I'm trying to tackle down a problem where the time limit is very low (1 second) and the number of cases is supposedly high.
You need to tell if a number is divisible by 3, but the problem is that you don't get the direct number, you get a number k, and then need to check if the concatenation of numbers from 1 to k (123...k) is divisible by 3.
Example input:
4 // The number of cases
2
6
15
130000000
Output:
YES // Because 12 is divisible by 3
YES // Because 123456 is divisible by 3
YES // Because 123456789101112131415 is divisible by 3
NO
I've found some topics about quickly checking the divisibility, but what most time takes I think is to build the number. There are cases where the initial number is as high as 130000000 (so the final is 1234...130000000) which I thinks overflows any numeric data type.
So, what am I missing here? Is there any way to know if something is divisible by 3 without concatenating the number? Any ideas?
PD: Someone also posted the triangular numbers formula which also is a correct solution and then deleted the answer, it was:
if ((1 + num) * num / 2) % 3 == 0 ? "YES" : "NO"
Take a look:
n digit sum mod 3
0 0
1 1
2 2
3 0
4 1
5 2
6 0
...
10 1
11 2
12 0
...
19 1
20 2
21 0
...
Say we have a string of digits constructed as you describe, and the number we just added was divisible mod 3. When we append the next number's digits, we are appending digits whose sum is congruent to 1 mod 3, and when added to those in our number, we will get a combined digit sum congruent to 1 mod 3, so our answer for the next one will be "no". The next one will add a number with digit sum congruent to 2 mod 3, and this causes the total to become congruent to 0 again, so the answer here is "yes". Finally, adding the next number which must be divisible by 3 keeps the digit sum congruent to 0.
The takeaway?
In particular, your example for n=15 is wrong; the digit string obtained represents a number that should be divisible by 3, and indeed it is (try it on a big enough calculator to verify).
All that is left is to find an implementation that is fast enough and handles all the required cases. If n is guaranteed to be under ~2 billion, then you are probably safe with something like
return (n % 3) != 1;
If n can be an arbitrarily large number, never fear; you can check whether the digit sum is congruent to 0 modulo 3 by adding up the digits in linear time. If not, you can add 1 from the number by coding addition like you do it by hand on paper and then check the result of that for divisibility by 3, again in linear time. So something like:
if (digit_sum_mod_3(n) == 0) return true;
else if (digit_sum_mod_3(add_one(n)) == 0) return false;
else return true;
Then you would have something like
digit_sum_mod_3(n[1...m])
sum = 0
for k = 1 to m do
sum = sum + n[k]
// keep sum from getting too big
if sum >= 18 then
sum = sum - 18
return sum % 3
add_one(n[1...m])
// work from right to left, assume big-endian
for k = m to 1 do
if n[k] < 9 then // don't need to carry
n[k] = n[k] + 1
break
else then // need to carry
n[k] = 0
if n[1] = 0 then // carried all the way to the front
n[1] = 1
n[m+1] = 0
return n