Question

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

What can be the most optimal way of solving this problem.

the approach i am using in naive i.e nlogn (easy to guess the approach by seeing complexity 🙂 )

Answer 1

The following algorithm computes the number of integers from 0 to (n-1) without “3” in their decimal representation quite efficiently. (I have modified the interval from 1 .. n to 0 .. n-1 only to simplify the following calculations slightly.)

(I am not an expert in complexity calculations, but I think the complexity of this algorithm is O(log n), because it does a fixed number of steps for each digit of n.)

The first observation is that the number of integers with at most d digits (i.e. the numbers in the interval 0 .. 10^d-1) not having the digit 3 in their decimal representation is exactly 9^d, because for each digit you have 9 possible choices 0,1,2,4,5,6,7,8,9.

Now let me demonstrate the algorithm with a 5 digit number n = a₄a₃a₂a₁a₀.

We compute separately the number of integers with no “3” in their decimal representation for the intervals

• I₀: a₄a₃a₂a₁ 0 <= i < a₄a₃a₂a₁a₀
• I₁: a₄a₃a₂ 0 0 <= i < a₄a₃a₂a₁ 0
• I₂: a₄a₃ 0 0 0 <= i < a₄a₃a₂ 0 0
• I₃: a₄ 0 0 0 0 <= i < a₄a₃ 0 0 0
• I₄: 0 0 0 0 0 <= i < a₄ 0 0 0 0

The number of integers in the interval I_j that do not have a “3” in the decimal representation is

• 0, if one of the higher valued digits a_j+1, a_j+2, … is equal to 3,
  otherwise:
• a_j * 9^j, if 0 <= a_j <= 3, (a_j choices for the
  j^th digit, 9 choices for all lower valued digits),
• (a_j – 1) * 9^j, if aj > 3 (because 3 is not a valid choice for the j^th digit) .

So we have the following function:

/*
 * Compute number of integers x with 0 <= x < n that do not
 * have a 3 in their decimal representation.
 */
int f(int n)
{
    int count = 0;
    int a;      // The current digit a_j
    int p = 1;  // The current value of 9^j

    while (n > 0) {
        a = n % 10;
        if (a == 3) {
            count = 0;
        }
        if (a <= 3) {
            count += a * p;
        } else {
            count += (a-1) * p;
        }
        n /= 10;
        p *= 9;
    }

    return count;
}