January 19, 2019

# Saturday Solutions #2

Today I will be breaking down different approaches for solving the code challenge in Sunday Studies #2. I will start with using fundamental JavaScript skills and then build up to more modern techniques with ES6+.

## Recap: The Problem

For each problem below, write out as many versions of the solution as possible for practice:

1. Double the value of every item in list
2. Transform each item in list to be a string " o'clock" appended to the current item (i.e., 1 becomes "1 o'clock")
3. Replace every item that is divisble by 3 with the string "triforce" Only loop through items that have an even index. In other words, the loop should only run 5 times.

## Solution

### Double the value of every item in list

#### for loop

``````/**
* ES5
*/
var doubledList = []

for (var i = 0; i < list.length; i++) {
var currentItem = list[i]

doubledList.push(currentItem * 2)
}
``````

#### forEach

``````/**
* ES5
*/
var doubledList = []

list.forEach(function(item) {
doubledList.push(item * 2)
})

/**
* ES6
*/
let doubledList = []

list.forEach(item => {
doubledList.push(item * 2)
})
``````

#### map

``````/**
* ES5
*/
var doubledList = list.map(function(item) {
return item * 2
})

/**
* ES6
*/
let doubledList = list.map(item => item * 2)
``````

### Transform each item in list to be a string " o'clock" appended to the current item (i.e., 1 becomes "1 o'clock")

#### for loop

``````/**
* ES5
*/
var timeList = []

for (var i = 0; i < list.length; i++) {
var currentItem = list[i]

timeList.push(currentItem + " o'clock")
}

/**
* ES6
*/
let timeList = []

for (let i = 0; i < list.length; i++) {
let currentItem = list[i]

timeList.push(`\${currentItem} o'clock`)
}
``````

#### forEach

``````/**
* ES5
*/
var timeList = []

list.forEach(function(item) {
timeList.push(item + " o'clock")
})

/**
* ES6
*/
let timeList = []

list.forEach(item => {
timeList.push(`\${item} o'clock`)
})
``````

#### map

``````/**
* ES5
*/
var timeList = list.map(function(item) {
return item + " o'clock"
})

/**
* ES6
*/
let timeList = list.map(item => `\${item} o'clock`)
``````

### Replace every item that is divisible by 3 with the string "triforce." Only loop through items that have an even index. In other words, the loop should only run 5 times.

This is a trick problem in that there is only one solution. If you use any other loop, you end up breaking the requirement of only running 5 iterations of the loop.

``````/**
* ES5
*/
var triList = []

for (var i = 0; i < list.length; i += 2)  {
var currentItem  = list[i]

if (currentItem % 3 === 0) {
triList.push("triforce")
} else {
triList.push(currentItem)
}
}

/**
* ES6
*/
let triList = []

for (let i = 0; i < list.length; i += 2)  {
let currentItem  = list[i]

if (currentItem % 3 === 0) {
triList.push("triforce")
} else {
triList.push(currentItem)
}
}
``````

### Extra Credit

#### What is the advantage of using a for loop over the other two methods?

The `for` loop allows you to choose how you iterate through a loop. In other words, you don't have to go through every item if you don't want to. You can increment the counter by however much you need it to! Every other loop will go through every item.

#### What is the difference between forEach and map?

`forEach` mutates the actual array whereas `map` returns a new array.

### Final Thoughts

As I continue to iterate on this series, please feel free to reach out with requests and/or feedback. You can either reach out to me directly on Twitter or open a new issue. Thanks for reading!