// 会问字符串判断某个字符串是否为回文字符串，譬如racecar与race car都是回文字符串：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("");
}

// 乱序同字母字符串给定两个字符串，判断是否颠倒字母而成的字符串，譬如Mary与Army就是同字母而顺序颠倒：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;
}

// 字符串颠倒字符串给定某个字符串，要求将其中单词倒转之后然后输出，譬如"Welcome to this Javascript Guide!" 应该输出为 "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);
}

// 数组交集给定两个数组，要求求出两个数组的交集，注意，交集中的元素应该是唯一的。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 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;
}