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Sup grg

You are doing really well. Thank you.

One pass? On a phone, didn't get to check all edge cases. def closest(arr): if len(arr) == 1: return arr[0] dist = abs(0 - arr[0]) res = arr[0] for i in range(1, len(arr)): curr = abs(0 - arr[i]) if curr == dist: res = max(res, arr[i]) elif curr < dist: dist = curr res = arr[i] return res arr = [-2, -4, 1, 2 , 8] # 1 # arr = [-2, 1, -1] print(closest(arr)) # 1

Memoize

If the abs values of nums[i] and abs(closest) are equal, closest can simply be considered as the abs(nums[i]) and returned. Here's my JS code: var findClosestNumber = function (nums) { let closest = nums[0]; for (var i = 1; i < nums.length; i++) { if (Math.abs(nums[i]) < Math.abs(closest )) { closest = nums[i]; } else if (Math.abs(nums[i] == Math.abs(closest ))) { closest = Math.abs(nums[i]); } } return closest ; }

This is very good. However, for dataframes with categorical / text data, you code will fail with a TypeError. I think this requires fitting only the columns of the dataframe which have numerical values.

Yet another thing you're asked in coding interviews and will never use in the job. How will a project manager react when he finds out an intern is coding a data structure from scratch?

Wow. Such a beautiful solution. Excellent work as always Greg!

Bro deserves an Oscar for explaining us time and space complexity just like that I a few minutes ❤❤

We can return abs (closest) and avoid using last condition?

Lets gooooo!

Im a little confused on what is happening here. From my understanding, the only thing that goes through the hash function is a key. Why would anyone use an array as a key? Doesnt using a key that can change defeat the purpose of a hash map? Can you explain why this is even important?

So what's the problem? Sounds easy You do Record/Map in JS or Dictionary in .net, get the result from modulo calculation, check if there is any value in the given array and insert it. But if we need to check whether value is already there, just create another dictionary with string as a key and int as a value. So firstly check if the value already exists in the second dictionary, if not then try to insert the item in the first dictionary (or find the free one) And write the second dictionary value. Easy.

algo comment drop. good content.

Hi grg

The expert dev is the one that realises that your prod code probably wouldn't notice the difference between the two solutions but broke because someone asked if 🤢💛🍇🍊🍌🍍🍎 is an anagram of 🤢🍎💛🍇🍊🍌🍍

am i the only one not understanding the question the same way ? "so that both splitted arrays don't contain duplicated numbers" i would interprete it as "the splitted array can't have common value" and such thing happens a lot with me in this kind of interviews. i just find the questions very ambiguous

As I understand it, if index 4 was occupied as well it would pick index 0, and when there was no empty spaces it would loop through the entire list to look for a value that isn't there, sound very complicated to me with lots of traps

How about bucket sorr

Hello, I'm from Bangladesh. My question, can a data analyst work as a software engineer?

Awesome

Can someone tell if this approach is correct or not? I think we can do it with 2 pointers approach. If there are only 3 integers that we have to account for theek we can have a starting index and a last index. So, 0's are supposed to be swapped to the starting index and the 2's are supposed to be swapped with the last index and then increment the starting pointer and decrement the ending pointer. If we just sort 0 and 2, 1 will automatically be sorted.

Isn't that used in bloomfilters?

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Love how simple and stupid Python with all these libraries can be.

𝘾𝙖𝙣 you give me data structure roadmap with example which I have to solve

Mergesort -> arr[odd] to arr1, arr[even] to arr2. No?

So it's the same as using HashMap, only with hashmap you get a universal solution any number of different colors

I think we can do it with 2 pointers approach. If there are only 3 integers that we have to account for theek we can have a starting index and a last index. So, 0's are supposed to be swapped to the starting index and the 2's are supposed to be swapped with the last index and then increment the starting pointer and decrement the ending pointer. If we just sort 0 and 2, 1 will automatically be sorted.

Now i understand mutable and immutable variables ❤❤❤

Just use .sort(), it's built in for a reason and it's exactly what every single company would use in production. Assuming you want the lower nunber colours at the start

Using three pointers we can solve this problem which is the most optimal solution

You can just change the original string to lowercase to avoid the problem with lowercase and uppercase vowels/consonants

never stop please, love you man.

now do it in place! (shifting zeros)

But why don't they/you use real programming languages? Python is bullshit

this is so basic omg, i instantly thought of this upon seeing this

Use the (φ^n - (1-φ)^n)/√5 technique

not match deduplicate duplicate

you could just get the answer with math

Basically it's a database with a unique id

Hi Greg, Thanks for the amazing video and explanation. I have a quick question on the code implementation. On line 12, I do not think the while loop runs more than once. Because we are flipping one zero at a time with the help of the condtion above it (Line 9). Example, if we have flipped 3 zeroes and k = 2 and our intention is to get back one zero (num_zeroes -= 1), we run through the while loop to decrement. But here, after the first iteration of while loop, the condition evaluates to false (2 > 2). This is because we incremented num_zeroes from 2 to 3 (We are flipping one zero at a time). So, my question is to know the intention of selecting a while loop on Line 12. Since, while loop doesn't work for more than one time, I replaced it with a simple if condition. So, Line 12 will become, if (num_zeroes > k): num_zeroes -= 1. The code will still work for all the cases. Here's my code: class Solution: def longestOnes(self, nums: List[int], k: int) -> int: left = 0 max_num = 0 num_of_zeroes = 0 for index, num in enumerate(nums): if num == 0: num_of_zeroes += 1 if num_of_zeroes > k: if nums[left] == 0: num_of_zeroes -= 1 left += 1 max_num = max(max_num, index - left + 1) return max_num

Omg my teacher made avl trees one of the hardest topics to understand in dsa wish I had discovered this short sooner

Why aren’t you a CEO then 😅

Just do in js nums.sort((a, b) =>a-b) it is o(n)

What kind of monster makes curly brackets like that?

@19:20 Can somebody explain to me what out[:, -1, :] does. I'm trying to learn the burn crate for Rust which is young and doesn't have enough documentation so I'm stuck referencing pytorch which is it's influence.

is he cracking his voice on purpose?