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Interview Algorithm Questions in Javascript() {...}

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

Table of Contents

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

Array

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

  var unsortedArray = [-10, 7, 29, 30, 5, -10, -70];
  
  computeProduct(unsortedArray); // 21000
  
  function sortIntegers(a, b) {
    return a - b;
  }
  
  // Greatest product is either (min1 * min2 * max1 || max1 * max2 * max3)
  function computeProduct(unsorted) {
    var sortedArray = unsorted.sort(sortIntegers),
      product1 = 1,
      product2 = 1,
      array_n_element = sortedArray.length - 1;
  
    // Get the product of three largest integers in sorted array
    for (var x = 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) return product1;
  
    return product2;
  }

  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 8
  var arrayOfIntegers = [2, 5, 1, 4, 9, 6, 3, 7];
  var upperBound = 9;
  var lowerBound = 1;
  
  findMissingNumber(arrayOfIntegers, upperBound, lowerBound); // 8
  
  function findMissingNumber(arrayOfIntegers, upperBound, lowerBound) {
    // Iterate through array to find the sum of the numbers
    var sumOfIntegers = 0;
    for (var i = 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 bound
  
    upperLimitSum = (upperBound * (upperBound + 1)) / 2;
    lowerLimitSum = (lowerBound * (lowerBound - 1)) / 2;
  
    theoreticalSum = upperLimitSum - lowerLimitSum;
  
    return theoreticalSum - 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 Implementation
  var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];
  
  Array.from(new Set(array)); // [1, 2, 3, 5, 9, 8]
  
  // ES5 Implementation
  var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];
  
  uniqueArray(array); // [1, 2, 3, 5, 9, 8]
  
  function uniqueArray(array) {
    var hashmap = {};
    var unique = [];
  
    for(var i = 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]);
      }
    }
  
    return unique;
  }

  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

  var array = [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);
  
  function findLargestDifference(array) {
    // If there is only one element, there is no difference
    if (array.length <= 1) return -1;
  
    // currentMin will keep track of the current lowest
    var currentMin = array[0];
    var currentMaxDifference = 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 (var i = 1; i < array.length; i++) {
      if (array[i] > currentMin && (array[i] - currentMin > currentMaxDifference)) {
        currentMaxDifference = array[i] - currentMin;
      } else if (array[i] <= currentMin) {
        currentMin = array[i];
      }
    }
  
    // If negative or 0, there is no largest difference
    if (currentMaxDifference <= 0) return -1;
  
    return currentMaxDifference;
  }

  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)

  var firstArray = [2, 2, 4, 1];
  var secondArray = [0, 0, 0, 2];
  var thirdArray = [-2, -2, -3, 2];
  
  productExceptSelf(firstArray); // [8, 8, 4, 16]
  productExceptSelf(secondArray); // [0, 0, 0, 0]
  productExceptSelf(thirdArray); // [12, 12, 8, -12]
  
  function productExceptSelf(numArray) {
  	var product = 1;
  	var size = numArray.length;
  	var output = [];
  
    // 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 + 1
  	for (var x = 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)
  	var product = 1;
  	for (var i = size - 1; i > -1; i--) {
  		output[i] = output[i] * product;
  		product = product * numArray[i];
  	}
  
    return output;
  }

  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!

  var firstArray = [2, 2, 4, 1];
  var secondArray = [1, 2, 0, 2];
  
  intersection(firstArray, secondArray); // [2, 1]
  
  function intersection(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.
  
    var hashmap = {};
    var intersectionArray = [];
  
    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 added
    secondArray.forEach(function(element) {
      if (hashmap[element] === 1) {
        intersectionArray.push(element);
        hashmap[element]++;
      }
    });
  
    return intersectionArray;
  
    // Time complexity O(n), Space complexity O(n)
  }

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

⬆ back to top

Strings

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

  var string = "Welcome to this Javascript Guide!";
  
  // Output becomes !ediuG tpircsavaJ siht ot emocleW
  var reverseEntireSentence = reverseBySeparator(string, "");
  
  // Output becomes emocleW ot siht tpircsavaJ !ediuG
  var reverseEachWord = reverseBySeparator(reverseEntireSentence, " ");
  
  function reverseBySeparator(string, separator) {
    return string.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"

  var firstWord = "Mary";
  var secondWord = "Army";
  
  isAnagram(firstWord, secondWord); // true
  
  function isAnagram(first, second) {
    // For case insensitivity, change both words to lowercase.
    var a = first.toLowerCase();
    var b = second.toLowerCase();
  
    // Sort the strings, and join the resulting array to a string. Compare the results
    a = a.split("").sort().join("");
    b = b.split("").sort().join("");
  
    return a === 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"); // true
  isPalindrome("race Car"); // true
  
  function isPalindrome(word) {
    // Replace all non-letter chars with "" and change to lowercase
    var lettersOnly = word.toLowerCase().replace(/\s/g, "");
  
    // Compare the string with the reversed version of the string
    return lettersOnly === 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'); // true
  isIsomorphic("paper", 'title'); // true
  isIsomorphic("kick", 'side'); // false
  
  function isIsomorphic(firstString, secondString) {
  
    // Check if the same lenght. If not, they cannot be isomorphic
    if (firstString.length !== secondString.length) return false
  
    var letterMap = {};
  
    for (var i = 0; i < firstString.length; i++) {
      var letterA = firstString[i],
          letterB = secondString[i];
  
      // If the letter does not exist, create a map and map it to the value
      // of the second letter
      if (letterMap[letterA] === undefined) {
        // If letterB has already been added to letterMap, then we can say: they are not isomorphic.
        if(secondString.indexOf(letterB) < i){
            return false;
        } else {
            letterMap[letterA] = letterB;            
        }
      } else if (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.
        return false;
      }
    }
    // If after iterating through and conditions are satisfied, return true.
    // They are isomorphic
    return true;
  }

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

⬆ back to top

Stacks and Queues

• 3.1 Implement enqueue and dequeue using only two stacks

  var inputStack = []; // First stack
  var outputStack = []; // Second stack
  
  // For enqueue, just push the item into the first stack
  function enqueue(stackInput, item) {
    return stackInput.push(item);
  }
  
  function dequeue(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 stack
    if (stackOutput.length <= 0) {
      while(stackInput.length > 0) {
        var elementToOutput = stackInput.pop();
        stackOutput.push(elementToOutput);
      }
    }
  
    return stackOutput.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

  var expression = "{{}}{}{}"