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! :)

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 Problems: Time Conversion

Back at it again today! My plan is to continue solving these practice problems till I feel that "Oh hell yes!" sense of accomplishment. Let's get on with it.

Your mission: Write a method that will take in a number of minutes and returns a string that formats the number into "hours:minutes"

In keeping with tradition,  think about how to solve this problem before doing any actual coding. I promise it's easier than Phoebe trying to teach Joey guitar cords without a guitar. #friends

The problem is asking us to convert any number of minutes given into the hours:minutes format. In order to do this there will have to be some math involved. Maybe set an hour counter equal to 0 and every time there's at least 60 that counter should go up by one. There should conversely be another counter that starts with the number passed in and decreases by 60 each time. Then it's just an issue of formatting, concatenating variables with strings and whatnot.

1. Let's go ahead and define our function with one argument. It is already provided in Ruby, but for Javascript it would be:

function time_conversion(minutes) {

}

2. I defined two new variables here: hours, which I set equal to 0, and min, which I set equal to the parameter, minutes. (This last step might be repetitive, but it was how I solved the problem and it worked so I'm keeping it for now.)

function time_conversion(minutes) {    
   var hours = 0;
   var min = minutes;
}

3. Every time min is greater than or equal to 60, I want hours to increase by 1 and I want min to decrease by 60. To do this, I used a while loop. Inside the while loop, I want it to do all the math I explained above. Remember to include the counter (i -= 60) or you'll find yourself in an infinite loop. The same three parts of a for-loop are contained in a while loop, just the formatting is different.

function time_conversion(minutes) {
   var hours = 0;
   var min = minutes;
   var i = min;
   while (i >= 60)
      hours++;
      min -= 60;
      i -= 60;
   }
}

4. Almost done! Just need to handle some special cases-- when there is less than ten minutes, you want it to read "06" or "0-whatever" not just "6". The second case you want it to handle is when there are 0 minutes, you want it to say "00". Sounds like an if statement.

function time_conversion(minutes) {
   var hours = 0;
   var min = minutes;
   var i = min;
   while (i >= 60)
      hours++;
      min -= 60;
      i -= 60;
   }
   if (min === 0) {
      min = "00" 
      return hours + ":" + min;
   } 
   else if (min < 10) {
      min = "0" + min;
      return hours + ":" + min;
   }
   else {
      return hours + ":" + min;
   } 
}

This problem sent me for a loop. Yes, pun intended. It literally sent me into an infinite loop multiple times. Despite having to come back to this problem on different days to give it another shot, that little victory kinda made it all worth it. It's a pretty awesome feeling of accomplishment figuring these challenges out, even if it may not be the most concise code or employ best practices.. I have to actually know how do it first, amirite? Here's to the small victories!

P.S. I move to New York in TEN DAYS. (!!!!!!!!!!!!!)

Check out past problems here:

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

a/A Practice Problems: Sum Nums

This one is pretty straight forward. The only thing to look out for: for loop starting with num and decreasing each time.

Problem: Write a method that takes in an integer 'num' and returns the sum of all integers between zero and num, up to and including num.


Game Plan: 

>>  function takes in one parameter, num.
>>  for loop to go through each number below (and including) num (i--).
>>  declare variable and set it equal to 0, will hold value from each iteration of loop.
>>  return total sum

Step 1: 

                                                        function sumNums(num) {
                                                        }

Step 2:

                                                       function sumNums(num) {
                                                          var sum = 0;
                                                       }

Step 3:

                                                        function sumNums(num) {
                                                           var sum = 0;
                                                           for (var i = num; i > 0; i--) {
                                                           }
                                                        }

Loop starts with num and stops when it is greater than 0, or 1.

Step 4: 

                                                         function sumNums(num) {
                                                            var sum = 0;
                                                            for (var i = num; i > 0; i--) {
                                                               sum += i;
                                                            }
                                                         }

You could also write sum = sum + i;

Step 5:

                                                          function sumNums(num) {
                                                             var sum = 0;
                                                             for (var i = num; i > 0; i--) {
                                                                sum += i;
                                                             }
                                                             return sum;
                                                          }

Check out past problems here:

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

a/A Practice Problems: Longest Word

This problem was a tricky one. Despite its "easy" rating, I struggled with it for a little thinking of the most roundabout ways of answering, including multiple nested for loops-- which as you can guess was just a joy ride. Maybe possible, but that way would have taken me forever to write and would have probably been very error prone. The hardest part for me turned out to be knowing how to set it up. How could I solve this so that I didn't need to go through every character of a string? Well, you don't want to go through every character necessarily (my first mistake), but rather every word. But I'm getting ahead of myself...

Problem:  Write a method that takes in a string. Return the longest word in the string. You may assume that the string contains only letters and spaces. You may use the 'split' method to aid you in your quest.

Game Plan: 

>>  Function that has one parameter (string) in this example.

>>  How can you quickly separate each word? Split method. The result of which will return an array of all the words. You'll want to set this equal to some variable for easy access later.

>>  For loop to go though each word to determine if it's the longest.

>>  This is where I got sort of lost-- I knew I needed to do all the above steps, but I didn't know how to keep track of which word was longest. If you had one independent variable to compare each word to, wouldn't that be easier than making some sort of March-Madness bracket system where you compare every word to each other? Methinks yes. So in order to do that, I'll set a variable equal to 0. The way the loop will work : compare the first word's length to 0, if the word is bigger (which it will be since it will be at least one letter character), i'll save the length of that word to the variable. As the loop progresses it will compare each word to this variable, saving the larger of the two.

>>  if statement to compare lengths

>>  Return longest word

Step 1:

                                             function longestWord(string) {

                                             }

Step 2:

                                             function longestWord(string) {

                                                var words = string.split(" ");

                                             }

Make sure there's a space between your quotes so that it knows where to split the string. If you didn't include the space, you would get each individual character at each index.

Step 3:

                                             function longestWord(string) {

                                                var words = string.split(" ");

                                                var longest = 0;

                                             }

We declare this variable, so that we don't need to compare each word to each other, but rather one independent variable.

Step 4:

                                            function longestWord(string) {

                                               var words = string.split(" ");

                                               var longest = 0;

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

                                               }

                                            }

Remember when you split the string, it results in an array of all the words so we can use the array's length.

Step 5:

                                             function longestWord(string) {

                                                var words = string.split(" ");

                                                var longest = 0;
                                                var longIndx = 0;

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

                                                   if (words[i].length > longest) {

                                                      longest = words[i].length;

                                                      longIndx++;

                                                   } 

                                                }

                                             }

The tricky part-- if the length of the word at index i is greater than longest (0), then we change the value of longest to the length of that word. AND since we want to keep track of which word we are on, we can set a marker of sorts at 0 (longIndx) that will increase if the word at position i is greater than longest. The value will let us know the index number of the longest word so that we can call it later.

Step 6:

                                             function longestWord(string) {

                                                var words = string.split(" ");

                                                var longest = 0;

                                                var longIndx = 0;

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

                                                   if (words[i].length > longest) {

                                                      longest = words[i].length;

                                                      longIndx++;

                                                   }

                                                 }

                                                 return words[longIndx];

                                              }

Try it yourself!

Check out past problems here:

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

a/A Practice Problems: Reverse a String

When I had first learned about App Academy I quickly tried to learn everything I could to solve those damned practice problems. Short story shorter, I could not wrap my mind around the first one. I came back to it the other night, however, and with all my new fandangled Javascript skills I was able to take on more than a couple. SHABOOYAH. What really helps me when I am learning something new is seeing the process someone went through to solve the challenge at hand, in this case reversing a string without using the reverse method. Because I found so few resources like that initially, I thought to do it myself and hopefully be that source for someone else.

Note, App Academy is a Ruby/Rails school, but if I read correctly they'll allow you to answer challenges in whichever language you choose -- obviously I'll be doing all these in JS.

Your mission: "Write a method that will take a string as input, and return a new string with the same letters in reverse order"

Disclaimer: This is my methodology, there are many other ways to do this; most likely there better ways to do this. But I'm a noob, so shut it.

Step 1: Write out what you need to do. It's like a game plan or blueprint to getting the answer. It helps me to really think about what the question is asking so that I can answer appropriately. This one is pretty straightforward.

     >> have to be able to accept input (function via arguments/parameters) 

     >> pinpoint last letter of string (1 less than length)

     >> run through & return letters from last character (loop with descending counter)

     >> return reversed string

Step 2: Define the function. Only rules that apply are the same for that of variables. (no numbers, symbols to start off; contains only letters, numbers, $ and _). Set up function to accept one parameter, "word" in this example.

  

function reverseString (word) {}

 Step 3: Find last character of string. I set the value equal to a new variable, lastChar so that it's easier to access. Remember indexes start with 0 (hence the minus 1).

                                                       function reverseString(word) {                                                  
                                                         var lastChar = word.length - 1;

                                                        }

Step 4: For loop with descending counter. Instead of setting i = 0, I set it to lastChar so that it would start with the last letter of the string; i >= 0 tells the loop to stop once it reaches the index 0, first letter of the string; i-- tells the loop to decrement by one each time.

                                              function reverseString(word) {

                                                 var lastChar = word.length - 1;

                                                 for (var i = lastChar; i >= 0; i--) {

  

                                                 }

                                              }

Step 5: Concatenate. If we were to give ourselves a test case, say reverseString("abc");, and if we were to console.log(word[i]) we would see that each letter would come back in reversed order. Hooray! But, (boo!) it wouldn't come back in one string, "cba". To fix this, I thought do declare another variable (empty this time) and then add each letter to it, aka concatenate it into one. The code block inside the for-loop is telling the function to add each letter to the variable newString. I'm not sure why you would use bracket notation for a string when they're reserved mostly for arrays, but I know it works and that they operate similarly and that's the best explanation I can give.

                                              function reverseString(word) {

                                                 var lastChar = word.length - 1;
                                                 var newString = "";

                                                  for (var i = lastChar; i >= 0; i--) {

                                                     newString += word[i];
                                                             // same as newString = newString + word[i];

                                                  }

                                              }

Step 6: Return newString. Now that we have all the working parts, the only thing left to do is to return the reversed string. Remember to put this after the function.

                                              function reverseString(word) {

                                                var lastChar = word.length - 1;
                                                var newString = "";

                                                 for (var i = lastChar; i >= 0; i--) {

                                                     newString += word[i]; 

                                                 }
                                                 return newString;

                                              }

If you were to type this into your console or whatever environment, you should get "cba". There you have it, guys! Comment with any questions or if you have any techniques that would solve this quicker, I'd love to read it. Good luck!

Check out other problems here:

• Factorial
• Longest Word
• Sum Num
• Time Conversion 
• Count Vowels 
• Palindrome
• Nearby AZ