# Algorithm Solutions

2016-06-19 Tiffany White

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.

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);``````

``````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);``````

## 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.