Count numbers that don't contain 3

Given a number n, write a function that returns count of numbers from 1 to n that don't contain digit 3 in their decimal representation. 
Examples: 
 

Input: n = 10
Output: 9 

Input: n = 45
Output: 31 
// Numbers 3, 13, 23, 30, 31, 32, 33, 34, 
// 35, 36, 37, 38, 39, 43 contain digit 3. 

Input: n = 578
Output: 385

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Solution: 
We can solve it recursively. Let count(n) be the function that counts such numbers. 
 

'msd' --> the most significant digit in n 
'd' --> number of digits in n.

count(n) = n if n < 3

count(n) = n - 1 if 3 <= n 10 and msd is not 3

count(n) = count( msd * (10^(d-1)) - 1) 
 if n > 10 and msd is 3

Let us understand the solution with n = 578. 
count(578) = 4*count(99) + 4 + count(78)
The middle term 4 is added to include numbers 
100, 200, 400 and 500.

Let us take n = 35 as another example. 
count(35) = count (3*10 - 1) = count(29)

Output: 

385

Time Complexity: O(log₁₀n)

Auxiliary Space: O(1),  since no extra space has been taken.