- 24ms
- 88%
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)
class Solution:
def countNumbersWithUniqueDigits(self, n: int) -> int:
if n==0: return 1
res = 10
availableNumber = 9
uniqueDigits = 9
while n>1 and availableNumber>0:
uniqueDigits *= availableNumber
res += uniqueDigits
availableNumber -=1
n-=1
return res