I attempted the D problem in JAVA 21 64bit with the simple logic of storing the remainder of a[i] in pairs along with their frequency in HashMap and check if (x-(a[i]%x)),(a[i]%y)) is already present, adding 1 to answer if it happens. I also make the remainder equal to x if it comes out to be zero.
Heres my solution
but for some reason it fails the 12th test. I don't know what on God's Earth it is, but I then just convert the same logic to C++ and its accepted.
Where am I going wrong? What seems to be the problem here? Is there an alternative? A friend of mine suggested using a 2D array instead of a map an hence saving the trouble of creating pair class in JAVA but I could not fully grasp his concept. Please help me out here.
Auto comment: topic has been updated by KiZamaDo17 (previous revision, new revision, compare).
Auto comment: topic has been updated by KiZamaDo17 (previous revision, new revision, compare).
I don't know much about Java but it looks like your "pair" implementation such return a 64-bit integer since to hash it in
a * M + bthenM should be greater than b.In addition, the answer may get larger than 32-bit integers so consider using 64-bit integers in this case.
Hey thank you for your reply. I am already using long data type. It's a 64 bit integer data type. For your hashing suggestion, let me try it real quick.
I implemented all your suggestions and this is what is happening. When I use M in hashing function anything less than 64, it gives me a wrong asnwer at test case 12
Using M=63 here
and when I use M=64, it gives me a TLE at M=64 at test case 4
When M is 64
What to do now? Vladosiya any clues you can give mate?
Line 16:
int ans = 0 ;it should be
long ans = 0;If that would actually be my only fault I swear to god I should quit. I will check it once CF allows me to. PS:- I saw your solution man. You actually just used a string instead of a pair. Why can't I think like you? That was so clever. Thank you for the help.
I just realized the maximum no. Of pairs can be n(n-1)/2 and not n/2 which I was previously thinking so this just might be it.