"Ariana ain't the only one that do it grande,
Knew I'd be his fiancé when he was broke on Causeway, yuh
And, and, and he ain't even hit his peak like Dante"
6:00pm ET in Classroom
(Mon-Fri 6pm EST)
Ready? Others?
Thank you!
Bootcamper!
"Just had a coffee chat was told they see so many applicants with empty githubs, no portfolio sites and no projects under their belt
We’re on the right track yall, Leon has definitely given us the cheat codes"
@Telescope_Thieves
let developer = true
let designer = false
if(designer === true){
console.log("Design your portfolio")
}else{
console.log("USE A TEMPLATE")
}
Sometimes you can get the first year free
OR
Just use netlify.app for free
Template: Bob Boberson
Once we have our hitlist, we'll make multiple versions
(Run your resume through to see what recruiters see don't pay)
Example: Bob Boberson
Once we have our hitlist, we'll make multiple versions
One of the most important things you will ever do for your career
NOW WITH TWO TABS!: Google Sheet
Custom everything plus, tweets, blog, and project
Sean, SOFTWARE FUCKING ENGINEER!
Works part time, gets paid full time!
Erica
Daddy_Anki
All computers, environments, ect... are different
For all inputs to our algorithm there is and will always be only one operation required
No matter how many inputs are located in num there will only ever be one operation needed!
const nums = [1,2,3,4,5]
const firstNumber = nums[0]For all inputs to our algorithm there will be one
operation per input
Here we sum the numbers in the array. We have to add each number to a running sum, so we have to operate on each one. This means one operation per input.
const nums = [1,2,3,4,5]
let sum = 0
for(let num of nums){
sum += num
}const sumContiguousArray = function(ary){
//get the last item
const lastItem = ary[ary.length - 1]
//Gauss's trick
return lastItem * (listItem + 1) / 2
}
const nums = [1,2,3,4,5]
const sumOfArray = sumContiguousArray(nums)Text
const hasDuplicates = function(num){
for(let i = 0; i < nums.length; i++){
const thisNum = nums[i]
for(let j = 0; j < nums.length; j++){
if(j !== i){
const otherNum = nums[j]
}
}
if(otherNum === thisNum) return true
}
return false
}
const nums = [1,2,3,4,5,5]
hasDuplicates(nums) //trueHere we’re iterating over our array, which we already know is O(n), and another iteration inside it, which is another O(n). For every item in our list, we have to loop our list again to calculate what we need.
Divide and Conquer?
Text
const findMaxBid = function(bids){
let maxBid = bids[0],
minBid = bids[0]
for(let i = 0; i < bids.length; i++){
for(let j = 0; j < bids.length; j++){
if(bids[i] > bids[j] && bids[i] > maxBid ){
maxBid = bids[i]
}else if(bids[i] < bids[j] && bids[i] < minBid){
minBid = bids[i]
}
}
}
return [minBid, maxBid]
}
const allBids = [2,7,3,1,4,5,5]
console.log(findMaxBid(allBids))Text
const findMaxBid = function(bids){
let maxBid = bids[0]
for(let i = 0; i < bids.length; i++){
if(bids[i] > maxBid){
maxBid = bids[i]
}
}
return maxBid
}
const allBids = [1,2,7,3,4,5,5]
console.log(findMaxBid(allBids))By Nora Smith
Penultimate Class of our Free Web Dev Bootcamp for folx affected by the pandemic. Join live T/Th 6:30pm ET leonnoel.com/twitch and ask questions here: leonnoel.com/discord
