วีดีโอ

EMO !!!

why i can't do something like this: while True: ans = rand7() + rand7() if ans <= 9: return ans + 1

give me your time machine

this is so helpful! love how you count the total subarrays using the left & right bracket!

yeah, I'd totally come up with that on the spot... :')) dp(n^2) solution though, I think so. But I will remember this in case I need it

Simple makes PERFECT!

Thank you so much kacy.

loved it, one of the easiest and best explanation. Thank you !

tqq for this stuff ^^

Thank you for the iterative explanation

Thanks! The editorial did not really explain it and the popular channels stopped at the iterative solution

solved it intuitively using max heap,, but this is more efficient.. alas less obvious to wrap my head around it. thankyou !

this guy is so cool

The proves were real nice. I was struggling with it. Thanks.

Love you solution (and your voice)

I got this question for amazon's OA and also a question similar to this for IBM's OA on the coding portion so this is a good question to review. Too bad i didnt know how to do it 😭

instead of adding only odd length why are u adding all ?

epic vid! thx!

Good job brother, you explained the whole algo withing the first 4th minute. thank you. keep it up.

omg, the prefix sum explanation was so good

good explanation

is it really hard to grasp or I'm dumb?

I missed why this algorithm is nlogn

Awesome!

Great Explanation! Thank you for such a beautiful explanation!😊😊

Bro u R me Fr

I put this exact code into my IntelliJ IDE and it doesn't work. The power function doesn't work. Any ideas? I am trying to write an asymmetric encryption program and need to figure this out.

How about using two pointers. The first pointer starts from the beginning of each range and increases by 1 until it finds the first pipe and is less than end of range. The second pointer if (left pointer +1) and increases by 1 until it finds a pipe. We add the difference between both the pointers -1 as the current count of the stars. Then, we make the first pointer to start from the second pointer. And the secondpointer starts from firstpointer +1. We always check if the pointers are less than right.

i like your hair, do not you get dandruff !!

Thanks man, it was very helpful and so much to learn form this video more than what was needed to solve question. Keep up the good work of spreading knowledge

Nice explanation and solution. appreciate your work.

Good solution. I like that you did it with regular arrays makes it easy to transfer the logic to other languages.

Incredible video! really good explanation thanks so much

Very clever explanation. Thank you so much!

Thank you brother beautiful explanation

awesome!

damn bro great explanation please continue doing vedioes

This guy gives off the vibes of a HERO !

Hey Kacy the explanation was so good. Can you teach monotonic queues as well?

Out of all the vides on youtube, I understood your explanation! Thank you!

You know it's a great explanation when you can code and submit the solution for yourself without even reading a single line of code from the instructor lol. Thank you!

2007 vibes

How did you come up with the idea of choosing the bottom left or top right for calculating the no. of elements <= mid?

Thanks for the clear explanation! Very helpful! 👍

Code Ninja

THANK YOU!!!!!!!

It is very useful. Thanks mate.

Great Kacy...

Great explanation in Java code. Better than some of the Python solution as they use built-in functions that let you "see" the solution as clearly.

Thanks for the explanation, able to understand the logic behind the solution for range addition and related problems, Thanks.