A mostly reasonable collection of technical software development interview questions solved in Javascript in ES5 and ES6

1. Array
2. Strings
3. Stacks and Queues
4. Recursion
5. Numbers
6. Javascript Specific
7. To Be Continued

• 1.1 Given an array of integers, find the largest product yielded from three of the integers

  varunsortedArray=[-10,7,29,30,5,-10,-70];computeProduct(unsortedArray);// 21000functionsortIntegers(a,b){returna-b;}// Greatest product is either (min1 * min2 * max1 || max1 * max2 * max3)functioncomputeProduct(unsorted){varsortedArray=unsorted.sort(sortIntegers),product1=1,product2=1,array_n_element=sortedArray.length-1;// Get the product of three largest integers in sorted arrayfor(varx=array_n_element;x>array_n_element-3;x--){product1=product1*sortedArray[x];}product2=sortedArray[0]*sortedArray[1]*sortedArray[array_n_element];if(product1>product2)returnproduct1;returnproduct2;}

  View on Codepen:https://codepen.io/kennymkchan/pen/LxoMvm?editors=0012

• 1.2 Being told that an unsorted array contains (n - 1) of n consecutive numbers (where the bounds are defined), find the missing number in O(n) time

  // The output of the function should be 8vararrayOfIntegers=[2,5,1,4,9,6,3,7];varupperBound=9;varlowerBound=1;findMissingNumber(arrayOfIntegers,upperBound,lowerBound);// 8functionfindMissingNumber(arrayOfIntegers,upperBound,lowerBound){// Iterate through array to find the sum of the numbersvarsumOfIntegers=0;for(vari=0;i<arrayOfIntegers.length;i++){sumOfIntegers+=arrayOfIntegers[i];}// Find theoretical sum of the consecutive numbers using a variation of Gauss Sum.// Formula: [(N * (N + 1)) / 2] - [(M * (M - 1)) / 2];// N is the upper bound and M is the lower boundupperLimitSum=(upperBound*(upperBound+1))/2;lowerLimitSum=(lowerBound*(lowerBound-1))/2;theoreticalSum=upperLimitSum-lowerLimitSum;returntheoreticalSum-sumOfIntegers;}

  View on Codepen:http://codepen.io/kennymkchan/pen/rjgoXw?editors=0012

• 1.3 Removing duplicates of an array and returning an array of only unique elements

  // ES6 Implementationvararray=[1,2,3,5,1,5,9,1,2,8];Array.from(newSet(array));// [1, 2, 3, 5, 9, 8]// ES5 Implementationvararray=[1,2,3,5,1,5,9,1,2,8];uniqueArray(array);// [1, 2, 3, 5, 9, 8]functionuniqueArray(array){varhashmap={};varunique=[];for(vari=0;i<array.length;i++){// If key returns undefined (unique), it is evaluated as false.if(!hashmap.hasOwnProperty(array[i])){hashmap[array[i]]=1;unique.push(array[i]);}}returnunique;}

  View on Codepen:http://codepen.io/kennymkchan/pen/ZLNwze?editors=0012

• 1.4 Given an array of integers, find the largest difference between two elements such that the element of lesser value must come before the greater element

  vararray=[7,8,4,9,9,15,3,1,10];// [7, 8, 4, 9, 9, 15, 3, 1, 10] would return `11` based on the difference between `4` and `15`// Notice: It is not `14` from the difference between `15` and `1` because 15 comes before 1.findLargestDifference(array);functionfindLargestDifference(array){// If there is only one element, there is no differenceif(array.length<=1)return-1;// currentMin will keep track of the current lowestvarcurrentMin=array[0];varcurrentMaxDifference=0;// We will iterate through the array and keep track of the current max difference// If we find a greater max difference, we will set the current max difference to that variable// Keep track of the current min as we iterate through the array, since we know the greatest// difference is yield from `largest value in future` - `smallest value before it`for(vari=1;i<array.length;i++){if(array[i]>currentMin&&(array[i]-currentMin>currentMaxDifference)){currentMaxDifference=array[i]-currentMin;}elseif(array[i]<=currentMin){currentMin=array[i];}}// If negative or 0, there is no largest differenceif(currentMaxDifference<=0)return-1;returncurrentMaxDifference;}

  View on Codepen:http://codepen.io/kennymkchan/pen/MJdLWJ?editors=0012

• 1.5 Given an array of integers, return an output array such that output[i] is equal to the product of all the elements in the array other than itself. (Solve this in O(n) without division)

  varfirstArray=[2,2,4,1];varsecondArray=[0,0,0,2];varthirdArray=[-2,-2,-3,2];productExceptSelf(firstArray);// [8, 8, 4, 16]productExceptSelf(secondArray);// [0, 0, 0, 0]productExceptSelf(thirdArray);// [12, 12, 8, -12]functionproductExceptSelf(numArray){varproduct=1;varsize=numArray.length;varoutput=[];// From first array: [1, 2, 4, 16]// The last number in this case is already in the right spot (allows for us)// to just multiply by 1 in the next step.// This step essentially gets the product to the left of the index at index + 1for(varx=0;x<size;x++){output.push(product);product=product*numArray[x];}// From the back, we multiply the current output element (which represents the product// on the left of the index, and multiplies it by the product on the right of the element)varproduct=1;for(vari=size-1;i>-1;i--){output[i]=output[i]*product;product=product*numArray[i];}returnoutput;}

  View on Codepen:http://codepen.io/kennymkchan/pen/OWYdJK?editors=0012

• 1.6 Find the intersection of two arrays. An intersection would be the common elements that exists within both arrays. In this case, these elements should be unique!

  varfirstArray=[2,2,4,1];varsecondArray=[1,2,0,2];intersection(firstArray,secondArray);// [2, 1]functionintersection(firstArray,secondArray){// The logic here is to create a hashmap with the elements of the firstArray as the keys.// After that, you can use the hashmap's O(1) look up time to check if the element exists in the hash// If it does exist, add that element to the new array.varhashmap={};varintersectionArray=[];firstArray.forEach(function(element){hashmap[element]=1;});// Since we only want to push unique elements in our case... we can implement a counter to keep track of what we already addedsecondArray.forEach(function(element){if(hashmap[element]===1){intersectionArray.push(element);hashmap[element]++;}});returnintersectionArray;// Time complexity O(n), Space complexity O(n)}

  View on Codepen:http://codepen.io/kennymkchan/pen/vgwbEb?editors=0012

⬆ back to top

• 2.1 Given a string, reverse each word in the sentence"Welcome to this Javascript Guide!" should be become "emocleW ot siht tpircsavaJ !ediuG"

  varstring="Welcome to this Javascript Guide!";// Output becomes !ediuG tpircsavaJ siht ot emocleWvarreverseEntireSentence=reverseBySeparator(string,"");// Output becomes emocleW ot siht tpircsavaJ !ediuGvarreverseEachWord=reverseBySeparator(reverseEntireSentence," ");functionreverseBySeparator(string,separator){returnstring.split(separator).reverse().join(separator);}

  View on Codepen:http://codepen.io/kennymkchan/pen/VPOONZ?editors=0012

• 2.2 Given two strings, return true if they are anagrams of one another"Mary" is an anagram of "Army"

  varfirstWord="Mary";varsecondWord="Army";isAnagram(firstWord,secondWord);// truefunctionisAnagram(first,second){// For case insensitivity, change both words to lowercase.vara=first.toLowerCase();varb=second.toLowerCase();// Sort the strings, and join the resulting array to a string. Compare the resultsa=a.split("").sort().join("");b=b.split("").sort().join("");returna===b;}

  View on Codepen:http://codepen.io/kennymkchan/pen/NdVVVj?editors=0012

• 2.3 Check if a given string is a palindrome"racecar" is a palindrome. "race car" should also be considered a palindrome. Case sensitivity should be taken into account

  isPalindrome("racecar");// trueisPalindrome("race Car");// truefunctionisPalindrome(word){// Replace all non-letter chars with "" and change to lowercasevarlettersOnly=word.toLowerCase().replace(/\s/g,"");// Compare the string with the reversed version of the stringreturnlettersOnly===lettersOnly.split("").reverse().join("");}

  View on Codepen:http://codepen.io/kennymkchan/pen/xgNNNB?editors=0012

• 2.3 Check if a given string is a isomorphic

  For two strings to be isomorphic, all occurrences of a character in string A can be replaced with another character to get string B. The order of the characters must be preserved. There must be one-to-one mapping for ever char of string A to every char of string B. `paper` and `title` would return true. `egg` and `sad` would return false. `dgg` and `add` would return true. 

  isIsomorphic("egg",'add');// trueisIsomorphic("paper",'title');// trueisIsomorphic("kick",'side');// falsefunctionisIsomorphic(firstString,secondString){// Check if the same lenght. If not, they cannot be isomorphicif(firstString.length!==secondString.length)returnfalsevarletterMap={};for(vari=0;i<firstString.length;i++){varletterA=firstString[i],letterB=secondString[i];// If the letter does not exist, create a map and map it to the value// of the second letterif(letterMap[letterA]===undefined){// If letterB has already been added to letterMap, then we can say: they are not isomorphic.if(secondString.indexOf(letterB)<i){returnfalse;}else{letterMap[letterA]=letterB;}}elseif(letterMap[letterA]!==letterB){// Eles if letterA already exists in the map, but it does not map to// letterB, that means that A is mapping to more than one letter.returnfalse;}}// If after iterating through and conditions are satisfied, return true.// They are isomorphicreturntrue;}

  View on Codepen:http://codepen.io/kennymkchan/pen/mRZgaj?editors=0012

⬆ back to top

• 3.1 Implement enqueue and dequeue using only two stacks

  varinputStack=[];// First stackvaroutputStack=[];// Second stack// For enqueue, just push the item into the first stackfunctionenqueue(stackInput,item){returnstackInput.push(item);}functiondequeue(stackInput,stackOutput){// Reverse the stack such that the first element of the output stack is the// last element of the input stack. After that, pop the top of the output to// get the first element that was ever pushed into the input stackif(stackOutput.length<=0){while(stackInput.length>0){varelementToOutput=stackInput.pop();stackOutput.push(elementToOutput);}}returnstackOutput.pop();}

  View on Codepen:http://codepen.io/kennymkchan/pen/mRYYZV?editors=0012

• 3.2 Create a function that will evaluate if a given expression has balanced parentheses -- Using stacks In this example, we will only consider "{}" as valid parentheses {}{} would be considered balancing. {{{}} is not balanced

  varexpression="{{}}{}{}"varexpressionFalse="{}{{}";isBalanced(expression);// trueisBalanced(expressionFalse);// falseisBalanced("");// truefunctionisBalanced(expression){varcheckString=expression;varstack=[];// If empty, parentheses are technically balancedif(checkString.length<=0)returntrue;for(vari=0;i<checkString.length;i++){if(checkString[i]==='{'){stack.push(checkString[i]);}elseif(checkString[i]==='}'){// Pop on an empty array is undefinedif(stack.length>0){stack.pop();}else{returnfalse;}}}// If the array is not empty, it is not balancedif(stack.pop())returnfalse;returntrue;}

  View on Codepen:http://codepen.io/kennymkchan/pen/egaawj?editors=0012

⬆ back to top

• 4.1 Write a recursive function that returns the binary string of a given decimal number Given 4 as the decimal input, the function should return 100

  decimalToBinary(3);// 11decimalToBinary(8);// 1000decimalToBinary(1000);// 1111101000functiondecimalToBinary(digit){if(digit>=1){// If digit is not divisible by 2 then recursively return proceeding// binary of the digit minus 1, 1 is added for the leftover 1 digitif(digit%2){returndecimalToBinary((digit-1)/2)+1;}else{// Recursively return proceeding binary digitsreturndecimalToBinary(digit/2)+0;}}else{// Exit conditionreturn'';}}

  View on Codepen:http://codepen.io/kennymkchan/pen/OWYYKb?editors=0012

• 4.2 Write a recursive function that performs a binary search

  functionrecursiveBinarySearch(array,value,leftPosition,rightPosition){// Value DNEif(leftPosition>rightPosition)return-1;varmiddlePivot=Math.floor((leftPosition+rightPosition)/2);if(array[middlePivot]===value){returnmiddlePivot;}elseif(array[middlePivot]>value){returnrecursiveBinarySearch(array,value,leftPosition,middlePivot-1);}else{returnrecursiveBinarySearch(array,value,middlePivot+1,rightPosition);}}

  View on Codepen:http://codepen.io/kennymkchan/pen/ygWWmK?editors=0012

⬆ back to top

• 5.1 Given an integer, determine if it is a power of 2. If so, return that number, else return -1. (0 is not a power of two)

  isPowerOfTwo(4);// trueisPowerOfTwo(64);// trueisPowerOfTwo(1);// trueisPowerOfTwo(0);// falseisPowerOfTwo(-1);// false// For the non-zero case:functionisPowerOfTwo(number){// `&` uses the bitwise n.// In the case of number = 4; the expression would be identical to:// `return (4 & 3 === 0)`// In bitwise, 4 is 100, and 3 is 011. Using &, if two values at the same// spot is 1, then result is 1, else 0. In this case, it would return 000,// and thus, 4 satisfies are expression.// In turn, if the expression is `return (5 & 4 === 0)`, it would be false// since it returns 101 & 100 = 100 (NOT === 0)returnnumber&(number-1)===0;}// For zero-case:functionisPowerOfTwoZeroCase(number){return(number!==0)&&((number&(number-1))===0);}

  View on Codepen:http://codepen.io/kennymkchan/pen/qRGGeG?editors=0012

⬆ back to top

• 6.1 Explain what is hoisting in Javascript

  Hoisting is the concept in which Javascript, by default, moves all declarations to the top of the current scope. As such, a variable can be used before it has been declared. Note that Javascript only hoists declarations and not initializations 

• 6.2 Describe the functionality of the use strict; directive

  the `use strict` directive defines that the Javascript should be executed in `strict mode`. One major benefit that strict mode provides is that it prevents developers from using undeclared variables. Older versions of javascript would ignore this directive declaration 

  // Example of strict mode"use strict";catchThemAll();functioncatchThemAll(){x=3.14;// Error will be thrownreturnx*x;}

• 6.3 Explain event bubbling and how one may prevent it

  Event bubbling is the concept in which an event triggers at the deepest possible element, and triggers on parent elements in nesting order. As a result, when clicking on a child element one may exhibit the handler of the parent activating. One way to prevent event bubbling is using `event.stopPropagation()` or `event.cancelBubble` on IE < 9 

• 6.4 What is the difference between == and === in JS?

  `===` is known as a strict operator. The key difference between `==` and `===` is that the strict operator matches for both value and type, as opposed to just the value. 

  // Example of comparators0==false;// true0===false;// false2=='2';// true2==='2';// false

• 6.5 What is the difference between null and undefined

  In Javascript, null is an assignment value, and can be assigned to a variable representing that it has no value. Undefined, on the other hand, represents that a variable has been declared but there is no value associated with it 

• 6.6 How does prototypal inheritance differ from classical inheritance

  In classical inheritance, classes are immutable, may or may not support multiple inheritance, and may contain interfaces, final classes, and abstract classes. In contrast, prototypes are much more flexible in the sense that they may be mutable or immutable. The object may inherit from multiple prototypes, and only contains objects. 

⬆ back to top