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I don’t wanna do shit, unless I'm with my partners"
Daddy_Anki
Sean, SOFTWARE FUCKING ENGINEER!
Works part time, gets paid full time!
Erica
"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
Custom everything plus, tweets, blog, and project
Ali Abdaal: https://youtu.be/ukLnPbIffxE
https://intelalearning.wordpress.com
https://www.pearsoned.com/three-simple-research-based-ways-to-ace-a-test
Less Memory, but Less Efficient
More Memory, but More Efficient
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) //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?