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You're the best

Amazing vid + i learned something new on how to work with queues. You got a subscriber

Underrated channel. I love these! Thank you :)

Brilliant. Thank you for putting it together.

This question was diffuclt for me, I understood we needed a prefix sum and I could see similiarties to two sum but when i began writing down the solution on paper my brain just cramped. The solutions on leetcode were not helping either. I'm glad I found your video! it was very clear, I may have to watch a few times to gain some of intution you have. Thanks!

Thank you so much for great content. I love watching your solutions, so easy to follow!

Absolute exaplanation !! Thank you

Amazing explanation thank you

Awesome explanation

Thank you!

Whoever made this problem I hope both sides of your pillow are warm

Hi, thanks for the solution. However, it doesn't seem to work any more for some test cases. For the two test cases below, if you change the order of the two for-loop, one of them will fail. [[2, 1, 1, 1, 1, 2, 2, 1, 0], [0, 0, 1, 2, 0, 0, 0, 2, 2], [2, 0, 2, 2, 1, 2, 1, 0, 0]] [[2, 3, 0, 0, 1, 0, 1, 2, 1], [0, 1, 3, 3, 0, 0, 2, 2, 0]]

Good One, keep this up

thank you for this

what happen if the two monkey swap there place in case of even number vertex, isn't other possibility to avoid a collision?

Nice explanation!

thank you

where does actual backtracking happens in code?

That was an incredible explanation

Superb Explanation

which drawing device you are using?

such an amazing soln. How did you come up with the last return statement. That line is magical

thank you and love from India

Hey Bro Youre Arab right? Nice Explaination by the way.

NICE SUPER EXCELLENT MOTIVATED

so it means i was able to solve 90% of this problem on my own but i rather than dividing c/e[0], i was doing every loop which gave me tle. and whole time during contest i was thinking how to even convert it into O(N) 🙂

thanks!!

No idea about the other solutions you've solved. But this one is really simplified and nice way to explain. You've made it piece of cake and I was able to solve few other hard Leetcode questions with simply by swaps. Keep it up don't worry about number of views or comments. Cheers

good explanation

best video for this problem ! thhanks

Thank you for this simple explanation!

Thanks, great description.

Thank you so much. I love the clean solution. Subscribed, looking forward to see more videos from you in coming days.

Hi ! Thanks for the awesome explanation The above code fails for one test case (might have been upated) words = ["A","B"] , result = "A". I think the what they mean by no leading zeors is valid if the number if of length >= 2. Here we can see that the requirement is that B has to be zero!. Hence no leadning zeroes is applicable only if len(word) > 1. 2 minor fixes fixed the code for me. One was this : here we can use 0 even if the column == len(word)-1 if and only if the word is itself of length 1 # digit assignment already exists -> use it if letter in letterToDigit and not (letterToDigit[letter] == 0 and column == len(word) - 1 and len(word) > 1): return search(column, row + 1, bal + sign * letterToDigit[letter]) The other fix was Here we skip zero assignment even if column == len(words)-1 if and only if len(words) > 1. for d, c in enumerate(digitToLetter): if c is None: # the second condition is to prevent leading 0s if d == 0: if len(word) > 1 and column == len(word)-1: continue Again, thanks for the awesome explanation.

Thank you!

Very important. ❤great

that's just awesome. I was confused over the prefix sum In 2d array

From minite around 22, I believe in the matrix, should be "(i+1, k +1)"

I could not understand why you did return min(money/7, children -1)? money/7 would occur when money/7< children and children-1would occur when money/7 > children. so how you put both condition in one line?

U explained the concept very well thank u for helping out

whoa! I had never noticed LT has a button that analyzes the time complexity for us!

Thanks for the explanation! I was unable to come up with a (sub O(N^2)) solution on my own, and reading LeetCode's explanation several times didn't help. This is good.

Thank you so much ❤

thank You

Welcome back 🎉

thank you, great explanation

You're Impressive. thank you <3

THANK YOU

Thank you so much for your nice explanation, it really is easy.

The cleanest and readable solution I've seen for this problem, thanks