Excellent one!

"another day another problem" -anyone who works in tech

Lot of time saved, Bcz of you. Thank you....

Please pause while you explain

It is better to solve in O(n) by simple getting the 3 max and the 2 min in one pass.

thhxxx

wow great solution

very precise solution. thanks

can i ask question in sliding window in excatly this part : # Expand sliding window if total < k: end += 1 if end == len(nums): break total += nums[end] this snippet of code should be like this: # Expand sliding window if total < k: if end == len(nums): break end += 1 total += nums[end] because imagine if we have start pointer anywhere in array and end on element previous last one and total < k: so we need to expand so we increase end pointer by one and if we check first the condition of end == len(nums) will be true and it will break loop and return counter for example = 1 but if we check first when end on element previous last one the condition will be false and we increase end pointer by one and total will change and it will loop again and check if total > k or total == k or total < k lets say it will be total becomes equal to k it will increase counter by one and shrink window and it will break the loop after this and correct for me if iam wrong

Thanks a lot for this, i just wanted to ask how do you find such patterns? Like how did you come to find out the step = 2 * rows -2 I just want to know the thought process so that i could practice more questions like this with the right approach

Thanks

There is a dynamic programming solution that’s more optimal. O(n) time in exchange for more space O(n)

Thank you very much

Thank you very much

not correct

Why are you starting from 0 and ending at n instead of (n-1)? No list/array starts from 0th element and ends at nth element. It always ends at (n-1)th element. so why have you done this? Going from 0 to n instead of 0 to (n-1) is not intuitive at all. I cannot believe this is intuitive to you. you must have memorized this.

Excuse me. I guess your code with stack doesn't work, because str objects does not support item assignment

good video

Thank you! very clear explain.

dude sounds depressed af

Thanks for your video! Just one thing, in the worse case scenario we would have four letters for digits 7 and 9, so the time comlexity would be O(4^n)

This solution is combination of art and magic

Retarded question but thanks for explaining.

The first solution is wrong. It will miss cases where the target could be the sum of the same number twice, if it isn't the first number. You got lucky with the test cases in your example because you are using j from 1..len(nums) instead of 0..len(nums). You should instead be doing j in range (i+1, len(nums)) to search all of the numbers further down in the array. This is in addition to the issues other people have brought up with the third solution.

I believe that complexity of the third one is O(n*(n/2)), as you have to iterate by 1 element until you get to the right number, and expected/average number of elements to investigate is n/2. To have O(n*log(n)) you would have to divide the search space by 2 on each iteration, while here you divide search space by 2 only on the first iteration.

Great way to promote a channel

i want to see more of your videos

Can you upload a divide and conquer approach?

Cool solution! Thank you!

Thanks a lot, videos are very useful!!!

OmG start posting again we need you😢

Nice this is probably the best explanation

Hi bro and thanks. Sliding Window Technique works also for an array with postive values and zeros ?

It was pretty easy, I don't undestand with it was to hard to me. Thanks for ypur explanation

你好冒昧請教你,這串代碼,他有個BUG,無法正確地回答出head=[1,1,2,1] 請問你能指出她錯誤的地方嗎 class Solution: def isPalindrome(self, head: Optional[ListNode]) -> bool: #空節點直接返回True if not head : return True #初始化reverse_link reverse_link=None cur = head while cur: #記錄下一個 next_node=cur.next cur.next=reverse_link #更新標頭 reverse_link=cur cur = next_node while reverse_link and head: if reverse_link.val != head.val: return False reverse_link = new_link.next head = head.next return True

You have a nested for loop. Your time complexity is certainly not O(n).

[3, 5, 5]

Thanks a lot!!

Thanks for the great job! Sometimes slides are rotated too fast kzbin.info/www/bejne/bpmVhoKGeah6hq8

Thanks a lot for your help!!!

Can we use sliding window on first and last string

Faced strange error , running it on Leetcode, cant fix it TypeError: Solution.expandAroundTheCenter() takes 3 positional arguments but 4 were given oddPalindrome = self.expandAroundTheCenter(s, i-1, i+1)

THANKS A LOT MAN!!!

poor solution, O(1) space solution also exists, you've just explained the most trivial brute force solution

Hello, can you explain why we check if root <= prevN? you said in the video that b/c stack uses LIFO(last in first out) that we can use that to our advantage by checking for root <= prevN. Sorry Im very new to this so I'm trying too understand the reason behind every step

nicely explained, thanks!

Good explanation, much appreciated

Sir pls make vedio specifically graph and dp

s[:] = s[::-1]

IT DOES NOT PASS THE USE CASE :- [0,0,0,0] OR [0,0,0,0,0]