Wow, this is the clearest explanation of the solution I have come across that explains constraints with the correct example, rather than just the default example, and I am left scratching my head at the final solution, like huh, why was this needed at all? Thanks so much!
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
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]:
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
@@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
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Wow, this is the clearest explanation of the solution I have come across that explains constraints with the correct example, rather than just the default example, and I am left scratching my head at the final solution, like huh, why was this needed at all? Thanks so much!
whoever gave the title to this problem knew what he was doing lol
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
I liked this solution a lot more than the previous one, keep up the amazing work!
Yeah this one is better. Thank you for being a paid member!!
Thank you Greg for the wonderful video, it is so helpful for me and my team, thanks again
Glad to hear it!
I'd recommend solving Two Sum and Two Sum II before approaching this problem
One of my favorite solutions for a problem so far 👍
Yeah it's pretty neat
That was a very good explanation!
Best explanation👌
is using sort in the spirit of these problems?
Good job
What about the time complexity that came from sorting? Wouldn't it be like nlog(n) times n²?
You sort only once, so its nlog(n) PLUS n^2, which simplifies to n^2
What about using set instead of filtration that we do?
Keep in course you'r good instructor
Elements in an a triplet can repeat, it's just the triplets themselves that have to be unique
@@TM40_AerialAce I have solved it with hash set it's a simple solutions more than this
Excellent .
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]:
Just before that part, high is decreased:
else:
res.append([offset, nums[low], nums[high]])
low += 1
high -= 1 #
8:51 What's the point of using a for loop if you do not intend to iterate through all elements?
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
@@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.
@@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
@@2EOGIY you can use either while or for loop for any use case they are literally just syntax differences.
greg i dont think hashmap 3sum video you've done
is this nums.sorting() allowed at interviews?