↳
const Sum = a => b => c => a + b + c;
↳
``` // Lexically nested function definitions defined within enclosing function function Sum(arg0) { function Inner1(arg1) { function Inner2(arg2) { return arg0 + arg1 + arg2; } return Inner2; } return Inner1; } console.log(Sum(3)(4)(5)); ``` Less
↳
var sum=0; // global variable function Sum(num){ sum=num+sum; return Sum; } Sum(3)(4)(5); Less
↳
#center{width: 100px; height: 50px; position: fixed; top: 0; left: 0; right: 0; bottom: 0; margin: auto;} Less
↳
wrong answer. u need to also margin it negative
↳
. container{ Position : absolute; Width : 200px; Top : 50℅; Left : 50℅; Transform : translate(-50℅) } Less
↳
What other types questions were on the written test?
↳
Given a scnerio, create a database model
↳
How to improve performance?
↳
Use a counter to track the status. Stack solution is no difference from this. Seeing an opening parenthesis increment the counter. Seeing a closing parenthesis decrement the counter. Check the counter is always non-negative. Counter has to be zero when string ends. Less
↳
Stack the parenthesis up. 1st of course the first parenthesis will go on to the stack. If you see a closing parenthesis, pop the stack and see if the parenthesis matches the one which is just read(closing one), if no match? error else keep going. Less
↳
To Aegis: If you take "[" as 1 and "]" as -1, then ][][ starts from -1. We can say that's invalid right away. Less
↳
They will mail the exact date. The joining date is between 6-16 aug. The difficulty of 2nd round was medium. One question was a bit difficult and the rest were medium.If you are good at coding and regularly doing it then you can easily crack it. Less
↳
Be sure you are good at coding.
↳
when is your joining??what is the difficulty of 2nd round?
↳
1. 1000 computers - parallel 2. the problem can be divided into 2 parts. (1) find the top 1000 numbers from that 1 billion numbers in one computer To do this. we just need to use merge sort. Everytime use the left part. With some calculation, the solution for this is actually O(N). n is 1 billion. (2) get that top 1000 from the 1k * 1k list. (easy) Less
↳
There is a very nice parallel sorting algorithm with a very good iso-effeciency. It is called Sample-sorting. I would use that. Less
↳
You could do a bunch of crazy stuff too... Map reduce and parallel stuff is kinda a given. You'd be reading in strings, so before converting them to numbers, you could just check the position of the comma/point, and skip if your comma/point counter is at a higher value. If there's a minus sign in the beginning ignore the number, provided you already have 1000 positive numbers. Do something crazy with a stream and a router. Write a HD driver that scans the sign bit and exponent bits. lol Find a way to scan vertically for the comma/point. And you wouldn't be sorting you'd be searching. Ideally, it would take as long as it takes to read 1 text file from disk, because you'd be kinda faux streaming it through your code, plus a little extra time, multiplied by the number of files. You'd get a nice boost from dividing your file by the number of processors. 32 cores = 32 pieces of file being processed at the same time. Less
↳
The most upvoted answer on this has a run time complexity of O(m * n) where m and n are the arrays. This is solvable in O(n) by using a hash map, by sacrificing space for complexity: var Array1 = ["a", "b", "c", "d", "e", "f", "c"]; var Array2 = ["c", "x", "y", "f", "c"]; let hash = {}; let result = []; // loop over the largest for (let i = 0; i < Array1.length; i++) { if (!hash[Array1[i]]) { hash[Array1[i]] = 1; } } for (let i = 0; i < Array2.length; i++) { if (hash[Array2[i]]) { result.push(Array2[i]); delete hash[Array2[i]]; } } console.log(result); Less
↳
var Array1 = ["a", "b", "c", "d", "e", "f"]; var Array2 = ["c", "x", "y", "f"]; Array1 = Array1.filter(function(val) { return Array2.indexOf(val) !== -1; }); Less
↳
I think that the last solution suggested might still return duplicates. For example if array1 = [1,2,3] and array2 = [2,2,2] we will get the result of [2,2]. This is another solution in O(n) - function findDuplicates(arr1, arr2) { var dict = {}; var duplicates = {}; for (i = 0; i < arr1.length; i++ ) { dict[arr1[i]] = 1; } for (i = 0; i < arr2.length; i++ ) { if (dict[arr2[i]]) { duplicates[arr2[i]] = arr2[i]; } } return Object.values(duplicates); } Less
↳
This ridiculous, senseless question was the first one asked. It set the tone for what was to follow. Less
↳
IDE, Build Tools, Version Control!
↳
Git, Codio, and Netflix
↳
A callback function is a piece of JavaScript code that executes after the main function that the callback is attached to executes successfully. Less
↳
udaykanth, I would say that a .forEach() would be the most common and most basic use of a callback function. I'm just writing this to help anyone that might have a hard time thinking up a quick example if the come across this question themselves. Example: var numArray = [ 1, 2, 3 ] ; numArray.forEach( function( i ) { console.log( arr[ i - 1 ] ) } ) ; // logs out // 1 // 2 // 3 Less
↳
I don't think Bloomberg is a very good company. I am an excellent web developer and have gotten multiple offers from other companies with big names, but was rejected by Bloomberg. They are too demanding during the job interview and it becomes a game of how well you can interview as opposed to how talented an employee you are and how much you can contribute to the growth of the company. Less
↳
#Assumes positive numbers def divide(num, divide_by) answer = 0 return answer if divide_by == 0 while(num >= divide_by) num = num - divide_by answer = answer + 1 end answer end puts divide(10,0) Less
↳
Simpler version (assuming you are allowed to use multiplication), just compute the sign at the end and multiply: function divide(a, b){ if(b == 0) throw "Cannot divide by zero"; var remainder = Math.abs(a); var dividend = Math.abs(b); var result = 0; while(remainder >= dividend){ result++; remainder -= dividend; } if(result > 0 && a*b < 0) result *= -1; return result; } Less
↳
http://www.bearcave.com/software/divide.htm