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Blessing to watch your videos. Keep going ....

The best video explanation for this problem is this video right here , trust me when i say that I've looked for other videos for this problem and this is the best !! Thank you

Probably the best there is when it comes to explaining this stuff

Thank you for keep posting problem solving videos! It helps a lot. 2 days back Ive searched for this problem.

but we have used list(string) and the question states use constant extra space .ain't that a violation

Can't stop loving your videos.

Great explanation❤❤thank you very much for making videos like this.... Please don't stop this i am learning a lot from your videos

Great explanation please continue this series❤❤

one of the best explanation , without getting bored i watched this video.

See there needs to be a pointer manipulation tag in leetcode. I had such a hard time visualizing O(1) memory. Great vid!

thankyou for sharing , even though i had woorked upon the solution was stuck with edge cases

Easy money i've ever made => return a+b 🐱‍🏍🐱‍🏍

3:47 what does last down right {s:S} mean ? If we a re talking about binary boolean, there should be only True (T) of False (F). in this case T for word "ts".

Your my hero, every time i stack i search for your tutorials direct, the way you can explain the problem in the leetcode IDE helps me a lot durign interviews

Hi

You are an inspiration Deepti

Super Quality Videos Deepti .

Under rated gem. Didnot expect such quality after seeing the subs count. Keep it up!

Your videos are gold. Thank you

thenks for the lecture misss it was really helpful!

Happy New Year! I really appreciate your content and you doing examples after writing out the code. Has helped my confidence out a lot!

clever solution. clear explanation. I don't think you need to slow down at all!

Nice approach! You did it easily and I could enhance my code with your explanation! Thank you 💚

Bravo this walkthrough was great!

Thank you for the explanation it was so clear! It is a great solution and It is too short i wrote more lines of code before I saw this video

Simply wow.

insane answer for this question, I've been locking in and watching multiple videos to find the best way and easiest answer to remember. Thanks to this video you taught me a new method in python.

Хелло

nice explanation

Can I make money in stocks using your Algo ?

Are you the star in the asteroidCollision func ?

Elegant Deepti with eloquent solution.

Vivacious girl Merry Christmas to you !

Amazing Dear Deepti. Salute to your passion.

This should honestly be a medium leetcode question😭

Thanks again for these fire content. During an interview setting would dry running / or stepping through it the way you’re demonstrating be good? I’m trying to iron out how to approach interviews

comments with code class Solution: def asteroidCollision(self, asteroids: List[int]) -> List[int]: # Initialize empty stack to store asteroids stack = [] # Iterate through each asteroid in the input array for ast in asteroids: # Enter while loop when: # 1. stack is not empty AND # 2. top asteroid is moving right (positive) AND # 3. current asteroid is moving left (negative) while stack and stack[-1] > 0 and ast < 0: # Get the last asteroid from stack last = stack.pop() # If last asteroid is bigger, it survives # Current asteroid is destroyed if last > - ast: stack.append(last) break # If both asteroids are equal in size # Both asteroids are destroyed elif last == -ast: break # If last asteroid is smaller # It gets destroyed and loop continues # to check against next asteroid in stack # This else clause executes when: # 1. Stack is empty OR # 2. No collision occurred OR # 3. Current asteroid destroyed all asteroids it collided with else: stack.append(ast) # Return final state of asteroids return stack

Happy New Year! Thanks for the content!

Happy new year deepti.. hope this year with your videos i land a FAANG job🤪😅

you said o(1) but at 8:27 you pop off two elements? still o(1)? Doesnt that mean it can be o(n)? thanks 👍

thank you deepti

you can write a gcd function to beat 100%

Such a crazy thing that you uploaded when I needed this the most😭

Very elegant solution indeed ! Here is one without using "math.gcd" (not sure which one is faster), edit: your solution uses Python's built-in math.gcd which is implemented in C and is highly optimized, while loop is slower. class Solution: def gcdOfStrings(self, str1: str, str2: str) -> str: # Helper function to compute GCD of two numbers def compute_gcd(a: int, b: int) -> int: while b != 0: a, b = b, a % b return a # 1. Check compatibility if str1 + str2 != str2 + str1: return "" # 2. Compute GCD of lengths gcd_length = compute_gcd(len(str1), len(str2)) # 3. Extract the candidate string candidate = str1[:gcd_length] # 4. Verify the candidate if (str1 == candidate * (len(str1) // gcd_length)) and \ (str2 == candidate * (len(str2) // gcd_length)): return candidate else: return "" Recursive solution with "if" instead of "while" loop: class Solution: def gcdOfStrings(self, str1: str, str2: str) -> str: # Recursive function to compute GCD of two numbers def compute_gcd(a: int, b: int) -> int: if b == 0: return a return compute_gcd(b, a % b) # 1. Check compatibility if str1 + str2 != str2 + str1: return "" # 2. Compute GCD of lengths gcd_length = compute_gcd(len(str1), len(str2)) # 3. Extract the candidate string candidate = str1[:gcd_length] # 4. Verify the candidate if (str1 == candidate * (len(str1) // gcd_length)) and \ (str2 == candidate * (len(str2) // gcd_length)): return candidate else: return ""

Great explanation, thanks.

Awesome clear explanation 💪

Merry Christmas Deepthi! Been learning from you for a while, your explanation through examples really help me understand things. Thank you so much for your help and happy holidays!

Hey there, How can we convert this number 3682 to Roman numbers? I am confused about how we will convert the number 6 to a Roman number according to your code.

Found this at the perfect time!Thanks, your explanation helped me to realise why BS can give optimal solution. Merry Christmas 💖 and Happy Holidays ⛄

Winter is coming !