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Count ‘d’ digit positive integers with 0 as a digit
• Difficulty Level : Easy
• Last Updated : 01 Apr, 2021

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

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## Javascript

 ``

Output :

```0
9
2439```

This article is contributed by Ashutosh Kumar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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