Continuous Subarrays | Leetcode 2762

God damn, my head hurts hahaha

Had the sliding window part down, the maximum and minimum threw me off a bit but the formula is the part where I'm thinking "how the heck??"

As you are moving pointer again to left when we found any fault
Isn't it is disturbing cpu cycle? Because it is making knot of pointers during computation
Can we optimize this knot ?

sir is it possible if you can explain the segment or fenwick tree approach for this problem?

Nice explanation, may I know which blackboard you are using?

Very beautifully explained !

Suddenly my brain is not braining😂😂

This is first time I see the window is going like a swing. That made me confusing(especially the double count & subtracton🥲). If I had never seen this algo my dumb brain could never figure out the soln.

sir can i use priority queues to maintain min and max values so as to avoid the left movement of L pointer where i can update L pointer to the index choosen from the priority queues after checking the condition?

Your python does not submit on leetcode and provides wrong output. Kindly, please check

could explain why the n*(n+1)/2 come about?

I was with you until you start talking about moving L pointer.

what an explanation!!!

Time to practice more sliding window problems 😢

Thank you

i think we can adjust the left pointer also simply by maintaining the index and updating the index if the same number occuring multiple times

Moving left pointer back to avoid duplicate was a bit tricky...

This guy ❤

Loving the explanation Hating the problem

would maintaining a index whenever I update max help last wala update will be stored and my next L would be ind+1;

good explanation as always, many thanks

Your Python code in the link need minor enhancement. n = len(nums) should be used instead of right pointer to calculate the final number of subarrays to be addes outside the loop on line number 126. It should be win_size = n - left instead of win_size = right - left.

the python solution will give you a wrong answer. After finishing the loop, you can not use the right variable because the value is n-1
win_size = len(nums) - left
count += (win_size * (win_size + 1)) // 2

why don't you move the left pointer to right man its striaght forward and less complicated for explaining!

Good explanation Sir.

And i was just counting subarrays

wonderful explain

Excellent expalination, I too comeup with the exact approach but struggle to code it. Thanks man.

Hi sir, I think exactly the time complexity is O(2N) in the worst case l and r is both come to the last of index. Please confirm if my comment is false.Thank you so much and have a good day

Nice explanation. Thanks. Just one tiny bug: when moving back left, we need to ensure we are not moving out of the array

thanks :)

great call to include the complexity analysis! i came up with a similar solution but rejected it because i thought moving left pointer back would cause the complexity to become N-squared!
thanks :)

using deque makes it o(n) both time and space