Thank u so much... Now I properly understood prefix sum topic

man this problem almost made me cry thank you for the explanation great video

@crackfaang

6 ай бұрын

Glad you found it useful. I really hate all the problems that are like "divisible by K", "subarray sum equals K"... they're simple to code but hard to understand

@firomsamt7642

6 ай бұрын

@@crackfaang yeah those questions that involve math and geometry are really hard to figure out

What helped me understand the motivation: Suppose we have two subarrays of sums, A, and B. Sum A has a remainder r and sum B doesn't have a remainder(divisible by k). We also know by the definition of divisibility that -- A = n1*k + r B = n2*k + 0 where n1 and n2 are some constants(not really important) adding them(pay attention to this part) A + B = n_1*k + r + n_2*k = (n_1 + n_2)*k + r WAIT THE REMAINDER IS THE SAME!! THIS HAS TO MEAN THAT ONE OF THEM DIDN'T HAVE A REMAINDER AND WAS DIVISIBLE BY K.

Very Clear and Helpful! Thanks a lot !

I've seen a lot of videos on this problem and this video is the best in explaining the solution. Thank you :)

Wow, that is very nice explanation. Thank you man!

Thanks great explanation

This was very useful for understanding the reasoning. thank you. I went through a bunch of other videos and your reasoning at 3:30-4:30 is what made it most useful to me. The "philosophical rant" was fine too, it's reassuring for those of us who couldn't figure it out from a cold start.

@AndrewCLC

Жыл бұрын

constructive criticism (and for anyone else coming to this that struggled) -- at 7:30 it gets a little confusing why you just added 3 to the total and didn't update the map. I reasoned this out: At -3, the remainder of 4 % 5 is 4. Map becomes {4=>4} Result is [-2,-3], [-2,-3,0], [-2,-3, 0, 5]. You add the current plus everything before. Result = 6. So if there was another X%5=4 down the line, it would be [x], [x,-2,-3], [x, -2, -3, 0] and [z, -2, -3, 0, 5],

Such a good example for the divisible by k (1,21).

Thank you!

Great video!!

amazing video only video i would prefer specially for this problem

Thank you so much

Very good explanation bro

good shit brh

Using defultdict(int), every remainder that is calculated will be 'present' in the remainders defaultdict, and it will have a default value of 0 if it hasn't been encountered before. So the if statement on line 11 will be true for all the keys. right? Maybe we can remove the if and just keep "count += remainders[running_sum % k]"?

Nice explanation. Thank you. The if statement is redundant though

the question is so dangerous that the police has to warn us

space would be O(k) cause the map would have max k elements since you're taking the modulo