splice(index, 1); const maxNew = Math.max(...arr); console.log(maxNew); } secLar(arr);

@rishabrajverma2883

Жыл бұрын

let arr=[5,4,8,2,3,6]; function Secondlargest(arr){ return arr.sort()[arr.length-2]; } console.log(Secondlargest(arr));

my sol: const inp1 = {a:1, b:2, c:3, d:10, e:12} const inp2 = {a:2, e:12, f:6, d:10} let common = {} Object.entries(inp1).forEach(([key,val])=> { Object.entries(inp2).forEach(([k,v])=>{ if(key == k && val ==v){ common[key] = val } }) }) console.log(common)

20:22, const arr = [2, 7, 11, 4, -2]; const arr1 = arr.splice(0,2); console.log([...arr, ...arr1]) how's this approach?

@coderdost

Жыл бұрын

it's fine to do it in a real life project. In that interview I think generally we want for-loop based without using in-built functions.. Just to check how one thinks about building logics

@SagarTakoresdt

Жыл бұрын

​@@coderdost Your KZread shorts are very helpful, in real project.

He already used sort He has to just print length-2

//Rotate array const input = [2, 7, 11, -2, 4]; const removeValue = input.splice(0, 2); input.push(...removeValue);

Sir this is my solution for the 1st question: ---> let input1 = { a: 1, b: 2, c: 3, d: 10, e: 12 }; let input2 = { a: 2, e: 12, f: 6, d: 10 }; let obj = {}; function findkeyValue(input1, input2) { for (let key in input1) { if (input1[key] == input2[key]) { if (obj[key] == undefined) { obj[key] = input1[key]; } } } return obj; } console.log(findkeyValue());

@coderdost

Жыл бұрын

its fine. if (obj[key] == undefined) {

@PHINIxXGaming

Жыл бұрын

@@coderdost oook sir thank you 😊

const returnCommonvalue=(input1,input2)=>{ let dict = {}; for(let key in input1) { if(key in input2 && input1[key] === input2[key]) { dict[key] = input1[key]; } } return dict }

lets asume that input1 has 100 entries and input2 has only 3 or 4 entries then above code wouldn't be appropriate below code might solve this issue function func(input1,input2){ let obj = {} if(Object.keys(input1).length > Object.keys(input2).length){ for(let i in input2){ if(input2[i]==input1[i]){ obj[i] = input2[i] } } }else{ for(let i in input1){ if(input1[i]==input2[i]){ obj[i] = input1[i] } } } return obj }

Solution 3: const input = [2,7,11,4,-2] let temp = [] for(let i=0; i if(i

let array = [9,1,4,3,7,2,11] let largest = 0; let secondLargest = 0; let i = 0; while(i array[i] && secondLargest secondLargest = array[i] } i++; }

where will i get this type of javascript question ?

@coderdost

Жыл бұрын

There are many sites like Geek for geeks.. leetcode etc

Sir please make a video on how to do async operations in redux/toolkit

@coderdost

Жыл бұрын

This will be covered in upcoming series of react JS. Thanks for suggestions.

@mobpsycho6600

Жыл бұрын

@@coderdost and one more thing please make it fully dependent on the redux 😁

Question 1 : your candidate is looping in only input 1 which will fail if input 2 is bigger than input 1 here is correct answer const input1 = { a: 1, b: 2, c: 3, d: 10, name: "dev" }; const input2 = { a: 2, name: "dev", f: 6, d: 10 }; function commonValues(input1, input2) { console.log("Input 1 :", input1); console.log("Input 2 :", input2); let obj = {}; for (let key1 in input1) { if (input2.hasOwnProperty(key1) && input1[key1] === input2[key1]) { obj[key1] = input1[key1]; } } return obj; } const output = commonValues(input1, input2); console.log("Output : ", output);

3 rd problem solution: const input=[2,7,11,4,-2] for(let i=0;i

solution to the second question let a=[1,4,7,2,4,7] let sortedArr=a.sort((a,b)=>a-b) let uniqueArr=[...new Set(sortedArr)] let secondLargest=uniqueArr[uniqueArr.length-2] console.log(secondLargest)

Sir how can give this type of mock interview?

@coderdost

Жыл бұрын

Post related to such mock interview dates are posted on channel community.

solved it in 15 seconds

Second question. const arr1 = [1, 2, -2, 11, 7, 1]; const arr2 = [1, 4, 7, 2, 4, 7]; function getSecondLargestNum(arr) { const filteredArr = []; const newArr = arr.sort((a, b) => a - b).map(num => filteredArr.includes(num) ? num : filteredArr.push(num)); return filteredArr[filteredArr.length - 2]; } console.log(getSecondLargestNum(arr2));

Question2: const SecondLargest = (arr)=>{ // function to store the freq in object. const obj = arr.reduce((acc,currentValue)=>{ acc[currentValue] = (acc[currentValue]|| 0)+1; return acc; },{}); // sort the object. const data = Object.entries(obj).sort((a,b)=>a[0]-b[0]); console.log(data) let n = data.length; if(n return data[n-2][0]; } console.log(SecondLargest(SecondLargest_Input));

@rishabrajverma2883

Жыл бұрын

let arr=[5,4,8,2,3,6]; function Secondlargest(arr){ return arr.sort()[arr.length-2]; } console.log(Secondlargest(arr));

@variant3283

Жыл бұрын

@@rishabrajverma2883 your code wont work when array has repated elements

const input1 = { a: 1, b: 2, c: 3, d: 10, e: 12, }; const input2 = { a: 2, e: 12, f: 6, d: 10, }; // Function to find common properties with the same values function findcommonInBoth(obj1 ,obj2){ const resultingObj = {}; for (const key in obj1) { if (obj2.hasOwnProperty(key) && obj1[key] === obj2[key]){ //construct a new object resultingObj[key] = obj1[key] } } return resultingObj } const result = findcommonInBoth(input1 ,input2) console.log(result);

//1 const input1 = { a: 1, b: 2, c: 3, d: 10, e: 12 }; const input2 = { a: 2, e: 12, f: 6, d: 10 }; const op={}; for(key in input1){ if(input1[key]===input2[key]) op[key] = input1[key]; } console.log(op) //2 const input = [2,7,11,4, -2]; const sorted = input.sort((a,b)=>a-b) console.log(sorted.reverse()) //or const cleaned = [...new Set(input)] console.log(cleaned.sort()[cleaned.length -2]) //3 const leftRotate = (arr,by)=>{ if(by>0){ arr[arr.length-1] = arr.shift() leftRotate(arr,by-1) } return arr; } const input = [2,7,11,4, -2]; console.log(leftRotate(input,3))

Question 3: Solutiion = [...input.slice(2),...input.slice(0,2)]

QUS 1 : const input1 = {a: 1, b: 2, d: 3, d: 10, e: 12} const input2 = {a: 2, e: 12, f: 6,d: 10 } let keys = Object.keys(input2) const obj = {} for(let i in input1 ){ if( keys.includes(i) && input1[i] === input2[i] ){ obj[i] = input1[i] } } console.log(obj)

@irajeshtailor

Жыл бұрын

This is correct answer

const input =[1,2,-2,11,7,1] const remDuplicate =[...new Set(input)] console.log(remDuplicate.sort((a,b)=>b-a)[1])

const input1 = { a: 1, b: 2, c: 3, d: 10, e: 12 }; const input2 = { a: 2, e: 12, f: 6, d: 10 }; const output1 = { b: 2, c: 3 }; const input1keys = Object.keys(input1); const output2 = {}; input1keys.forEach((key) => { if (input1[key] === input2[key]) { Object.assign(output2, { [key]: input2[key] }); } }); console.log(output2);

How many years of experience this guy ?

@coderdost

Жыл бұрын

Fresher non IT background

Problem 1: function intersection (obj1,obj2){ let output={} if (Object.keys(obj1).length===0 || Object.keys(obj2).length===0){ return {} } Object.keys(obj1)?.forEach(ele=>{ if(obj1[ele]===obj2[ele]){ output[ele]=obj1[ele] } }) return output }

Question 2 : const arr1 = [1,2,-2,11,7,1]; const findMax = Math.max(...arr1); const secondLargest = arr1.filter(num => num console.log(Math.max(...secondLargest));

We can rotate the any array by k position either left or right most in just O(N) with space Complexity O(1) . we just need to observe the pattern of the output for at least 2 to 4 different input arrays.

@coderdost

Жыл бұрын

How to do in O(1) space ?

@ArunKumar-gx8iv

Жыл бұрын

@@coderdost create a reverse function for reverse // Reverse first n-k elements Reverse(arr, 0, n - k - 1); // Reverse last k elements Reverse(arr, n - k, n - 1); // Reverse whole array Reverse(arr, 0, n - 1); this is for Rotate array by K elements to right

const i1={a:1,b:2,c:3,d:10,e:12} const i2={a:2,e:12,f:6,d:10} let output={}; for (let i in i1){ for (let j in i2){ if (i==j && i1[i]==i2[j]){ output[i]=i1[i] } } } console.log(output)

let obj = { a:1, b:2, c:3, d:4, e:5 } let obj1 = { a:10, b:20, c:3, d:4, } let result = {} Object.keys(obj).forEach((val)=>{ if(obj[val] === obj1[val]){ result[val] = obj[val] } })

let arr = [1,3,4,1,264,2311,2311,6]; let [max,secMax] = [-Infinity,-Infinity] for(let i = 0 ; i if(arr[i]>max){ secMax = max max = arr[i] }else if(arr[i]>secMax && arr[i] secMax = arr[i] } } console.log(secMax)

Sir can I get a chance of getting a Mock interview by you?

@coderdost

Жыл бұрын

Please fill the Form. weekend - we will again try to have some rounds

@rupbhaiya

Жыл бұрын

@@coderdost Sir would you mind give me the link? :(

@coderdost

Жыл бұрын

@@rupbhaiya www.geeksforgeeks.org/practice-for-cracking-any-coding-interview/

Sir, is this real or fake interview? Please reply 🙏🏼🙏🏼🙏🏼

@chandrabhanrahangdale1534

9 ай бұрын

It's is a mock interview

@coderdost

9 ай бұрын

practice only

function getCommonProps(input1,input2){ const Input1Keys= Object.keys(input1) const Input2Keys= Object.keys(input2) return Input1Keys.reduce((acc, curr)=>{ if(Input2Keys.includes(curr) && input1[curr] === input2[curr]){ acc[curr]= input1[curr] return acc } return acc },{}) }

function commonKeys(a, b) { let myObj = {}; Object.keys(a).filter(function (key) { if (a[key] == b[key]) { myObj[key] = b[key]; return b.hasOwnProperty(key); } }); return myObj; } var obj1 = { a: 10, b: 30, c: 60, d: 90, }; var obj2 = { a: 20, b: 40, c: 60, d: 90, }; const data = commonKeys(obj1, obj2); console.log("data", data);

How to find second largest number in native javascript: nums=[2,45,3,4,2,9,45] max=nums[0] secondmax = Number.NEGATIVE_INFINITY //finding largest number nums.map((num)=>{ if(num > max){ max=num } }) //removing largest number const arr = nums.filter(function(item) { return item !== max }) //finding second largest number arr.map((num)=>{ if(num > secondmax){ secondmax=num } }) console.log(arr) console.log(max) console.log(secondmax)