Questions d'entretiens - Front end developer

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Questions d'entretien pour Front End Developer partagées par les candidats

Principales questions d'entretien

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Accenture
On a demandé à un Senior Front End Developer...14 juillet 2017

What are your greatest technical strengths?

8 réponses

Management

Social media experience

I am technically sound in using R script for R programming language, base SAS, SAS analytics and Advanced SAS. Moins

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LinkedIn

// Given var endorsements = [ { skill: 'css', user: 'Bill' }, { skill: 'javascript', user: 'Chad' }, { skill: 'javascript', user: 'Bill' }, { skill: 'css', user: 'Sue' }, { skill: 'javascript', user: 'Sue' }, { skill: 'html', user: 'Sue' } ]; getSkills = (endorsements) => { // Result // [ // { skill: 'javascript', user: ['Chad', 'Bill', 'Sue'], count: 3 }, // { skill: 'css', user: ['Sue', 'Bill'], count: 2 }, // { skill: 'html', user: ['Sue'], count: 1 } // ]; } see this image: http://i.imgur.com/UIeB3n4.png

7 réponses

let endorsements = [ { skill: 'css', user: 'Bill' }, { skill: 'javascript', user: 'Chad' }, { skill: 'javascript', user: 'Bill' }, { skill: 'css', user: 'Sue' }, { skill: 'javascript', user: 'Sue' }, { skill: 'html', user: 'Sue' }, ]; let x = endorsements.reduce((acc, { skill, user }) => { if (skill in acc) { acc[skill] = { user: [...acc[skill]['user'], user], count: acc[skill]['count'] + 1, skill: skill, }; } else { acc[skill] = { user: [user], count: 1, skill: skill, }; } return acc; }, {}); console.log(Object.values(x)); Moins

I did in a different way let outPut = []; let skillMap = {}; for (let char of endorsements) { if (char['skill'] in skillMap) { skillMap[char['skill']] = [...skillMap[char['skill']], char['user']]; } else { skillMap[char['skill']] = [char['user']]; } } for (let key in skillMap) { let skillObj = {}; skillObj['skill'] = key; skillObj['user'] = skillMap[key]; skillObj['count'] = skillMap[key].length; outPut.push(skillObj); } return outPut.sort((a, b) => b.count - a.count); Moins

getSkills = endorsements => { const skillMap = {}; // assumption - each object has all the properties, nothing is missing endorsements.forEach((item, index) => { const { skill, user } = item; if (!skillMap[skill]) { skillMap[skill] = {}; skillMap[skill]["user"] = []; // skillMap[skill]["user"].push(user); skillMap[skill]["count"] = 0; } skillMap[skill]["user"].push(user); skillMap[skill]["count"] += 1; // } }); return Object.keys(skillMap).map(key => [ { skill: key, ...skillMap[key] } ]); }; Moins

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Booking.com

In your opinion, what is our company's single most important metric?

7 réponses

Reservations per day I guess

It's always revenue my man/woman

"It's always revenue my man/woman." I hope you understand how irrelevant you observation is in the context of this position. So no, it's not "always revenue." Moins

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PE International

"If you were stranded on a desert island, what three development tools would you bring with you?"

5 réponses

This ridiculous, senseless question was the first one asked. It set the tone for what was to follow. Moins

IDE, Build Tools, Version Control!

Git, Codio, and Netflix

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Bloomberg L.P.

What is a JavaScript callback function?

5 réponses

A callback function is a piece of JavaScript code that executes after the main function that the callback is attached to executes successfully. Moins

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 Moins

Is there a front end role at bloomberg. I guess your position must have been labelled software dev right? altho ur a dront end dev Moins

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IBM

1. To add an HTTP request to the search engine, then log a result depending on the# of pages and requirements. 2. Use DP to implement a Knapsack-like problem. Restrictions on time complexity. 3. Implement a to-do list's delete method on website.

5 réponses

To X: Get rejected

Did they request to do a background screening on you before rejecting?

No

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Meta

Given an array where each entry can be another array, and so forth, flatten the array. [4, [3, 6, [9, 1, 9, [5, 1]]], 8] -> [4, 3, 6, 9, 1, 9, 5, 1, 8]

5 réponses

Assuming that we are solving this in Java, I am treating an array as an array list // Use Hash Set to prevent infinite loops HashSet hs = new HashSet(); ArrayList flatten_list(ArrayList arr, HashSet hs) { ArrayList result = new ArrayList; int len = arr.size(); if(len < 1) return result; // Empty List. for(int i = 0 ; i Moins

function serialize(a, start){ if(start >= a.length) return a; if(Array.isArray(a[start])) return serialize([...a.slice(0,start),...a[start], ...a.slice(start+1,a.length)], start+1); return serialize(a, start+1); } Moins

const flatten = array = > { return array.reduce((elem, acc) = > { return elem.concat(Array.isArray(acc) ? flatten(acc) : acc) }, []) } Moins

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LinkedIn

// How can I know which radio was clicked or selected? <h2>Monstrous Government Form</h2> <form id="myForm" name="myForm"> <fieldset> <legend>Do you live in an:</legend> <p><input type="radio" name="home" value="apartment" id="apartment" /> <label for="apartment">Apartment</label></p> <p><input type="radio" name="home" value="house" id="house" /> <label for="house">House</label></p> <p><input type="radio" name="home" value="mobile" id="mobile" /> <label for="mobile">Mobile Home/Trailer</label></p> <p><input type="radio" name="home" value="coop" id="coop" /> <label for="coop">Co-op</label></p> <p><input type="radio" name="home" value="none" id="none" /> <label for="none">None</label></p> </fieldset> <fieldset> <legend>Your income is:</legend> <p><input type="radio" name="inc" value="0-50K" id="0-50K" /> <label for="0-50K">$0-50,000 USD</label></p> <p><input type="radio" name="inc" value="50-100K" id="50-100K" /> <label for="50-100K">$50,000-100,000 USD</label></p> <p><input type="radio" name="inc" value="100K+" id="100K+" /> <label for="100K+">$100,000+ USD</label></p> </fieldset> <fieldset> <legend>Your status is:</legend> <p><input type="radio" name="status" value="single" id="single" /> <label for="single">single</label></p> <p><input type="radio" name="status" value="married" id="married" /> <label for="married">married</label></p> <p><input type="radio" name="status" value="partner" id="partner" /> <label for="partner">domestic partner</label></p> </fieldset> <p>This form goes on with another 97 questions....</p> <input type="submit" value="Submit" /> </form>

5 réponses

Attach the event listener to the form id="myForm"...

document.querySelector('input[name="house"]:checked').value;

you can add oncl1ck event to assign event target value to a variable for each fieldset, since there are 3 different field sets, and we want to know each of them separately. Moins

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Yahoo

Design a data structure to store sparse 2D matrix which contains only 0 or 1. then write function to add 2 such matrix.

4 réponses

use run-length encoding.

store each row as a decimal for ex: if the row is 1011 -&gt; store it as 13!

Since all values are mod 2 you can pack 64 entries together into on int64_t. You can then add two matrices by XORing each entry. Moins

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Pinterest

How many characters in the front of a string would need to be added in order to make it a palindrome.

4 réponses

Here's a python solution: def min_add_for_pal(word): drow = word[::-1] # Reverse string if drow == word: return 0 for shift in range(len(word)): if word[:-shift] == drow[shift:]: return shift return None # Shouldn't reach here Moins

A better asnwer is to do it in-place with pointers. O(N) speed and O(N) space. function checkPalindrome(i, input){ var leftPointer = 0; var rightPointer = 0; if(i===0){ //base case of single char return true; } var odd = (i+1)%2; //we know it's not 0 if(odd){ leftPointer = i/2; //not 0 rightPointer = i/2; //not 0 while(input[leftPointer] === input[rightPointer] &amp;&amp; leftPointer &gt; 0 &amp;&amp; rightPointer 0 &amp;&amp; rightPointer &lt; input.length){ leftPointer = leftPointer - 1; rightPointer = rightPointer + 1; } } if(input[leftPointer] === input[rightPointer] &amp;&amp; leftPointer === 0){ return true; }else{ return false; } } var improvedMinPalindromeAddition = function(input){ var inputArray = input.split(""); //max string size is greater then max array size(4.29 billion elements)..we could also do it inline w/string manipulation var largestPalindrome = 0; for(var i=0;i Moins

function isPallindrom(str) { if (!str) return; let start = 0; let end = str.length -1; while (start start) { console.info(str.substring(0, end)) if (isPallindrom(str.substring(0, end))) { return count; } else { count++; } end--; } return count; } Moins

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