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
Ouput: 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

count(n) = count(msd) * count(10^(d-1) - 1) + 
           count(msd) + 
           count(n % (10^(d-1)))
           if 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)
#include <stdio.h>

/* returns count of numbers which are in range from 1 to n and don't contain 3 
   as a digit */
int count(int n)
{
    // Base cases (Assuming n is not negative)
    if (n < 3)
        return n;
    if (n >= 3 && n < 10)
       return n-1;

    // Calculate 10^(d-1) (10 raise to the power d-1) where d is
    // number of digits in n. po will be 100 for n = 578
    int po = 1;
    while (n/po > 9)
        po = po*10;

    // find the most significant digit (msd is 5 for 578)
    int msd = n/po;

    if (msd != 3)
      // For 578, total will be 4*count(10^2 - 1) + 4 + count(78)
      return count(msd)*count(po - 1) + count(msd) + count(n%po);
    else
      // For 35, total will be equal to count(29)
      return count(msd*po - 1);
}

// Driver program to test above function
int main()
{
    printf ("%d ", count(578));
    return 0;
}

Output:

385

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