Count 'd' digit positive integers with 0 as a digit - GeeksforGeeks

Count 'd' digit positive integers with 0 as a digit - GeeksforGeeks
Given a number d, representing the number of digits of a positive integer. Find the total count of positive integer (consisting of d digits exactly) which have at-least one zero in them.

Input : d = 2
Output : 9
The numbers are, 10, 20, 30, 40, 50, 60, 
                 70, 80 and 90.

One Simple Solution is to traverse through all d digit positive numbers. For every number, traverse through its digits and if there is any 0 digit, increment count (similar to this).

Following are some observations:

1. There are exactly d digits.
2. The number at most significant place can’t be a zero (no leading zeroes allowed).
3. All the other places except the most significant one can contain zero .

We can place any of {1, 2, ... 9} in D1
Hence D1 can be filled in 9 ways.

Apart from D1 all the other places can be  10 ways. 
(we can place 0 as well)
Hence the total numbers having d digits can be given as: 
Total =  9*10d-1

Now, let's find the numbers having d digits, that
don't contain zero at any place. 
In this case, all the places can be filled in 9 ways.
Hence count of such numbers is given by:
Non_Zero = 9d

Now the count of numbers having at least one zero 
can be obtained by subtracting Non_Zero from Total.
Hence Answer would be given by:
9*(10d-1 - 9d-1 ) 

int findCount(int d)

{

    return 9*(pow(10,d-1) - pow(9,d-1));

}

http://www.geeksforgeeks.org/count-numbers-0-digit/

Count how many integers from 1 to N contains 0’s as a digit.

Input: n = 155
Output: 24
The numbers having 0 are 10, 20,..90, 100, 101..110,
120, ..150.

// Returns 1 if x has 0, else 0

int has0(int x)

{

    // Traverse througn all digits of

    // x to check if it has 0.

    while (x)

    {

        // If current digit is 0, return true

        if (x % 10 == 0)

          return 1;

 

        x /= 10;

    }

 

    return 0;

}

 

// Returns count of numbers from 1 to n with 0 as digit

int getCount(int n)

{

    // Initialize count of numbers having 0 as digit

    int count = 0;

 

    // Travers through all numbers and for every number

    // check if it has 0.

    for (int i=1; i<=n; i++)

        count += has0(i);

 

    return count;

}

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