"But that was the shit that I prayed about I thank God I made it out
Damn it feels good to go pick up my momma and take her out
Lil Tony got paper now all in my bank account, and it ain't shit to debate about"
!templates
Building Hitlists & How To Get A Job
6:00pm ET in Classroom
(Mon-Fri 6pm EST)
Ready? Others?
let stockTrader = {
firstName: 'Bob',
lastName: 'Wehadababyitsaboy',
location: 'New York, New York, USA',
jobTitle: 'Trading Associate',
yoe: 1,
recentTrades: ['tsla','gme','amc']
}
let tslaPurchasePrices =
[208.17, 205.05, 217.57, 220.48]
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 * (lastItem + 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) //true
Here 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))
Do: Take your first application step!
Do: https://frontendmasters.com/courses/practical-algorithms