3Sum (Updated Solution) - Leetcode 15 - Two Pointers (Python)

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[@GregHogg]

Master Data Structures & Algorithms for FREE at AlgoMap.io/

[@user-jm6gp2qc8x]

whoever gave the title to this problem knew what he was doing lol

[@anna-plink]

One of my favorite solutions for a problem so far 👍

[@GregHogg]

Yeah it's pretty neat

[@2EOGIY]

8:51 What's the point of using a for loop if you do not intend to iterate through all elements?

[@TM40_AerialAce]

You do intend to iterate through almost all elements (length of array - 2, so you have space for the two pointers). the i in `for i in range(n)` acts as a boundary that squeezes the two pointers closer together every iteration. Sort of like the trash compactor scene from A New Hope

[@2EOGIY]

@@TM40_AerialAce, if there is an intent for an early exit, then a for loop is a wrong choice. Just add those conditions to a while loop.

[@wilderclover3354]

​@@2EOGIYi don't think there is some rule written in python documentation that u should not used for loop in those situations. Its easier to explain by using for loop than while

[@Redaxi]

@@2EOGIY you can use either while or for loop for any use case they are literally just syntax differences.

[@WatchNoah]

instead of the two while loops in the summ == 0, you can actually also simply use a set as answer and add sorted tuples. this also has a runtime of 558ms

[@user-jm6gp2qc8x]

greg i dont think hashmap 3sum video you've done

[@pravinprince3221]

Thank you Greg for the wonderful video, it is so helpful for me and my team, thanks again

[@GregHogg]

Glad to hear it!

[@malakggh]

That was a very good explanation!

[@flynnblu6992]

Love your videos! But puzzled... after finding the first answer, isn't nums[hi + 1] out of range on the following, given hi starts out as (n - 1), hence we are checking nums[n]?
while hi > lo and nums[hi] == nums[hi + 1]:

[@brianwkinyua]

Just before that part, high is decreased:
else:
res.append([offset, nums[low], nums[high]])
low += 1
high -= 1 #

[@sirojiddinSoftwareEngineer]

Good job

[@venkateshkannan7398]

Best explanation👌

[@thomasthedankengine5400]

What about the time complexity that came from sorting? Wouldn't it be like nlog(n) times n²?

[@endless9142]

You sort only once, so its nlog(n) PLUS n^2, which simplifies to n^2

[@christianjt7018]

I liked this solution a lot more than the previous one, keep up the amazing work!

[@GregHogg]

Yeah this one is better. Thank you for being a paid member!!

[@mehdismaeili3743]

Excellent .

[@mahmoudelazb8028]

What about using set instead of filtration that we do?
Keep in course you'r good instructor

[@TM40_AerialAce]

Elements in an a triplet can repeat, it's just the triplets themselves that have to be unique

[@mahmoudelazb8028]

@@TM40_AerialAce I have solved it with hash set it's a simple solutions more than this