The code give TLE for n=8. How to improve it ?
Code:
Count Number With Unique digit
go for a O(1) solution
This is a digit combination problem. Can be solved in at most 10 loops.
When n == 0, return 1. I got this answer from the test case.
When n == 1, _ can put 10 digit in the only position. [0, … , 10]. Answer is 10.
When n == 2, _ _ first digit has 9 choices [1, …, 9], second one has 9 choices excluding the already chosen one. So totally 9 * 9 = 81. answer should be 10 + 81 = 91
When n == 3, _ _ _ total choice is 9 * 9 * 8 = 684. answer is 10 + 81 + 648 = 739
When n == 4, _ _ _ _ total choice is 9 * 9 * 8 * 7.
…
When n == 10, _ _ _ _ _ _ _ _ _ _ total choice is 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
When n == 11, _ _ _ _ _ _ _ _ _ _ _ total choice is 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 * 0 = 0
see this: