POV: *you are struggling on a leetcode problem* then you found neetcode have solve it the BEST feeling ever
@aadil4236
Жыл бұрын
I feel much safer doing daily leetcode challenges by your return. Thank you! Suggestion: We would love to see explanations of weekly contest as well. After it ends of course.
@carsonfreeman6955
Жыл бұрын
These explanations are amazing!
@polycrylate
Жыл бұрын
A small improvement: If nums[l] == nums[m] and nums[l] != nums[r] It's guaranteed that the pivot is right of middle i.e. you are on the higher/right part of array Because if the pivot is on the left of middle, it means that middle -> right must all be the same number that loops back to the left, however as middle != right (as left != right and left == middle) this isn't the case So only l += 1 in the case of nums[m] == nums[l] and nums[l] == nums[r], and extend the other case to be
@sumitraj6878
6 ай бұрын
in the case of nums[m] == nums[l] and nums[l] == nums[r] why only l++, do e- - as well. more efficient.. eliminate the same start and end elements because they aint our target. this will help shorten up the search space.
@metarus208
Жыл бұрын
glad to having you post regularly again.
@Scarsofevil
4 ай бұрын
If this problem worstcase senario is o(n). Can't we just integrate through the array and return True in the worstcase scenario?
@panmacabre9895
Жыл бұрын
thank you daddy
@josepadilla4024
6 ай бұрын
He sounded so mad at this problem haha
@rhugvedbhojane4387
Жыл бұрын
Good to see you back buddy.
@michelle_tsai_drums
Жыл бұрын
Thorough explanation as always!
@MP-ny3ep
Жыл бұрын
Great explanation as always . Thank you
@uptwist2260
Жыл бұрын
Thanks for the daily
@peskovdev
10 ай бұрын
sometimes you can understand is that left or right portion not only by left & mid pointers, but also by mid & right pointers, so you will omit some linear operations. Here is the code: ``` class Solution: def search(self, nums: List[int], target: int) -> bool: lp, rp = 0, len(nums) - 1 while lp nums[rp] or nums[mp] > nums[lp]: # left sorted portion if nums[lp]
@MrLeyt1125
4 ай бұрын
On #equal you should move both lp and rp
@uttamkumarreddygaggenapall2070
Жыл бұрын
Thank You
@ievgen6209
4 ай бұрын
imo, a little bit overcomplicated... the only case we need to handle in N time - skip duplicates if nums[0] == nums[-1] otherwise it is original solution: l = 0 while l < (len(nums) - 1) and nums[l] == nums[-1]: l += 1 # Code for Search in Rotated Sorted Array I problem Beats 86.73% of users with Python3 right or I'm missing something?
@MrLeyt1125
4 ай бұрын
Why not move both left and right pointers?
@floatingpoint7629
Жыл бұрын
the question mentions to decrease the overall operation steps. how does this algo do that?
@polycrylate
Жыл бұрын
It doesn't force a linear search, it always tries to do a binary until it's stuck and removes elements 1 by 1 until it can again What they meant by that line I think is even tho worst case is O(n) they wanted a better AVG case
@floatingpoint7629
Жыл бұрын
@@polycrylate got it, thanks
@lesterdelacruz5088
4 ай бұрын
Not the best problem. Because eliminating left pointer 1 by 1 in the worst case would still be O(n) so it doesn't improve anything if you simply just linearly search.
@aaditya_87
4 ай бұрын
2:08 which day? which problem
@SASA_maxillo
Жыл бұрын
why not just doing: return target in nums EASYYYYYYYYY
@flamendless
Жыл бұрын
Would "target in numbers" in Python work?
@NeetCodeIO
Жыл бұрын
That is basically a linear scan, it may get accepted but i think it's not the intended solution.
@sumitsharma6738
Жыл бұрын
the only catch in this problem is when you don't know which part is sorted so you just compare mid value with s and e and if (s and mid) are equal then s++ or if (mid or e) are equal then e--
@aniruddhachaki7165
Жыл бұрын
Wouldn't it be easier if we just sort the input first and then apply traditional Binary Search? in this problem, we don't need the target's index anyway. It works in this case where we just need to enter true or false.
@panmacabre9895
Жыл бұрын
sorting would take O(nlogn)
@aniruddhachaki7165
Жыл бұрын
@@panmacabre9895 Yeah. That might be the issue. So, the above solution is the best, if we have to return index, we can with just do a small change
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