2 Pointers Algorithm - DSA Course in Python Lecture 12

[@GregHogg]

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[@TJ-jh6xx]

Clean solution, thanks! As an improvement, I think you can skip the reverse step completely if you just use insert instead of append: array.insert(0,value). So, basically insert at the front of the array each time instead of append which inserts at the end, requiring the reverse later.

[@docsmate2umass]

simple and clear explanation makes this video more valuable .

[@BigStupidTech]

Spitting straight value. Thanks for these

[@GregHogg]

It's what I do

[@Schadenfreude596]

I like to watch your videos non-decreasing order

[@Za3DoRzX]

I love your explanations Greg, keep up the good work.

[@majestif]

Many thanks for a great explanation!

[@GregHogg]

You're very welcome 🥰

[@christianjt7018]

Great explanation, thanks a lot for sharing

[@leandrormor]

Thanks, your content is brilliant. I hope you'll touch monotonic stack, double heap, prefix sum...

[@nurbekmalikov3467]

for space comp just create one iteration index like => iterator = len(num) -1 and every greater value puts original array which is num and iterator-- this is O(1) space!

[@MrTo1223]

Hi greg can I just change th3 conditions to append the smallest of the squares of the left and right pointers?

[@helloworldcsofficial]

Thanks! Will there be more easy and medium problems where we can I apply the two pointer (indices) technique?

[@RaviPrakash-dz9fm]

Is this good? from typing import List def squares(array : List[int]) -> List[int]: n = len(array) l = 0 r = n - 1 for i in range(len(array)): array[i] = array[i]**2 while l < r: start = array[l] end = array[r] if end < start: array[l] , array[r] = array[r] , array[l] r -= 1 return array squares([-4, -1, 0, 3, 10]) I tried not to use extra space. I am just learning so I might have missed some edge case

[@ryanlalliss5977]

Sorry if this is stupid but why not use the insert method instead of append rather than reverse the array at the end?

[@GregHogg]

That will be inefficient because inserting at beginning is slow

[@izzya1282]

Perfectooo amigo

[@timkvernen4852]

It's not accurate to say that the biggest values are on the outside. The second biggest value could be the second or second-to-last element. It would be accurate to say that the biggest value will always be on the outside.

[@abodora532]

hmm, but what if they weren't in non decreasing order

[@divyanshsingh6668]

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