After moving past Basic JavaScript and Object-Oriented Programming (OOP) parts in FCC, the fun part is Basic Algorithm. I have completed half of it so far. I am going to review each completed challenges. They should be brief and easy to understand.

Reverse a String

• Reverse the provided string
• May need to turn the string into an array
• Result must be a string

function reverseString(str) {
  var arraized = str.split("");
  var backward = arraized.reverse("");
  var nuStr = backward.join("");
  return nuStr;
}

reverseString("hello");
// 'olleh'

What I did was following the suggestions in FreeCodeCamp. Using .split("") to turn str into an array, and then no brainer, using .reverse() to turn array backward. Finally I used .join("") to turn array back into a string again. It is important to assign each function with a variable. A shortcut could be done like this below:

function reverseString(str) {
  var reversed = str.split("").reverse("").join("");
  return reversed;
}

reverseString("hello");
// 'olleh'

Factorialize a Number

• Return the factorial of provided integer
• If the integer is represented with the letter n, a factorial is the product of all positive integers less than or equal to n
• Factorials are often represented with the shorthand notation “n!”

For example: 5! = 1 * 2 * 3 * 4 * 5 = 120

Challenge

function factorialize(num) {
  return num;
}

factorialize(5);

It might be challenging at first, if out of touch with math. The for loop is a great way of pulling numbers magically out of a number or num by using indexes i. We want to multiply all indexes i from top down to one. So why not do the for loop backward? We will need to assign a variable to get final result. product came to my mind since it is mathematic. Let’s set var product = 1. The product will multiply i and becomes the new product every times the loop goes through until it hits the bottom.

function factorialize(num) {
  var product = 1;
  for (var i = num; i > 0; i--) {
    product *= i;
  }
  return product;
}

factorialize(5);
// 120

Palindrome

• Return true if the given string is a palindrome. Otherwise, return false
• A palindrome is a word or sentence that’s spelled the same way both forward and backward, ignoring punctuation, case, and spacing
• We’ll also pass strings with special symbols, such as “2A33a2”, “2A3 3a2”, and “2_A33#A2”
• Given hints: use .replace() and .toLowerCase()

function palindrome(str) {
  // Good luck!
  return true;
}

palindrome("_eye");

Woo… okay in this problem, I had spent awhile. Need to ignore special characters, and lowcase all letters. The first part was bit hard, but the second part is easy. I kept on testing how to use .replace() with console.log() to make sure it functions correctly.

In .replace() I used regular expression (regex) to either remove value, and or replace value with a new value. .replace() has two parameters, could use only one parameter if wanted to. .replace(searchValue, newValue), so I am going to apply regex in searchValue. Regex is little confusing and weird, it is recommended to search through stackoverflow to match your desire. [a-zA-Z0-9_*]) includes all alphanumeric characters, underscore, and asterisk, but there is another way and is better which lays out like this /[\W_]/. Captialized W means any non-word character and important to add an underscore to it too. [] to include everything inside, /.../ for start and ends, and \ to escape. Don’t forget to apply a new variable, I am going to name the variable clean. Test with .console.log(clean).

function palindrome(str) {
  var clean = str.replace(/[\W_]/g, '');
  console.log(clean);
}

palindrome("_eye");
// eye

Notice g at the end of the regex, it stands for global search, or search for ALL occurences. I added space for .replace() second parameter as a newValue. Special characters are now out of the way. Next step is pretty easy. Just type in .toLowerCase(). String will turn into lowcases automatically, apply new varaible to it, I am calling it strLo.

The third step is going to require some brain work. How am I going to have the computer to return true if the word is palindrome? Conditional statement is encouraged since it is a boolean. Palindrome means the first half of the word has to match with the second half of word in mirror manner looking at each letters. I am going to use the for loop to scrutinize each letter, and word only needs to be half so I will divide the .length by half.

function palindrome(str) {
  var clean = str.replace(/[\W_]/g, '');
  var strLo = clean.toLowerCase();
  for (var i = 0; i < strLo.length/2; i++) {
  }
}

palindrome("_eye");

How do I determine if letter at first index matches with the letter at last index and so on looping inwardly toward fowardly or backwardly to the half of the word? The for loop is set. Since palindrome focues on element, I am going to put it like this strLo[i]. Looping through string backward has to use strLo.length -1 - i but adding [ ] to fetch elements instead of indexes. It will be easier to say if it doesn’t match, then return false first.

function palindrome(str) {
  var clean = str.replace(/[\W_]/g, '');
  var strLo = clean.toLowerCase();
  for (var i = 0; i < strLo.length/2; i++) {
    if (strLo[i] !== strLo[strLo.length-1-i]) {
      return false;
    }
  } return true;
}

palindrome("_eye");
// eye

It is important to set return true outside of the loop so that it inspects all letters. We will continue with Basic Algorithm Part 2, 3, and 4 in different posts.