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

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

// The output of the function should be 8var 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];  }
  // 以高斯求和公式計(jì)算理論上的數(shù)組和  // 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 Implementationvar array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];
Array.from(new Set(array)); // [1, 2, 3, 5, 9, 8]

// ES5 Implementationvar 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;}

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) {
  // 如果數(shù)組僅有一個(gè)元素，則直接返回 -1
  if (array.length <= 1) return -1;
  // current_min 指向當(dāng)前的最小值
  var current_min = array[0];  var current_max_difference = 0;
  // 遍歷整個(gè)數(shù)組以求取當(dāng)前最大差值，如果發(fā)現(xiàn)某個(gè)最大差值，則將新的值覆蓋 current_max_difference  // 同時(shí)也會(huì)追蹤當(dāng)前數(shù)組中的最小值，從而保證 `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 emocleWvar reverseEntireSentence = reverseBySeparator(string, "");
// Output becomes emocleW ot siht tpircsavaJ !ediuGvar reverseEachWord = reverseBySeparator(reverseEntireSentence, " ");
function reverseBySeparator(string, separator) {  return string.split(separator).reverse().join(separator);}

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;}

isPalindrome("racecar"); // trueisPalindrome("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 stackvar outputStack = []; // Second stack
// For enqueue, just push the item into the first stackfunction 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); // trueisBalanced(expressionFalse); // falseisBalanced(""); // 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); // 11decimalToBinary(8); // 1000decimalToBinary(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 '';  }}

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); // trueisPowerOfTwo(64); // trueisPowerOfTwo(1); // trueisPowerOfTwo(0); // falseisPowerOfTwo(-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);}