Following the hint. Let f(n) = count of number with unique digits of length n.

f(1) = 10. (0, 1, 2, 3, ...., 9)

f(2) = 9 * 9. Because for each number i from 1, ..., 9, we can pick j to form a 2-digit number ij and there are 9 numbers that are different from i for j to choose from.

f(3) = f(2) * 8 = 9 * 9 * 8. Because for each number with unique digits of length 2, say ij, we can pick k to form a 3 digit number ijk and there are 8 numbers that are different from i and j for k to choose from.

Similarly f(4) = f(3) * 7 = 9 * 9 * 8 * 7....

...

f(10) = 9 * 9 * 8 * 7 * 6 * ... * 1

f(11) = 0 = f(12) = f(13)....

any number with length > 10 couldn't be unique digits number.

The problem is asking for numbers from 0 to 10^n. Hence return f(1) + f(2) + .. + f(n)

As @4acreg suggests, There are only 11 different ans. You can create a lookup table for it. This problem is O(1) in essence.

```
public int countNumbersWithUniqueDigits(int n) {
if (n == 0) return 1;
int res = 10;
int uniqueDigits = 9;
int availableNumber = 9;
while (n-- > 1 && availableNumber > 0) {
uniqueDigits = uniqueDigits * availableNumber;
res += uniqueDigits;
availableNumber--;
}
return res;
}
```