Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

[WDP - PT - 2024] Victor Mamani #3327

Open
wants to merge 1 commit into
base: master
Choose a base branch
from
Open
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
46 changes: 46 additions & 0 deletions index.js
Original file line number Diff line number Diff line change
@@ -1,7 +1,53 @@
// Iteration 1: Names and Input
let hacker1 = "pedrot"
let hacker2 = "pedros"

console.log(`The driver's name is ${hacker1}`)
console.log(`The navigator's name is ${hacker2}`)

// Iteration 2: Conditionals

if (hacker1.length > hacker2.length) {
console.log(`The driver has the longest name, it has ${hacker1.length} characters.`)
} else if (hacker2.length > hacker1.length){
console.log(`It seems that the navigator has the longest name, it has ${hacker2.length} characters.`)
} else {
console.log(`Wow, you both have equally long names, ${hacker1.length} characters!.`)
}

// Iteration 3: Loops

//3.1
let nameArr = hacker1.toUpperCase().split('')
let nameUpperCase = ''

for(i=0; i<nameArr.length; i++){
nameUpperCase += ` ${nameArr[i]}`
}
console.log(nameUpperCase)

//3.2
let reverseName = ""
for(i=hacker1.length -1 ; i>=0; i--){
reverseName += `${hacker1[i]}`
}

console.log(reverseName)

//3.3

let arr = [hacker1, hacker2]

arr.sort(function(a,b) {
return a.localeCompare(b)
})

console.log(arr)

if(arr[0] === hacker1 && arr[0] !== hacker2) {
console.log("the driver's name goes first.")
} else if(arr[0] === hacker2 && arr[0] !== hacker1){
console.log("Yo, the navigator goes first, definitely.")
} else {
console.log("What?! You both have the same name?")
}