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Common interview questions completed in JavaScript

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[前端(javascript) 所属分类 前端(javascript) | 发布者 店小二04 | 时间 2017 | 作者 红领巾 ] 0人收藏点击收藏
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 Array Strings Stacks and Queues Recursion Numbers Javascript Specific To Be Continued Array

1.1Given 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 };

1.2Being 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) }

1.3Removing 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; }

1.4Given 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; } 1.5Given 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

2.1Given 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); }

2.2Given 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); }

2.3Check 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

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