Domande poste durante i colloqui per Front End Engineer

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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; }, []) } Meno

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This solution will give much faster and immutable result. You can test with performance.now items.reduce((allItems, item) => { if(!excludes.some(exclude=>item[exclude.k] === exclude.v)){ allItems.push(item); } return allItems; }, []); Meno

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

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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; }; Meno

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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' Meno

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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 ``` Meno

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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(); } Meno

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

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Social media experience

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

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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. Meno

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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. Meno

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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); Meno

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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] } ]); }; Meno

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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)); } } Meno

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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) Meno

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It's always revenue my man/woman

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"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." Meno