I struggled quite a lot with problem E. Hope this explanation helps you.

Let’s first look at the constraints of the problem, here the value of (n = 100) and the value of (a <= 10^4) and (1 <= b <= 10000). We have to find the value of the expression (n * a — b). So, what can be the max value of this expression? If n = 100 and a = 10^4 and assuming b = 1 (because the minimum value of b can be 1) we get that the max value of the expression becomes (n * a — b = 10^2 * 10^4 — 1) = 999999. From this deduction what can we say? There will be 6 digits maximum in the final solution we want to find. If there are more than 6 digits we can simply ignore those answers because we found out the max value to be of 6 digits. So, the problem has been reduced to a minimized version and can be solved with brute force on the value of 'a' only.

If we run a loop for a from 1 to 10^4, and inside that loop we firstly find digits(n) * 'a', which is basically the number of digits in the number (n * a) for each value of 'a'. Now as we know this value (digits(n) * a) must not be greater than 6. Because as we discussed earlier those values will not contribute to our answer. So, we run a loop of digits(d) from 1 to 6 and can find from that b = (digits(n) * a — d). In case you didn't get this point what was the original equation supposed to be? (digits(n) * a — b = d).

From this, you got the value of both a and b. Now you can calculate the original value of n * a — b and the value digits(n) * a — b and compare if both of them are equal. If both of them are equal then these (a, b) pairs will contribute to your answer. And as we are running a loop for the value of a only from 1 to 10^4 and n = 100(max) and for the digits 1 to 6.

The total time complexity reduces to 100 * 10000 * 7 ~ 10^7(roughly). Which is fast enough to pass the test cases.

I'm glad to hear that.

You are right. I will test this idea.