# Count ‘d’ digit positive integers with 0 as a digit

• Difficulty Level : Easy
• Last Updated : 13 Jun, 2022

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.
Examples:

```Input : d = 1
Output : 0
There's no natural number of 1 digit that
contains a zero.

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

## We strongly recommend that you click here and practice it, before moving on to the solution.

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 . So considering the above points, let’s find the total count of numbers having d digits:

```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) ```

Below is the program for the same.

## C++

 `//C++ program to find the count of positive integer of a` `// given number of digits that contain atleast one zero` `#include` `using` `namespace` `std;`   `// Returns count of 'd' digit integers have 0 as a digit` `int` `findCount(``int` `d)` `{` `    ``return` `9*(``pow``(10,d-1) - ``pow``(9,d-1));` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `d = 1;` `    ``cout << findCount(d) << endl;`   `    ``d = 2;` `    ``cout << findCount(d) << endl;`   `    ``d = 4;` `    ``cout << findCount(d) << endl;` `    ``return` `0;` `}`

## Java

 `// Java program to find the count` `// of positive integer of a` `// given number of digits ` `// that contain atleast one zero` `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Returns count of 'd' digit ` `    ``// integers have 0 as a digit` `    ``static` `int` `findCount(``int` `d)` `    ``{` `        ``return` `9` `* ((``int``)(Math.pow(``10``, d - ``1``)) ` `                 ``- (``int``)(Math.pow(``9``, d - ``1``)));` `    ``}` `    `  `    ``// Driver Code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `d = ``1``;` `        ``System.out.println(findCount(d));` `        `  `        ``d = ``2``;` `        ``System.out.println(findCount(d));` `        `  `        ``d = ``4``;` `        ``System.out.println(findCount(d));` `        `  `    ``}` `}`   `// This code is contributed by Nikita Tiwari.`

## Python3

 `# Python 3 program to find the` `# count of positive integer of a` `# given number of digits that` `# contain atleast one zero` `import` `math`   `# Returns count of 'd' digit` `# integers have 0 as a digit` `def` `findCount(d) :` `    ``return` `9``*``((``int``)(math.``pow``(``10``,d``-``1``)) ``-` `(``int``)(math.``pow``(``9``,d``-``1``)));`     `# Driver Code` `d ``=` `1` `print``(findCount(d))` `    `  `d ``=` `2` `print``(findCount(d))`   `d ``=` `4` `print``(findCount(d))`     `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to find the count` `// of positive integer of a` `// given number of digits ` `// that contain atleast one zero.` `using` `System;`   `class` `GFG {` `    `  `    ``// Returns count of 'd' digit ` `    ``// integers have 0 as a digit` `    ``static` `int` `findCount(``int` `d)` `    ``{` `        ``return` `9 * ((``int``)(Math.Pow(10, d - 1)) ` `                 ``- (``int``)(Math.Pow(9, d - 1)));` `    ``}` `    `  `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `d = 1;` `        ``Console.WriteLine(findCount(d));` `        `  `        ``d = 2;` `        ``Console.WriteLine(findCount(d));` `        `  `        ``d = 4;` `        ``Console.WriteLine(findCount(d));` `        `  `    ``}` `}`   `// This code is contributed by nitin mittal.`

## PHP

 ``

## Javascript

 ``

Output :

```0
9
2439```

Time Complexity : O(logn)

Auxiliary Space: O(1)