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Algorithm Solutions

2016-06-19 Tiffany WhiteAlgorithms

 

I’ve completed three more Free Code Camp algorithms since my last go on May 30th.

I needed less guidance on these particular algorithms, all but one. The algorithm solutions weren’t too hard to come up with, however on Slasher FLick I really overcomplicated my solution. I was thinking way too hard about how to solve it— new array? Should I return newArray as a part of the function call? Do I push the result of the cut off part of the array into newArray?

The instructions were:

Return the remaining elements of an array after chopping off n elements from the head.

The head means the beginning of the array, or the zeroth index.

Remember to use Read-Search-Ask if you get stuck. Write your own code.

Here are some helpful links:

Array.slice()

Array.splice()

Seems simple enough. But I was thinking too much and making it more complex than it needed to be because of my recent previous solutions.

I used splice instead of slice as splice returns the chopped off part as a new array. For these instructions, here was the initial code:

function slasher(arr, howMany) {
// it doesn't always pay to be first

}
slasher(["burgers", "fries", "shake"], 1);

Originally I had this:

function slasher(arr, howMany) {
// it doesn't always pay to be first
  var newArray = [];
  arr.splice(0, howMany);
   return newArray;
}
slasher(["burgers", "fries", "shake"], 1);

I would get a double array, because like I said, splice returns a new array from the chopped off part. So I tried to use a non-initialized variable— var newArray;— that returned a TypeError.

I went to the Free Code Camp wiki to look at the explanation in more detail. I finally came up with an Aha! Moment. I only needed to return the array that was resulting in the splice method.

I settled on my final solution here:

function slasher(arr, howMany) {
// it doesn't always pay to be first
  arr.splice(0, howMany);
   return arr;
}

slasher(["burgers", "fries", "shake"], 1);

Not too bad.

Getting Easier?

That one was easier to do. I hardly needed any help from Gitter. But then…

Mutations.

The instructions:

Mutations

Return true if the string in the first element of the array contains all of the letters of the string in the second element of the array. For example, ["hello", "Hello"], should return true because all of the letters in the second string are present in the first, ignoring case. The arguments ["hello", "hey"] should return false because the string “hello” does not contain a “y”. Lastly, ["Alien", "line"], should return true because all of the letters in “line” are present in “Alien”. Remember to use Read-Search-Ask if you get stuck. Write your own code. Here are some helpful links: String.indexOf()

I thought, okay. I got this.

The code:

function mutation(arr) {

}

mutation(["hello", "hey"]);

I started out with this:

function mutation(arr) {

  for (var i = 0; i < arr.length; i++) {
    if (arr === arr.indexOf(i)) {
    return true;
   } else {
    return false;
  }

}

mutation(["hello", "hey"]);

That didn’t work. I went to Gitter.

A guy there told me I needed to compare two arrays. I couldn’t figure out what he meant by that. To the wiki.

There I found out I should use toLowerCase and think about turning the array strings into an array of chars.

So then I came up with this:

var arr1 = arr.toLowerCase();
var arr2 = arr.toLowerCase();
var chars = arr1.split(" ");

as part of the equation. That also didn’t work. By this time, I am tired, it is late and I just wanted this to work. I went back to the wiki and found the solution.

I wouldn’t have ever came up with this yesterday night:

function mutation(arr) {
  var arr1 = arr[1].toLowerCase();
  var arr2 = arr[0].toLowerCase();
  for (var i = 0; i < arr1.length; i++) {
    if (arr2.indexOf(arr1[i]) < 0)
      return false;
  }
return true;
}

mutation(["hello", "hey"]);

I am studying this. I am trying to figure out what is going on here and I will probably go to Code Newbie Slack to ask around.

Back At It

I am doing the Falsy Bouncer algorithm after this post. It looks easy enough, and if I get stuck I will go to Gitter instead of the wiki as the solutions are there. You don’t have to look, but I was tired and frustrated last night which is never a good mix.

I will also be reading all the algorithm books I have and the Coursera courses I downloaded from Stanford.

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