You are the best explainer for leetcode problems bar none on the internet !!!! Wow .... !!! Thank you.
@smartsoothing776
Жыл бұрын
This is definitely not an easy problem!!
@JustinK0
7 ай бұрын
it definitely it easy, i got the O(n^2) solution on the first try, even if the O(n) is a bit harder it doesnt really matter if an easy solution still exists.
@chrischika7026
5 ай бұрын
@@JustinK0 they are obviously referring to the O(n) solution and it does matter because it is more efficient and that is what the interviewer would want.
@berserker556
4 ай бұрын
Yeah it did not feel easy to me
@thebigpart783
4 ай бұрын
@@JustinK0 if you show up with the n^2 solution they won't hire you so no it is not easy ...
@cicakmuhamed
Ай бұрын
@@JustinK0 Justin, solving a problem in suboptimal solution is easy for some of the hardest problems on leetcode. The whole point of leetcode is finding an optimal solution...
@lingxu9697
2 жыл бұрын
Always enjoyable to watch your video solutions, thanks!
@paul_tee
11 ай бұрын
i don't really see why the monotonic solution is O(n) time, where n = len(nums2). suppose nums1 =[1,2,...,n] and nums2 = [n, n-1, n-2,...,1]. in this case, don't you need to perform 1+2+...+n checks, which is on the order of n^2? EDIT: ok i got it, you don't need to perform 1+2+...+n checks, because of the decreasing nature of the stack. instead, you perform 1+1+...+1 checks because if you can't pop the first guy out, you surely can't pop any other the guys before it out. neat idea using monotonicity!
@ingluissantana
2 жыл бұрын
Such a great explanationnnnnnn thankssssss 🙏🏼🙏🏼🙏🏼🙏🏼
@chankwongyin7455
2 жыл бұрын
hey Neetcode, could you summarize all the questions you have done and make a leetcode list link to us? thx!
@lex-zt6uc
2 жыл бұрын
That would be amazing
@transgenicznyogorek
6 ай бұрын
Good explanation! My only nitpick is at 10:24 that the stack is monotonically *increasing, not decreasing* - the rightmost element is considered the top of the stack, and every following element is going to be greater than the top of the stack. That confused me for a while since R to L solution uses a decreasing stack :P
@shalinisangal84
4 ай бұрын
Great explanation, with ur explanation feeling like it was so easy. Thank u so much
@taiwoadebisi9315
Жыл бұрын
How is this easy?!!
@ajain2603
5 күн бұрын
Same question
@gauravmasand
Ай бұрын
I was stuck in understanding of problem what to do nice and simple easy to get explanation of question i got solution in mind at 2:30 Very nice
@manjultripathi8309
9 ай бұрын
Any better way to explain a LC problem than above can't be thought of, ever.Period!
@andreytamelo1183
2 жыл бұрын
Thanks!
@tamarehrenreich7428
Жыл бұрын
Perfect explanation. Very helpful. Thanks so much.
@ninjacloud4748
2 жыл бұрын
Thank you so much for all the Amazing videos you have made so far. Could you please add videos for some of HARD questions for Google, for example "Guess Word" and other such Leetcode questions. That would be a great help!! Thank you once again!!!
@zhonglingsun6743
2 жыл бұрын
Love all the videos on your channel. Could you possibly cover #31. Next Permutation which is similar to this one but a bit more complex?
@akhilr94
Жыл бұрын
still can't wrap my head around an algorithm like this, ie the O(n + m) one. obv I can understand the explanation. but how can it come naturally during interviews?
@alfamatter12
Жыл бұрын
It won't 😢
@polasumanth9826
2 жыл бұрын
Congratulations for 100k subscribers🎉🥳👏👏👏
@cc-to2jn
2 жыл бұрын
def not an easy lol, great job as always
@NeetCode
2 жыл бұрын
Agreed!
@iliauk1
2 жыл бұрын
Awesome video! Could we also have something like so (not sure if same complexity since seems shorter) res = {} stack = [] for v in nums2: while stack and stack[-1]
@avinashtiwari4025
Жыл бұрын
How did you calculate the space complexity for the stack part?
@jasonl.7466
2 жыл бұрын
tbh this should be a medium question (without knowing the pattern), at least for the optimum solution.
@niharikkatyagi4089
Жыл бұрын
The line ->while cur > stack[-1] is giving me the error: '>' not supported between instances of 'int' and 'list' I dont understand why, what do I do??
@memeproductions4182
2 жыл бұрын
Why do you need an hashmap in brute force? wouldn't just iterate nums1 and for each iterate nums2 to find the corrispective number and then go on and find the successor still be O(n *m) but without extra memory?
@akhilr94
Жыл бұрын
as you have mentioned, there are 2 find operations in nums2, leading to (n * m * m).
@faithcyril513
Жыл бұрын
@@akhilr94 the iteration over nums2 is the finding process so it will be O(n*m) not O(n*m*m)
@whathappened2872
2 жыл бұрын
How old are you now?I'm learning js right now should I do this practice parallel in Python I have a knowledge of Python but not in dsa exactly
@orangethemeow
2 жыл бұрын
I brute forced it but somehow beat 97.62% of python 3 submissions For the second solution, do we need to check if cur is less than the top of the stack when adding it in?
@alexmercerind
2 жыл бұрын
lmao
@lingyuhu4623
2 жыл бұрын
Why cannot I first append element in stack, then while loop? The order makes a big difference, but I dont know the reason
@sanooosai
7 ай бұрын
thank you sir
@junkim4323
2 жыл бұрын
I love your videos! Can you possibly cover problem #979?? It’s an interesting tree problem
@hasferrr
Күн бұрын
i love you neetcode
@leventoz9530
Жыл бұрын
First calculate the next greater element (NGE) of each element in nums2 starting from the right, such that NGE(nums2.length-1) = -1 and NGE(i) = max(nums[i+1], NGE(i+1)). Store the values in a hashmap. This is O(nums2.length). Next, iterate over the elements of nums1 from left to right, accessing the values in the hashmap. This is O(num1.length).
@realoctavian
Жыл бұрын
You can't calculate the next greater like that. Take for instance this input: 5 4 2 1 3 7. At some point the next greater is neither nums[i + 1], nor NGE(i + 1). Cool idea though, I was trying to find something similar.
@tonyz2203
2 жыл бұрын
what software do you use to draw?
@NeetCode
2 жыл бұрын
Paint3d
@droft1312
2 жыл бұрын
Great video! Do you think you could cover #828 - Count Unique Characters?
@kirillzlobin7135
9 ай бұрын
Great video
@marlieemam216
2 жыл бұрын
case if num2 array is 5, 4, 3, 2, 6 doesn't this make the second solution O(m * m ) where m is size of num2 ?
@zergenzerg6853
2 жыл бұрын
At first I was initially thinking you can probably solve this with a stack instead of a hashmap. You can just push the values onto the stack from the end and pop them off as they're greater? Let me finish the video
@eknathyadav8744
2 жыл бұрын
The time complexity is O(m + n) amortized right ? btw great explanation as usual.
@ameydhimte_c5956
Жыл бұрын
Nope its worst case think about it a bit Initially even I thought it would be O(n*m) for worst case but no... every element out of the m elements can be pushed/poped just once at most
@thewanderingguy123
Жыл бұрын
Worst case scenario m = n so time complexity is O(n+n) = O(n)
@Vishal_84_k
2 жыл бұрын
2nd bro💓💓 lots of respect💥💥
@ibrahimmalik4155
11 ай бұрын
In the for loop, you don't need to check if cur is present in nums1. As the question states that nums1 is a subset of nums2. Just a little nuance that could help out your solution. Great job!
@olawaleojodu6704
Ай бұрын
Without the checks, I think it will throw a key error if you're trying to lookup an element not present in an num1 hashMap??
@samagrasinghtomar79
Ай бұрын
The stack push op needs to be done unconditionally. The O(m+n) solution won't work as it is in the video.
@greatestever2914
2 жыл бұрын
the easy are harder than some of the hards
@JasonAhn-u5u
Жыл бұрын
@12:59 Can anyone please explain to me line 4, where it's written as { n: i for i, n in enumerate(nums1)}? Shouldn't it be {i:n for i, n in enumerate(nums1)} to match both i and n? Thanks!
@sivaprakash_prabakaran
Жыл бұрын
we want the numbers as keys and their index as values, to quickly look up and find the index of the numbers; thats why we are doing n:i; { n: i for i, n in enumerate(nums1)} is a short hand for creating dictionaries on the go; like we have list comprehension, this is dictionary comprehension. enumerate will result in a list of [index, value] pairs; we match the index with i and numbers with n (i, n) and we use those i and n as key:value pairs of dictionary but in reverse; we use numbers n as keys and index i as values; so i:n
@yuniorsanchez8578
2 жыл бұрын
how can we code this "num1index = {n:i for i, n in enumerate(nums1)}" lets pythonic and more simple?
@0Mynameisearl0
Жыл бұрын
For anyone else that still needs it: {key: value for value, key in enumerate(nums)}
@JasonAhn-u5u
Жыл бұрын
@@0Mynameisearl0 Mind if I know why we're switching it around? Should it be {value: key for value, key in enumerate(nums)}?
@neoncold2679
2 жыл бұрын
Is this question supposed to be easy category? =(
@atifzia124
Ай бұрын
n2 is fine, 😢 it ain't easy to lower complexity at easy level
@ruchitagarde4642
2 ай бұрын
What would be the solution if nums2 = [5,1,6] nums1 = [5,1] This result=[6,6] or result = [-1,6] ?
@sachidanandanradhakrishnan6630
21 күн бұрын
[6,6] r8?
@davyroger3773
2 жыл бұрын
All you gotta do is make a sorted copy of nums2, then for every i in nums1, slice the ordered list at the ith position to get all of the possible values greater than i in nums 2. Then call a separate method that takes the sliced ordered list , the i value , and the og nums2, make another slice at pos i this time in the og nums2. def find_elem(self,ordered_slice, nums2, value): pos = nums2.index(value) unordered_slice = nums2[pos+1:] now we can do a quick list comp to keep all of the values in the unordered slice list if they are in the ordered slice possible = [i for i in unordered_slice if i in ordered_slice] At this point there will be two possible outcomes, 1 the list is empty indicating that we couldn't find any value greater than i to the right of i, 2 that we were able to find such a value(s), and have stored them in the order that we found them. So all thats left to do is if possible == [ ]: self.tracker.append(-1) else: self.tracker.append(possible[0])
@akhilr94
Жыл бұрын
bro sorting is O(nlogn)
@0yustas0
Жыл бұрын
@@akhilr94 Counting Sort is O(n+m)
@nikhil199029
2 жыл бұрын
Ut should be medium difficulty i think
@ujjvalw2684
Жыл бұрын
how is that an "Easy" Question... maybe coding isnt for me lol
@mrlectus
Жыл бұрын
How is this Easy?
@ivanwen8335
2 жыл бұрын
can you do "Next Greater Element II"?
@__________________________6910
2 жыл бұрын
Third bro
@abhishaiwinston9794
2 жыл бұрын
First view
@lb9gx
Жыл бұрын
Python brute force solution without using Hashmap. The idea I used here is to begin at the end of nums2 and iterate from there. class Solution: def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]: res = [1]*len(nums1) g = -1 for i in range(len(nums1)): for j in range(len(nums2) - 1, -1, -1): if nums2[j] > nums1[i]: g = nums2[j] if nums1[i] == nums2[j]: res[i] = g break g = -1 return res
@necx5510
Жыл бұрын
my solution was O ( n * m) but still had the same runtime as your optimized code lmao
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