a/A Practice Problem: Third Greatest pt.1

This is the first part of this question/answer series because the answer I came up with was very easy only because I was able to use methods to help me solve it quicker than I think they (a/A) wants. They didn't explicitly say it in this problem, but in previous challenges it said to refrain from using methods like .reverse, which I used here. SO because of that I'm going to show you my short code and talk about .sort, and then another day I'll post a longer more confusing answer that shows your know how to harness the power of loops and variables with the wrath of Aphrodites. Or something. Idfk.

The problem:
# Write a method that takes an array of numbers in. Your method should
# return the third greatest number in the array. You may assume that
# the array has at least three numbers in it.

My Solution:

SO. Let's talk about .sort! Here's my wacky explanation, but please avail yourself to some of the links at the end of this post.

.sort is an array method. It'll work beautifully on arrays containing words and will alphabetize it the way you would. However, it will not do that with numbers. the computer would consider 100 before 2 because 1 is less than 2. So when it comes to numbers, we use this comparison function that you see in there.   return a-b; will sort your numbers from least to greatest, while return b-a will sort your numbers from greatest to least. To avoid saying the wrong thing and causing someone out there (even if it's me at a later date) a massive headache because what I said was gobbledegook, this is as far as I'll go.

Next we want to reverse the array, because when we do that we can easily pull out the third number which by then should be the third greatest!

Make sure to check out these links:
http://www.w3schools.com/jsref/jsref_sort.asp
http://www.javascriptkit.com/javatutors/arraysort.shtml


How would you solve this?!

a/A Practice Problem: Is Power of 2?

Truth be told: I definitely looked for help on this problem. And that's OK!! (At least I think it is.) Part of the whole teaching yourself how to code (or anything really!) is learning from others. That means looking at solutions and understanding how that person got there. Not copy-and-pasting their code into the answer box and saying "DONE!", but writing it out yourself, a hundred times if you have to. I looked at a/A's answer to this (in Ruby) and was able to translate it into a working JS program.

Here's the problem:

# Write a method that takes in a number and returns true if it is a
# power of 2. Otherwise, return false.
#
# You may want to use the `%` modulo operation. `5 % 2` returns the
# remainder when dividing 5 by 2; therefore, `5 % 2 == 1`. In the case
# of `6 % 2`, since 2 evenly divides 6 with no remainder, `6 % 2 == 0`.

My Solution

That first if statement, tells us what to reject right away -- if the number passed in isn't divisible by 2, then we know that two things: that it's definitely an odd number, and that it cannot be a power of 2.

The while loop portion was something I picked up from Codewars. Let me try to dissect and explain: The (!num) portion only, excluding the first exclamation mark, is basically saying num is false. The exclamation mark acts like negation (I think-- please feel free to correct me!). So the outer exclamation mark (!(!num)) is a double negative and is basically saying while num is true.

A while loop still needs the same components as a for-loop: the initiation (num), the condition(while num is true) and termination (num is false).

• The first if within the loop would break us out of the loop if and when num is equal to 1 (which would be our 2 to the 0 power case. If we get to this case, that means it is a power of 2 and we should return true, which the code states!
• The else if scenario checks if the remainder of num / 2 is equal to 0. If it is, divide it by 2 again.
• The last else case will handle the other cases and if the number hasn't passed the first two cases, it's not a power of two, so we should return false.

What do you think? How would you solve this problem?!

a/A Practice Problems: Two Sum

It's been a minute. A New York minute, if you will. Cheeeese.  Today we're back with another App Academy practice problem.  Here it is:

# Write a method that takes an array of numbers. If a pair of numbers
# in the array sums to zero, return the positions of those two numbers.
# If no pair of numbers sums to zero, return `nil`.

A couple things beforehand: nil in javascript would be null, so I return null. I came to a working answer through trial and error, but I'm pretty sure there is a much better way to solve this. I tried a couple other ways, but nothing worked so any advice out there would be more than welcome!!

I started the first loop off at 0 and the second loop at 1 because there's no need to add the first index to itself.  Just remember the way nested loops work : the second loop will finish an entire iteration (from 1 to less than nums.length before the first loop moves on to i = 1.

That last line there was my test case, so don't mind that! :) Usually when I solve problems, I write everything out. I'm working on being more concise with my code, which is why I didn't declare an array and then push the values to it (even though I definitely tried this a billion times). But I gotta say, I kinda liked just returning i and j in an array.

That's all for today.
Happy coding! :)

Codewars as a Training Tool

In this post I wanted to stick up for CodeWars and help explain to the novice coder why this site is so fucking awesome.

First, a little backstory. My boyfriend and I are both just starting out in this whole programming world. He was the one who told me about it first and when I first went to the site, I could barely get past the mandatory (but simple) coding challenges you have to get right in order to gain access to the full site.

Fast forward to today and I am on the site at least once a day training. It has become the best learning tool for me thus far. It challenges me every day, reminding me why I like coding. My boyfriend on the other, while he can build out an entire rails app, absolutely hates the site. I think it's because he hasn't given it enough of a chance (which is true), but he says his learning style is different (which also is true). Either way, I happen to think that anyone who uses the system the right way will soon too become a true believer. So, herein this post I present to you my methodology to using the site so that even you can conquer and earn your honor.

1. Select your level.
On the left hand side you will find everything you need to make sure you are doing problems within your range. The levels are from easiest, 8, to hardest, 1. Truth be told, it's best to have gone through the basics so that you can attempt the problems, but even if you are going along as you learn you can customize which problems to do (see below). Start off with 8, and move at your own pace. Some of the problems within are going to be a lot more difficult than some of the others. But give yourself some patience and work through them, absorbing everything you can.

2. Select what areas you want to train.
Once you've selected which level you want to compete in, you can then choose problems that have been tagged with certain areas of concentration. On the left hand side, under Tags, you can select Arrays, Numbers, Bugs, Fundamentals, etc. When I was starting out I found that the best thing that helped me was completing level 8 "bug" problems. They re-familiarized me with syntax and some other basic concepts that helped me to solve some other problems without the worry of that. If you're working through the problems while trying to master one area of a certain language, this will be one of the best ways to use CodeWars to do just that.

3. Don't be ashamed to forfeit honor.
No one likes to lose, obviously. But sometimes, you need to suck it up. There are problems that for some reason will not pass the tests, but once you click submit will actually work. Save for these cases, if you can't figure out why something won't work, it might be time to just unlock the solutions to that problem. It could be something simple you overlooked, but if it's something you can't figure out on your own, unlock the solutions and see how other people solved it. Think of it as learning, not cheating if you happen to be so honorable.

4. Read other peoples' answers.
Whether I submitted my answer or forfeited honor, I always look at other peoples' answers. There will be things in there that look like straight up gibberish. Even when I finished my Javascript book, there were things I had never seen before. Methods that would produce the right code with a fraction of the lines. It was insane. But it's also insanely insightful. Whenever there's something I don't understand I look it up, and you can get the general idea of the way something works once you see it.

Going through the site without any sort of order can be pretty depressing as a noob. You can easily get that "omg-I'm-never-going-to-understand-this-and-I-better-get-back-to-my-dead-end-job" feeling, but I swear with time, patience and a little order CodeWars can and will open all sorts of doors.

a/A Practice Problems: Nearby AZ

Practice Problem 8 of 21 and the first medium! It's exciting stuff happening 'round these parts. Without further adieu...

Write a method that takes a string in and returns true if the letter "z" appears within three letters **after** an "a". You may assume that the string contains only lowercase letters.

This is a little trickier, but still employs the same methods and logic that we've already seen. A couple loops and if statements.

1.  Declare our function with one parameter, string.

function nearbyAZ (string) {
}

2.  I wanted to declare a variable, index1, that would hold the place of the position in which the letter "a" appears. This will be helpful later when determining if there's a "z" within the next three letters.

function nearbyAZ (string) {

   var index1 = 0;

}

3. We want to loop through all the letters in order to see if there's an "a".  The code block within the loop will be an if/else statement. If there's an "a", store it's index number in the variable, index1, and break out of the loop. Else, I want to increase index1 by one with each iteration of the loop. This way we will store the correct position number.

function nearbyAZ (string) {

   var index1 = 0;

   for (var i = 0; i < string.length; i++) {

      if (string[i] === "a") {
         index1 = i;

         break;

      }

      else {

         index1++;

      }

   }

}

4.  After the previous loop, I want another variable, index2, which will store the value of index1 + 1. This will be where our loop will start, and it should end at the third letter after the first a: index1 + 3. 

function nearbyAZ (string) {

   var index1 = 0;

   for (var i = 0; i < string.length; i++) {

      if (string[i] === "a") {

         break;

      }

      else {

         index1++;

      }

   }

   var index2 = index1 + 1;

   for (var j = index2; j <= index1 + 3; j++) {

   }

}

5. Start the loop at index2 because we are only concerned with the first three letters after the letter "a". Likewise, we'll want to stop the loop once it reaches that third letter. An if statement is the last part required to return true or false if there's a "z". 

function nearbyAZ (string) {

   var index1 = 0;

   for (var i = 0; i < string.length; i++) {

      if (string[i] === "a") {

         break;

      }

      else {

         index1++;

      }

   }

   var index2 = index1 + 1;

   for (var j = index2; j <= index1 + 3; j++) {

      if (string[j] === "z") {

         return true;

      }

   }

   return false;

}

Check out past problems here:

• Reverse a String
• Factorial
• Longest Word
• Sum Num
• Time Conversion 
• Count Vowels 
• Palindrome

a/A Practice Problems: Palindrome

We have made it to the last of the "easy" problems. Rejoice!  Here we go...

Write a method that takes a string and returns true if it is a palindrome. A palindrome is a string that is the same whether written backward or forward. Assume that there are no spaces; only lowercase letters will be given.

 We know that we need a function that will take in a single parameter. We also want to declare an empty string that we can use to store the new word we will form by running a loop adding each letter from the string in reverse order. Lastly, we need a check to see if the original and the new string are the same, a palindrome, and return either true or false. Easy peasy.

1. Construct your function.

function palindrome(string) {
}

2. I declared a string that would hold the index position of the last letter of the string so as to use the variable in the loop instead. This isn't necessary. But you will want to declare a variable newString, which will be the original string in reversed order.

function palindrome(string) {
   var lastIndex = string.length - 1;
   var newString = "";
}

3. The loop starts with the index of the last letter, lastIndex, which we defined previously. It will continue so long as i is greater than or equal to 0, and will decrease by one each time. The code block inside the loop will concatenate each letter and form the new string in newString.

function palindrome(string) {
   var lastIndex = string.length - 1;
   var newString = "";
   for (var i = lastIndex; i >= 0; i--) {
      newString += string[i];
   }
}

 4. Lastly, an if statement that says if the original string and newString are the same will return true, else it will return false.

function palindrome(string) {
   var lastIndex = string.length - 1;
   var newString = "";
   for (var i = lastIndex; i >= 0; i--) {
      newString += string[i];
   }
   if (string === newString) {
      return true;
   } else {
      return false;
   }
}

Next up the mediums! **tingles**

Check out past problems here:

• Reverse a String
• Factorial
• Longest Word
• Sum Num
• Time Conversion 
• Count Vowels 
• Nearby AZ

a/A Practice Problem: Factorial

Challenge Description: "Write a method that takes an integer 'n' in; it should return
n*(n-1)*(n-2)*...*2*1. Assume n >= 0. As a special case, factorial(0) == 1"

Quick refresher on factorials: say n = 5, a factorial would be the product of 5*4*3*2*1 or 120.

Step 1: Write out your plan of action:

                         >> make function with one parameter
                         >> declare variable and set equal to 1
                         >> for loop to take each number less than n and multiply to variable
                         >> if statement for if n = 0
                         >> return product

Step 2: Declare function. Same rules for naming variables.

                                                  function factorial(n) {
                                                  }

Step 3: Set variable and set equal to 1.

                                                   function factorial(n) {
                                                      var product = 1;
                                                   }

Step 4: For loop with descending counter: Set i equal to n because you want to make sure it's the first number to start with and is included. The loop should stop once it hits one, or when i is greater than 0. Decrement by one each time to include each number below n (and over 0).

                                                    function factorial(n) {
                                                      var product = 1;
                                                      for (var i = n; i > 0; i--) {
                                                      }
                                                    }

Step 5: If statement for if n is 0; else multiply i to the variable product. The loop will start with n and every time it will be multiplied to the variable product.

                                                     function factorial(n) {
                                                       var product = 1;
                                                       for (var i = n; i < 0; i--) {
                                                           if (n === 0) {
                                                              product = 1;
                                                           } else {
                                                              product *= i;
                                                           }
                                                       }
                                                     }

Step 6: Return the variable product: because you have to return the answer...

                                                     function factorial(n) {
                                                        var product = 1;
                                                        for (var i = n; i < 0; i--) {
                                                           if (n === 0) {
                                                              product = 1;
                                                           } else {
                                                              product *= i;
                                                           }
                                                        }
                                                        return product;
                                                     }

There you have it! Any comments or questions are all appreciated.

Check out past problems here:

• Reverse a String
• Longest Word
• Sum Num
• Time Conversion 
• Count Vowels 
• Palindrome
• Nearby AZ
• 99 Bottles of Beer