


Detailed explanation of common algorithm questions in JavaScript interviews
JavaScript Specification
Explain the variable promotion in JavaScript
The so-called promotion, as the name suggests, means that JavaScript will promote all declarations to the top of the current scope. This means that we can use a variable before it is declared, but although JavaScript will raise the declaration to the top, it will not perform the actual initialization process.
Explain the role of use strict;
##use strict;As the name suggests, JavaScript will be executed in the so-called strict mode. One of the main The advantage is that it forces developers to avoid using undeclared variables. For older versions of browsers or execution engines, this instruction will be automatically ignored.
// Example of strict mode "use strict"; catchThemAll(); function catchThemAll() { x = 3.14; // Error will be thrown return x * x; }
event.stopPropagation() or below IE 9, use
event.cancelBubble. What is the difference between
=== is the so-called strict comparison. The key difference is
===Will compare types and values at the same time, instead of only comparing values.
// Example of comparators 0 == false; // true 0 === false; // false 2 == '2'; // true 2 === '2'; // false
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 };
O(n) complexity to find the missing number.
// 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]; } // 以高斯求和公式计算理论上的数组和 // 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) }
// 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; }
[7, 8, 4, 9, 9, 15, 3, 1, 10]The calculated value of this array is 11(15 – 4) instead of 14(15 – 1), because 15 The subscript is less than 1.
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) { // 如果数组仅有一个元素,则直接返回 -1 if (array.length <= 1) return -1; // current_min 指向当前的最小值 var current_min = array[0]; var current_max_difference = 0; // 遍历整个数组以求取当前最大差值,如果发现某个最大差值,则将新的值覆盖 current_max_difference // 同时也会追踪当前数组中的最小值,从而保证 `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; }
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; }
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) }
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); }
Mary and
Army is the same letter but in reverse order:
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; }
racecar and
race car are all palindrome strings:
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(""); }
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();
}
Copy after login
Determine whether the braces are closed Create a function to determine whether the braces in a given expression are closed: 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(); }
var expression = "{{}}{}{}" var expressionFalse = "{}{{}"; isBalanced(expression); // true isBalanced(expressionFalse); // false isBalanced(""); // true function isBalanced(expression) { var checkString = expression; var stack = []; // If empty, parentheses are technically balanced if (checkString.length <= 0) return true; for (var i = 0; i < checkString.length; i++) { if(checkString[i] === '{') { stack.push(checkString[i]); } else if (checkString[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; }
decimalToBinary(3); // 11 decimalToBinary(8); // 1000 decimalToBinary(1000); // 1111101000 function decimalToBinary(digit) { if(digit >= 1) { // If digit is not pisible 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 ''; } }
function recursiveBinarySearch(array, value, leftPosition, rightPosition) { // Value DNE if (leftPosition > rightPosition) return -1; var middlePivot = Math.floor((leftPosition + rightPosition) / 2); if (array[middlePivot] === value) { return middlePivot; } else if (array[middlePivot] > value) { return recursiveBinarySearch(array, value, leftPosition, middlePivot - 1); } else { return recursiveBinarySearch(array, value, middlePivot + 1, rightPosition); } }
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);
}
Copy after login The above is JavaScript Detailed explanations of common algorithm questions in interviews. For more related content, please pay attention to the PHP Chinese website (www.php.cn)!
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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