Raise hand if you still didn't understood it by now 🤚
@theruler10969 ай бұрын
If you face interview question like this, you're not unlucky. You're just dead.
@tryundel9 ай бұрын
Sorry a noob here... how bad is it to solve it with O(n^2) on an interview?
@anti-tankartur6779 ай бұрын
@@tryundel you're basically showing that you have a grasp on the basics but you're not good enough to optimize solutions. For prestigious companies, I think it's generally perceived as red flag.
@ducthinh24129 ай бұрын
@@tryundel The obvious brute force approach is n^2 so you will need to come up with a O(n) solution. The interview is at least 45 mins and if it's just 1 question, after you provide the brute foce, you will be expected to come up with the optimal solution
@haoyu12479 ай бұрын
you will not be moving forward to next step, even it is an internship position (my experience with one of the FAANG)@@tryundel
@chisomedoka4019 ай бұрын
what we call "your village people" over here
@davi_singh9 ай бұрын
Everytime I think I have a handle on leetcode, I get a question like this one and I wonder why the heck am I so dump. Thanks @NeetCodeIO you have the best explanations
@abrahamlincoln57249 ай бұрын
Same reason why I hesitate to apply to jobs in big tech where LC questions are common. How do you solve a question if you never encountered similar question(or pattern) before?
@davi_singh9 ай бұрын
@@abrahamlincoln5724 same boat brother, I am shit scared cause of the same thing. Have been grinding LC for 3 months gone through 4 study plans but don't feel ready at all
@soumyajitganguly25935 ай бұрын
@@abrahamlincoln5724 you practice enough problems so that its unlikely to be a new pattern
@reckyru9 ай бұрын
Disgustingly hard problem IMO
@shreehari25899 ай бұрын
Exactly
@elyababakova21259 ай бұрын
crazy question
@hengyulee43199 ай бұрын
For the modified code, because when calculating left and right we are computing the distance of two elements and appending/prepending does not affect the distance. Two -inf solve some edge cases and empty the stack (except for the first -inf) in the last iteration of the for loop. Correct me if I am wrong.
@m.varunreddy73659 ай бұрын
couldnt understand, this q is definitely not medium
@iliadmitriev019 ай бұрын
you could also add -inf only to the back off the arr, this absolute minimum will purge the stack without involving extra code at the end of the algorithm
@coolgamertm44119 ай бұрын
Secret POV : Leetcode daily hard problem help this channel grow more
@codingoak47019 ай бұрын
The reason that optimization works is, for all the small values that dont get removed from the stack, the result will be dynamically updated for the distance to the ends of the array.
@chiragsrivastava42179 ай бұрын
class Solution: def sumSubarrayMins(self, arr: List[int]) -> int: MOD=10**9+7 stack=[-1] res=0 arr.append(0) for i,val in enumerate(arr): while stack and val
@alxolr9 ай бұрын
Gosh if you get one of this during an interview you are done.
@arunrajput10079 ай бұрын
hey @NeetcodeIO i am not sure for any new medium level problem i would be able to come up with a solution in 25 mins and this includes explanation + logic + coding + edgecases + dry run. This is not possible if one has already solved this question before. Just need your thoughts whether this is possible for everyone.
@sankalppatil29949 ай бұрын
Rip I still dont get it
@sproutboot9 ай бұрын
Same
@quirkyquester19 күн бұрын
great one! Thank you!!
@jianhuahe40669 ай бұрын
I am wondering for n elements the total combination would be n square. Should it be multiple 2 n times and minus 1? Taking an example: n = 2, there are 3 combo, n =3: there are 7.
@kimmyliu55096 ай бұрын
for the second for loop, when you iterate over a stack. are you iterating it like an array? it seems that you assume you can iterate from the earliest added element all the way to the last added element, which is a queue not a stack. in stack you can only iterate over it via the last added element no?
@sk_41429 ай бұрын
A humbling problem indeed
@coffeebytes32579 ай бұрын
💀 why cant my brain figure this stuff out without help 😭
@lingyundai9649 ай бұрын
more practice more practice we got this
@shawncodinghouse7 ай бұрын
It's too complexity for me, though I have solved 243 questions.
@MohammadSohail-j6x2 ай бұрын
real
@rohitudamale40912 ай бұрын
Good Explanation!!
@KhyatiSatija9 ай бұрын
Hello i have been following you since really long , I usually understand what you teach, but this video was a bit complex for me. But I have done this question. Here's my Python code. Its the same approach as yours, just more simple and brute forced. class Solution: def sumSubarrayMins(self, arr: List[int]) -> int: arr = [float('-inf')] + arr + [float('-inf')] previousSmaller = {} preStack = [] nextStack = [] nextSmaller = {} #hashmap result = 0 MOD = 10 ** 9 + 7 for i in range(len(arr) - 1, -1, -1): while preStack and preStack[-1][1] > arr[i]: j, m = preStack.pop() previousSmaller[j] = i #prev smaller element preStack.append((i, arr[i])) for i in range(len(arr)): while nextStack and nextStack[-1][1] >= arr[i]: j, m = nextStack.pop() nextSmaller[j] = i nextStack.append((i, arr[i])) for i in range(1,len(arr) - 1): minTimes = (i - previousSmaller[i]) * (nextSmaller[i] - i) #number of subarrays for which the element at index i is the minimum result += arr[i] * minTimes result = result % MOD return result # arr - 0 3 4 5 6 1 4 6 7 0 # index - 0 1 2 3 4 5 6 7 8 9 #for handling duplicates,we do >= while calculating for nextSmaller and > in prevSmaller #vice versa will also work
@36saurabh2 ай бұрын
Around 14:13, if I understand it correctly, does left always evaluate to 1 on line 10? Correct me if I'm wrong. Since all the elements in the stack before the popped element m at index j will be less than m, hence number of subarrays with m as the smallest element on the left of m will always be 1? Is this correct?
@vaishnavejp92472 ай бұрын
i had the understanding that monotonic stacks come in handy for problems like NGE because we dont just want to keep track of minimum element youve seen, but how close it is to the current element. that is, the closeness matters as well. but for problems like this, i dont understand why a monotonic stack is even needed? cant i just keep track of the minimum most element seen so far? this and trapping rainwater problem as well. someone pls explain why a stack is needed instead of just using mini pointer
@SickleM9 ай бұрын
Can't believe I did this in first try!
@upendrakumar-bw1jd6 ай бұрын
Actually i am big fan of neetcode because he is the only one who gives better intitution but this time really want to say this time neetcode fail to explains this problem
@chien-yuyeh93869 ай бұрын
Thanks for the amazing video
@dingus23329 ай бұрын
Could this be solved with Segment Tree ?
@sbera879 ай бұрын
Stack grows upward, but I get the point
@HeroHunter079 ай бұрын
Hard one
@XEQUTE9 ай бұрын
wohoo , you did it!! ( hello from yesterday)
@soumithjavvaji33104 ай бұрын
Either getting rejected or giving a O(n^2) solution is predefined , if the interviewer asks this!
@rajchinagundi74989 ай бұрын
class Solution: def sumSubarrayMins(self, arr: List[int]) -> int: mod=10**9+7 n=len(arr) @lru_cache(maxsize=None) def solve(i, n, mn): if i == n: return 0 curr = arr[i] if curr < mn: mn = curr return (mn + solve(i + 1, n, mn)) % mod sumTotal = 0 for i in range(len(arr)): sumTotal = (sumTotal + solve(i, len(arr), inf)) % mod return sumTotal. Hi Neetcode I wrote this recursive approach and it passes 82/87 test cases, and then fails with memory limit exceeded error, I was wondering it this question is only possible via iterative approach?
@CrabGuyy9 ай бұрын
every iterative code can be done recursively, i would suggest trying to implement tail recursion because python optimizes it for memory
@rajchinagundi74989 ай бұрын
Doesnt make a differnce if recursive stack overflows@@CrabGuyy
@s016_aviratshambharkar29 ай бұрын
Subarrays should be n (n + 1) / 2 right ??
@ahmedbenromdhane-q9s9 ай бұрын
there are n(n+1)/2 sub arrays not n²
@kartikhegde5339 ай бұрын
he probably mentioned about overall complexity. which is still n^2
@pdjeowudjx9 ай бұрын
i was waiting thx
@Logeshwar-s7m9 ай бұрын
could you also upload videos for leetcode biweekly contests
@akialter9 ай бұрын
I just solved this and the video uploaded 😅. Still watching tho, I like your explanation
@krishnakanthati85109 ай бұрын
How did you solve this?
@shellingford55349 ай бұрын
I still don't understand why we are doing j + 1 in the else portion. Can someone please explain this to me like I am 5...
@anti-tankartur6779 ай бұрын
Basically, else is only done when the stack is empty. If the stack is empty after popping the last element, that means it is the smallest number in that array upto THAT index, which means that it is going yo be present in every subarray and will also be the minimum of all these subarrays. For example, suppose the stack is empty and the element popped is 1 at the index of 2. At index 0 and 1, you have 3 and 2. So for all the subarrays from the index 0 to 2, 1 is going to be the minimum nunber and hence j+1 (2+1 =3 ) as 3 subarrays will have the minimum as 1. Hope this helps.
@ducthinh24129 ай бұрын
Let's say we have: nums = [2, 3, 5, 1] index = [0, 1, 2, 3] Let's say we are currently at index 3 (value = 1). Our stack (index, value) is: stack = [(0,2), (1,3), (2,5)] Since 1 is less than all of the values in the stack, we pop everything from the stack. When we get to (0,2), the top of the stack, the right count is 3. For the left count, since the stack is now empty, there is only a single value from the beginning of the array till this point, which is 2. Its index is 0, hence we need to add 1 to the index to get the count of subarrays from the beginning up to and including 2
@AdenGolden9 ай бұрын
+ 1 is include the jth element itself if the stack is empty is 0 + 1(itself) = 1
@JosephMuturi-j4d4 ай бұрын
@@anti-tankartur677 This is an amazing explanation,,,,,only few can understand how figuring that out almost solves the problem,,,even neetcode couldn't explain that one,,,,it's genius
@jamjam34485 ай бұрын
but total number of subarrays is n*(n+1)/2 instead of n^2.
@harshitsharma26394 ай бұрын
thats order n^2 tho
@jamjam34484 ай бұрын
@@harshitsharma2639 You're talking of complexity, not the number of subarrays. They are different
@harshitsharma26394 ай бұрын
@@jamjam3448 i know that dude. The time to traverse and build the result will still be O(n^2) thats what neet was trying to say.
@jamjam34484 ай бұрын
@@harshitsharma2639 yeah i think that's what he was trying to say also but he kept saying number of times instead of the time complexity as the two are different.
@randomisedstrength50509 ай бұрын
I didn't understand anything related to the use of stack in the code 😶🌫
@ehm-wg8pd4 ай бұрын
if you get this interview question, omaewa mou shindeiru
@pastori26729 ай бұрын
too much intuition
@haoli89834 ай бұрын
this question is so hard == why just medium ?
@Moch1179 ай бұрын
This question made me question Leetcode smh Did you solve it on your own
@vinsin46199 ай бұрын
can you explain the last code
@vinsin46199 ай бұрын
the negative inf added to front and back of the array
@harryliu7999 ай бұрын
The neg inf at the back ensures that the elements in the monotonic increasing stack are processed, as everything inside is smaller than neg inf (same as the original loop over the stack).@@vinsin4619
@aaditya_875 ай бұрын
hey man , its time for u to answer here why u did that
@sumitrohilla14946 ай бұрын
how can I continue watching this video, when 1 min in you are giving wrong explaination.... no. of subarrays are not n^2 its n(n+1)/2
@anandp74825 ай бұрын
its O(n ^ 2) . n * (n + 1) /2 is still O (n ^ 2)
@sumitrohilla14945 ай бұрын
Ohh, sry for the wrong explanation, I am not talking about time complexity.. He said no. of subarrays is array are n^2..but no of subarrays are n(n+1)/2
@eyesgotshowyo78006 ай бұрын
i understood nothing
@shreehari25899 ай бұрын
Def hard not med
@rishujeetrai57803 ай бұрын
😭👍
@johnthompson13279 ай бұрын
first
@jaatharsh9 ай бұрын
checking solution in Python is useless for understanding, when iterating ur iterating from top of stock to bottom or reverse. Python is such a sloppy choice of language for DSA