/interview-questions-in-javascript

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

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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 unsorted_array = [-10, 7, 29, 30, 5, -10, -70];
    
    computeProduct(unsorted_array); // 21000
    
    function sortIntegers(a, b) {
      return a - b;
    }
    
    // greatest product is either (min1 * min2 * max1 || max1 * max2 * max3)
    function computeProduct(unsorted) {
      var sorted_array = unsorted.sort(sortIntegers),
        product1 = 1,
        product2 = 1,
        array_n_element = sorted_array.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 * sorted_array[x];
      }
      product2 = sorted_array[0] * sorted_array[1] * sorted_array[array_n_element];
    
      if (product1 > product2) return product1;
    
      return product2
    };

  • 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 array_of_integers = [2, 5, 1, 4, 9, 6, 3, 7];
    var upper_bound = 9;
    var lower_bound = 1;
    
    findMissingNumber(array_of_integers, upper_bound, lower_bound); //8
    
    function findMissingNumber(array_of_integers, upper_bound, lower_bound) {
    
      // Iterate through array to find the sum of the numbers
      var sum_of_integers = 0;
      for (var i = 0; i < array_of_integers.length; i++) {
        sum_of_integers += array_of_integers[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
    
      upper_limit_sum = (upper_bound * (upper_bound + 1)) / 2;
      lower_limit_sum = (lower_bound * (lower_bound - 1)) / 2;
    
      theoretical_sum = upper_limit_sum - lower_limit_sum;
    
      //
      return (theoretical_sum - sum_of_integers)
    }

  • 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 null (unique), it is evaluated as false.
        if(!hashmap.hasOwnProperty([array[i]])) {
          hashmap[array[i]] = 1;
          unique.push(array[i]);
        }
      }
      return unique;
    }

  • 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;
    
      // current_min will keep track of the current lowest
    
      var current_min = array[0];
      var current_max_difference = 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] > current_min && (array[i] - current_min > current_max_difference)) {
          current_max_difference = array[i] - current_min;
        } else if (array[i] <= current_min) {
          current_min = array[i];
        }
      }
    
      // If negative or 0, there is no largest difference
      if (current_max_difference <= 0) return -1;
    
      return current_max_difference;
    }

  • 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;
    }

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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 reverse_entire_sentence = reverseBySeparator(string, "");
    
    // Output becomes emocleW ot siht tpircsavaJ !ediuG
    var reverse_each_word = reverseBySeparator(reverse_entire_sentence, " ");
    
    function reverseBySeparator(string, separator) {
      return string.split(separator).reverse().join(separator);
    }

  • 2.2 Given two strings, return true if they are anagrams of one another "Mary" is an anagram of "Army"
    var first_word = "Mary";
    var second_word = "Army";
    
    isAnagram(first_word, second_word); //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);
    }

  • 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 letters_only = word.toLowerCase().replace(/\s/g, "");
    
      // Compare the string with the reversed version of the string
      return (letters_only === letters_only.split("").reverse().join(""));
    }

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Stacks and Queues

  • 3.1 Implement enqueue and dequeue using only two stacks
    var input_stack = []; // first stack
    var output_stack = []; // second stack
    
    // For enqueue, just push the item into the first stack
    function enqueue(stack_input, item) {
      return stack_input.push(item);
    }
    
    function dequeue(stack_input, stack_output) {
      // 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 (stack_output.length <= 0) {
        while(stack_input.length > 0) {
          var element_to_output = stack_input.pop();
          stack_output.push(element_to_output);
        }
        return stack_output.pop();
      }
    }

  • 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 = "{{}}{}{}"
    var expression_false = "{}{{}";
    
    isBalanced(expression); // true
    isBalanced(expression_false); //false
    isBalanced(""); // true
    
    function isBalanced(expression) {
      var check_string = expression;
      var stack = [];
    
      // If empty, parentheses are technically balanced
      if (check_string.length <= 0) return true
    
      for (var i = 0; i < check_string.length; i++) {
        if(check_string[i] === '{') {
          stack.push(check_string[i]);
        } else if (check_string[i] === '}') {
          // pop on an empty array is undefined
          if (stack.length > 0) {
            stack.pop();
          } else {
            return false;
          }
        }
      }
    
      // If the array is not empty, it is not balanced
      if (stack.pop()) return false;
      return true;
    }

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Recursion

  • 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) // 11
    decimalToBinary(8) //1000
    decimalToBinary(1000) //1111101000
    
    function decimalToBinary(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 digit
        if (digit % 2) {
          return decimalToBinary((digit - 1) / 2) + 1;
        } else {
          // recursively return proceeding binary digits
          return decimalToBinary(digit / 2) + 0;
        }
      } else {
        // exit condition
        return '';
      }
    }

  • 4.2 Write a recursive function that performs a binary search

    function recursiveBinarySearch(array, value, leftposition, rightposition) {
    
      // Value DNE
      if (left > right) return -1;
    
      var middle_pivot = Math.floor((leftposition + rightposition) / 2);
      if (array[middle_pivot] === value) {
        return middle_pivot;
      } else if (array[middle_pivot] > key) {
        return recursiveBinarySearch(array, value, leftposition, middle_pivot - 1);
      } else {
        return recursiveBinarySearch(array, value, middle_pivot + 1, rightposition);
      }
    }

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Numbers

  • 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); // true
    isPowerOfTwo(64); // true
    isPowerOfTwo(1); // true
    isPowerOfTwo(0); // false
    isPowerOfTwo(-1); // false
    
    // For the non-zero case:
    function isPowerOfTwo(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)
      return (number & (number - 1) === 0)
    };
    
    // For zero-case:
    function isPowerOfTwoZeroCase(number) {
      return (number != 0) && ((number & (number - 1)) === 0);
    };
    

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Javascript

  • 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();
    function catchThemAll() {
      x = 3.14; // error will be thrown
      return x * 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 comparators
    0 == false; // true
    0 === false; // false
    
    2 == '2' // true
    2 === '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.
    

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