So I came across this problem and I would like to implement and idea that after removing values that are directly divisible by 3 and the pairs that add upto sum (divisible by 3), the rem. ones may have some kind of combination of elements with sum adding upto 3
Example:
arr = [1,1,1,1,1,2,2]
after two operations , pairs gets out [1,2],[1,2] with sum(%3==0)
rem = [1,1,1]
Now, how to know that rem array may have some combinations sum divisible by 3?
After all your operations, there is a subsequence with sum divisible by 3, if one of these conditions works:
There are at least three elements with the same mod 3
There's at least 1 element mod 3 == 1 and 1 element mod 3 == 2
Don't you think it is getting lengthy? Its problem A, maybe there is a simple solution than that.
Lengthy???
gosh, your solution is so long, I could barely read till the end :rofl: 141667428 {just kidding}
potter btw this is problem B so it's okay
The idea is very simple , just replace all the values of the array with a[i]%3 and then count the number of 0s , 1s and 2s and with some small equation u ll end up solving the problem.