Fragen im Vorstellungsgespräch: Front end engineer

Der Schwerpunkt der Tätigkeit eines Front End Engineers liegt auf der Benutzerfreundlichkeit einer Software oder App. Beim Vorstellungsgespräch müssen Sie zeigen, dass Sie mit den Grundlagen des UX/UI-Designs vertraut sind und sich zu sauberem und für das jeweilige Produktangebot optimiertem Code bekennen. Außerdem müssen Sie die Bereitschaft haben, gemeinsam mit Back End Engineers an der Lösung von Problemen zu arbeiten. Rechnen Sie mit Fragen nach Ihrer technischen Erfahrung sowie nach Ihren Fähigkeiten bei Mitarbeiterführung und Design.

9 522Fragen aus Vorstellungsgesprächen für „ Front End Engineer“, von Bewerbern geteilt

Typische Bewerbungsfragen als Front End Engineer und wie Sie diese beantworten

Das sind die drei häufigsten Fragen im Vorstellungsgespräch als Front End Engineer und wie Sie diese am besten beantworten:

Frage 1: Was ist Ihr bevorzugter Arbeitsablauf oder Managementstil?

So beantworten Sie die Frage: Beschreiben Sie, welche Tools und Methoden Sie bei der Entwicklung eines Produkts verwenden. Sprechen Sie über die Strategien, die Sie bei der Zusammenarbeit mit verschiedenen Stakeholdern wie Kunden, Vertriebs- und Marketingmitarbeitern sowie Back End Engineers anwenden. Zeigen Sie anhand konkreter Beispiele auf, dass Ihre Arbeitsabläufe erfolgreich sind, und bekunden Sie auch Ihre Bereitschaft, sich anzupassen und zu verändern, wenn dies erforderlich ist.

Frage 2: Wie verwalten Sie Tests, Reviews und die Versionskontrolle?

So beantworten Sie die Frage: Ein Großteil der Aufgaben eines Front End Engineers besteht darin, sich um die kleinen Details zu kümmern, die für eine nahtlose Benutzererfahrung unentbehrlich sind. Unterstreichen Sie, dass Sie um die Bedeutung von Konzepten wie sauberem Code, Testprotokollen und Versionsverwaltung wissen. Geben Sie Beispiele für Methoden, die Sie bereits verwendet haben, und für Probleme, die Sie damit bearbeitet und ggf. gelöst haben.

Frage 3: Was reizt Sie am meisten am Bereich UX/UI?

So beantworten Sie die Frage: Eine Frage wie diese ist Ihre Gelegenheit zu zeigen, dass Sie mit Leib und Seele Front End Engineer sind. Erläutern Sie, wie Sie nutzerzentriertes Design in Ihre Projekte integrieren und welche Philosophien Sie dabei verfolgen. Geben Sie an, welche Bücher oder Artikel Sie gelesen haben und was Sie für richtig halten. Sprechen Sie ggf. darüber, welche Veränderungen Sie erwarten und wie Design und Technologie sich Ihrer Meinung nach daran anpassen werden.

Top-Fragen in Vorstellungsgesprächen

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Front End Engineer wurde gefragt...7. April 2013

Given an input array and another array that describes a new index for each element, mutate the input array so that each element ends up in their new index. Discuss the runtime of the algorithm and how you can be sure there won't be any infinite loops.

25 Antworten

Essentially the same as Anh's answer but less code, assuming ES5 is available var arr = ["a","b","c","d","e","f"]; var indices = [2, 3, 4, 0, 5, 1]; arr = indices.map(function (item, index) { return arr[indices.indexOf(index)]; }); Weniger

function reposition(arr, indices) { var newArr = []; // I'm not sure if extra space is allowed. If it is, the solution should be this simple. for(var i = 0; i < arr.length; ++i) { var newIndex = indices[i]; newArr[newIndex] = arr[i]; } return newArr; } var arr = ["a", "b", "c", "d", "e", "f"]; var indices = [2, 3, 4, 0, 5, 1]; reposition(arr, indices); // returns: ["d", "f", "a", "b", "c", "e"] Weniger

function repositionElements(arr, indices) { // assert(arr.length === indices.length) var moved = []; for (var i = 0; i < arr.length; i++) { moved.push(false); } var moveFrom, moveTo, itemToMove; for (moveFrom = 0; moveFrom < arr.length; moveFrom++) { itemToMove = arr[moveFrom]; while (!moved[moveFrom]) { moveTo = indices[moveFrom]; var tmpItem = arr[moveTo]; arr[moveTo] = itemToMove; itemToMove = tmpItem; moved[moveFrom] = true; moveFrom = moveTo; } } return arr; } var arr = ["a", "b", "c", "d", "e", "f"], indices = [2, 3, 4, 0, 5, 1]; repositionElements(arr, indices); // returns: ["d", "f", "a", "b", "c", "e"] Weniger

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Given input: // could be potentially more than 3 keys in the object above items = [ {color: 'red', type: 'tv', age: 18}, {color: 'silver', type: 'phone', age: 20} ... ] excludes = [ {k: 'color', v: 'silver'}, {k: 'type', v: 'tv'}, .... ] function excludeItems(items, excludes) { excludes.forEach(pair => { items = items.filter(item => item[pair.k] === item[pair.v]); }); return items; } 1. Describe what this function is doing... 2. What is wrong with that function ? 3. How would you optimize it ?

24 Antworten

I agree with converting the excludes to an object, but in order to get linear performance that doesn't depend on the number of excluded things, you have to concatenate the k and v into one value to be used as the key in the object: let excludesObject = {}; excludes.forEach(pair => excludesObject[`${pair.k}_${pair.v}`] = true); Then you can check if an item should be excluded in O(k) time where k is the number of keys in an item. And the whole thing will run in O(nk) where n is the number of items. // if there is some key which is found in the excludesObject, the filter will return false items = items.filter(item => !Object.keys(item).some(key => excludesObject[`${key}_${item[key]}`]); ); Facebook, hire me! lol Weniger

function excludeItems(items, excludes) { let excludesMap = excludes.reduce((entry, result)=>{ entry[result.k + result.v] = true; return entry; },{}); return items.reduce( (result, item) => { let updatedObject = Object.keys(item).reduce( (result,key) => { if(!excludesMap[key + item[key]]){ result[key] = item[key] } return result; }, {}) result.push(updatedObject) return result; }, []) } Weniger

Time complexity O(n*k) where k is the number of excludes. Nothing change if you change from: for(const item of items){ for(const exclude of excludes) { ... } } to for(const exclude of excludes){ for(const item of items) { ... } } time complexity will be the same. Main problems in this code I see next: 1. typo in the line ```item[pair.k] === item[pair.v]``` should be ```item[pair.k] === pair.v``` 2. overriding incoming parameter, it does not mutate the global items, but still it's bad practice in my opinion. 3. we can rewrite the code without using extra variables. my try: const excludeItems = (items, excludes) => items.filter(item => !excludes.some(({k,v}) => item[k] === v)) Weniger

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1. In JavaScript, write a function that takes an array as input that can contain both ints and more arrays (which can also contain an array or int) and return the flattened array. ex. [1, [2, [ [3, 4], 5], 6]] => [1, 2, 3, 4, 5, 6] 2. Using HTML and CSS, show how you would create an image that would display another image (aligned to the bottom, right) when the user hovers over the image. ex. The Facebook "edit profile picture" icon

12 Antworten

ary.join().split(',');

array.toString().split(',').map(Number), map need because input array has only integer types. Weniger

var flatten = function(arr, resultArr) { var result = resultArr || []; for(var i = 0; i < arr.length; i++) { if(Array.isArray(arr[i])) { flatten(arr[i], result); } else { result.push(arr[i]); } } return result; }; Weniger

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Given a grid of characters output a decoded message. The message for the following would be IROCKA. (diagonally down right and diagonally up right if you can't go further .. you continue doing this) I B C A L K A D R F C A E A G H O E L A D

11 Antworten

as there are 21 characters and you describe a grid, I'm assuming it's 3x7: I B C A L K A D R F C A E A G H O E L A D Starting from the upper left and moving SE till a wall, then moving NE, then moving SE yields IROCLED, not IROCKA. However my human-powered fuzzy search is telling me it's much more likely that the answer is IROCKED, so I'm wondering whether the 1. author misremembered the original grid 2. the author misremembered the correct answer and the grid 3. I misunderstood the question. Weniger

var arr = [ ['I','B','C','A','L','K','A'], ['D','R','F','C','A','E','A'], ['G','H','O','E','L','A','D'] ]; var row = 0, col = 0; var totalCols = arr[0].length; var totalRows = arr.length; var msg = ''; while (col < totalCols) { msg += arr[row][col]; // row++ if less than total rows // row-- if back at row 0 row = (row === 0 || row < totalRows - 1) ? row + 1 : row - 1; // always go forward in column col++; } // returns 'IROCLED' Weniger

All solutions in here are wrong as they never go back up to the 0th row after first descent. ``` var arr = [ ['I','B','C','A','K','E','A'], ['D','R','F','C','A','E','A'], ['G','H','O','E','L','A','D'] ]; var row = 0, col = 0; var totalCols = arr[0].length; var totalRows = arr.length; var goingDown = false; var msg = ''; while (col < totalCols) { msg += arr[row][col]; // row++ if less than total rows // row-- if back at row 0 if (row === 0 || (row < totalRows - 1 && goingDown)) { row += 1; goingDown = true; } else { row -= 1; goingDown = false; } // always go forward in column col++; } console.log(msg) // IROCKED ``` Weniger

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Meta

Given an array, return it's flattened structure(skip objects)

10 Antworten

function flatten(input) { return (input || []).reduce(function (prev, current) { if (Array.isArray(current)) { return prev.concat(flatten(current)); } else { prev.push(current); return prev; } }, []); } Weniger

function flattenRecursive( a, ret ) { return a.reduce((ret, cur) => { if (Array.isArray(cur)) { flattenRecursive(cur, ret); } else if(typeof cur !== 'object'){ ret.push(cur); } return ret; }, ret || []) } function flattenIterative( a ) { let ret = []; while( a.length ) { let cur = a.pop(); if (Array.isArray(cur)) { a = [...a, ...cur]; } else if ( typeof cur !== 'object' ){ ret.push(cur) } } return ret.reverse(); } Weniger

let flat = (z = []) => { return !Array.isArray(z) ? [].concat(z) : (z.length > 0) ? flat(z.splice(0, 1)[0]).concat(flat(z)) : z } Weniger

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Accenture

What are your greatest technical strengths?

8 Antworten

Management

Social media experience

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

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Implement a simple store class with set(Node, value), get(Node) and has(Node) methods, which store a given Nodes with corresponding values.

7 Antworten

Since node is object we can store value directly on the node itself. But it become complex and may mess with original object. Luckily we have Symbol come to rescue: class DOMStore { constructor () { this.DOMStoreSymbol = Symbol('DOMStore'); } has (node) { return node[this.DOMStoreSymbol] !== undefined; } set (node, value) { node[this.DOMStoreSymbol] = value; } get (node, defaultValue) { return node[this.DOMStoreSymbol] || defaultValue; } } const e1 = document.createElement('A'); const e2 = document.createElement('P'); const e3 = document.createElement('DIV'); const e4 = document.createElement('TABLE'); const e5 = document.createElement('A'); const map1 = new DOMStore(); const map2 = new DOMStore(); console.assert(map1.has(e1) === false, 'map1 should be empty'); map1.set(e1, 'E1 in map1'); console.assert(map1.has(e1) === true, 'map1 must have e1'); map2.has(e1); console.assert(map2.has(e1) === false, 'map2 should be empty'); map1.set(e2, 1234); map1.set(e3, "String"); map1.set(e4, [1,2,3,4]); console.assert(map1.get(e5, null) === null); console.assert(map1.get(e5) === undefined); console.assert(map1.get(e5, 5) === 5); map1.set(e5, {1: 2, 3: 4}); console.assert(map1.get(e1) === 'E1 in map1'); console.assert(typeof(map1.get(e5)) === 'object'); Weniger

Are you able to elaborate on this question? Can each node have multiple values or just 1? The way I interpret what you have written, it just sounds like they are asking for an ES6 Map? ES6 Maps can have objects as keys, so you can use the Node as the key. If you had to code it from scratch without the use of an ES6 Map, something like: class CachedNode { constructor(node, value) { this._node = node; this._value = value; } getNode() { return this._node; } getValue() { return this._value; } setValue(value) { this._value = value; return this; } } class SimpleStore { constructor() { this._container = []; } set(node, value) { let cachedNode; if (this.has(node)) { cachedNode = this.get(node); cachedNode.setValue(value); } else { cachedNode = new CachedNode(node, value); this._container.push(cachedNode); } return this; } // you might want to change this method so it returns the Node's value not the CachedNode. // If you did, that would mean adding another method to get the CachedNode. Easy enough. get(node) { return this._container.find((cachedNode) => { return cachedNode.getNode() === node; }); } has(node) { return !!this.get(node); } } If it needed to store multiple values against a node, then just change the single value for a Set or Array. Weniger

The tricky part of the question was on how to store a DOM Node which is an object as an old javascript object key. I can't recall was it about 1 to 1 relationship or 1 to many, but it is really doesn't matter because the later gives just a small overhead. I suppose your solution is pretty what they were expected from the task, starting from explaining ES6 Map and ending up with an old javascript solution. Weniger

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Write an emitter class: /* emitter = new Emitter(); // 1. Support subscribing to events. sub = emitter.subscribe('event_name', callback); sub2 = emitter.subscribe('event_name', callback2); // 2. Support emitting events. // This particular example should lead to the `callback` above being invoked with `foo` and `bar` as parameters. emitter.emit('event_name', foo, bar); // 3. Support unsubscribing existing subscriptions by releasing them. sub.release(); // `sub` is the reference returned by `subscribe` above */

7 Antworten

var Emitter = function() { this.events = {}; }; var Subscription = function(event, callback, event_listeners, key) { this.event = event; this.callback = callback; this.event_listeners = event_listeners; this.key = key; }; Subscription.prototype.release = function () { var ret = false; if (this.event_listeners[this.key]) { delete this.event_listeners[this.key]; ret = true; } return ret; }; Emitter.prototype.subscribe = function(event_name, callback) { if (!this.events.hasOwnProperty(event_name)) { this.events[event_name]; this.events[event_name] = []; } var subscription = new Subscription(event_name, callback, this.events[event_name], this.events[event_name].length); this.events[event_name].push(subscription); return subscription; }; Emitter.prototype.emit = function(event_name, param1, param2) { var subs = this.events[event_name]; return subs.forEach(function(sub) { return sub.callback.call(sub, param1, param2); }); } Weniger

class Emitter { constructor() { this.events = {}; } on(name, handler) { (this.events[name] || (this.events[name] = [])).push(handler); return this.off.bind(this, name, handler); } off(name, handler) { this.events[name] && this.events[name].filter(handle => handle === handler); } emit(name, ...payload) { this.events[name].map(handler => handler(...payload)); } } Weniger

class Emitter { constructor() { this.cb = {}; } subscribe = (name, cb) => { this.cb[name] = cb; return {release: () => delete this.cb[name]}; } emit = (name, ...arges) => { this.cb[name](arges); } }; e = new Emitter(); rm1 = e.subscribe("f1", p => console.log("f1111 ", p)); rm2 = e.subscribe("f2", p => console.log("f222 ", p)); e.emit("f1", "GGG", "ssskoko") e.emit("f1", "GGG", "ssskoko", 1,5,7,2,3) Weniger

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Booking.com

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

7 Antworten

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." Weniger

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Meta

How would you ensure clicking on this picture would go to a specific link?

6 Antworten

tags, note: don't nest these.

Sorry but why listeners? Can't you just wrap an with ?

Sorry but why listeners? Can't you just wrap an img tag with a tag?

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