Minimum Difference Between Largest and Smallest Value in Three Moves - Leetcode 1509 - Python 

That was so bloody clever. I just learnt heap a week ago and did not even think that we can use a heap here

doesn't heapifying an array of size n, itself is a process of O(n*log(n)), you perform sink and swim operations that both take O(log(n)) for n elements?

Awesome man. Learnt something new today. Nsmallest and nlargest. Thank you!

Thanks for showing us previous attempts! Good explanation & 2nd solution was interesting...

Python is such a cheatcode it is ridiculous, and it isn't even fair 🤕

by remove, he means setting it to the min value right? coz you cant actually remove, you can only change it to any val

Logically, the solution was obvious but it would take me a lot of time to come up with this: int right = nums.length - 4 + left; btw is it really necessary to sort Heaps? It is supposed to be in the right order out of the box.

Great explanation as always. Thank you

is heap actually a N operation? i feel like its not and when i view the time complexity on leetcode its n lgn

the video and the explaination is awsome..but could you try to code in java or c++

similar idea : class Solution { public: int minDifference(vector& nums) { int n=nums.size(); if(n

Why not just? fun minDifference(nums: IntArray): Int { if(nums.size

can someone clarify further why the sorted in the heaps solutions? doesn't adding the sorting will end us at nlogn ?

Not entirely sure I even understand the problem, but it seems to me using a vector in C++ can do it all with sort and erase followed my just accessing the first and last elements for the difference.

"Solution to the most hardest part of this problem is simple, just use a built-in python library" Amazing explanation 👏 -.-

Interesting that if you try to do in java with heaps - it will be more than twice slower. Because with sorting it can be done on int[], without using a collections with all their overhead

na bro i am so done i cant fucking solve these questions no matter how much i try shit

The question is not removing could you pleaSe fix that. I know it is still the same but just confuses

It's a bit unfortunate that you jumped right in with the intention to REMOVE elements. The description says "change to any value". The observation that this is effectively the same as removing was not made in the video, but beforehand.

*no need to sort bud* min_four = nsmallest(4, nums) max_four = nlargest(4, nums) res = float("inf") for i in range(4): res = min(res, max_four[4 - i - 1] - min_four[i]) return res

oh first haha. I appreciate you Neet

brilliant

Nice🎉🎉

I think the question should be more clear and provide more details

multiple choice problem detected brute force selected 🗿

I guess I'm the only one that found this explanation confusing sigh

if you think neet's way of using hardcoded numbers is hard to wrap your mind around, try this: n = len(nums) i = 0 j = 3 if len(nums) i will go from 0 to 3, j will go from 3 to 0 -> i will be the left pointer, ie number of values cut off from the left -> j will be the right pointer, ie number of values cut off from the right -> thru this, we can get 4 combinations of removals: => 0 values from left, 3 values from right => 1 value from left, 2 from right => 2 from left, 1 from right => 3 from left, 0 from right return the min among these calculations hope this helps those who couldnt grasp neet's explanation peace, x edit: i had no clue heap could be used here im dumb asf

difficult to come up with the intuition here.

@NeetCodeIO can you make a video on giving tips regarding job search and some websites where we can apply. I think this would be very helpful to many people like me in the middle of a job hunt.

Just wanted to say I love you bro you're awesome

The video is speeded-up by default

I am waiting for you from 1 hour

Man coding in a language that doesn't have min or max heap is tricky, probably just sort the array and call it a day Edit: can implement quickselect which average O(n), and a good segway to LC 215 kth largest element in array

What's the Java equivalent fo heapq.nsmallest and largest?