Hello! I came up with the following problem, tried to solve it (faster than $$$O(N^2)$$$) and failed. I asked some other people (all much smarter than me) but neither of them could solve it.
The problem statement:
There is an array of positive integers. Count the number of even length subarrays such that the first half has the same sum as the second half.
Is it possible to solve this faster than $$$O(N^2)$$$? Maybe if we add additional constraints like the values are small or randomly generated (this one does not seem to help but I included it anyways). If it cannot be done faster than $$$O(N^2)$$$, could you provide a proof? Thank you for your time.
If the sum of all numbers is bounded by $$$C$$$, we can solve the problem in $$$O\left(\frac{nC} {64}\right) $$$ using bitsets.
nvm, I don't take the sizes of the subarrays into account.
Maybe it can be done using binarysearch+hashing
What do you binary search? What do you hash?
Hash to the array like hashing of string, and for every i as mid i will bs on how long j such that i…j=i-j..i using hash in O(1) ,
You cannot binary search that, if some j is good, that does not mean that any j smaller is also good. Also, how much of the array do you hash? If you hash too much you still don't get good results.
Umm sry u r right I thought it as palindrome
https://en.wikipedia.org/wiki/3SUM
i think it's related (or i guess check "is answer = 0" is equal hard than 3SUM)
How could one prove that 3SUM can be reduced to this problem? (i.e. solving this fast would imply solving 3SUM fast)