Count of N digit numbers with at least one digit as K

Given a number N and a digit K, The task is to count N digit numbers with at least one digit as K.

Examples:

Input: N = 3, K = 2
Output: 252
Explanation: 
For one occurrence of 2 –
In a number of length 3, the following cases are possible:
=>When first digit is 2 and other two digits can have 9 values except ‘2’. 
Thus 9*9 = 81 combination are possible.
=> When second digit is 2 and first digit can have 8 values from 1 to 9 except ‘2’ 
and the third digit can have 9 value from 0 to 9 except ‘2’.
Thus 8*9 = 72 valid combination.
=>When third digit is 2 the first digit can have 8 values from 1 to 9 except ‘2’ 
and the second digit can have 9 values from 0 to 9 except ‘2’ thus 8*9 = 72. 
Hence total valid combination with one occurrence of 2 = 72 + 72 + 81 = 225.
For two occurrence of 2 –
First and second digit can be 2 and third digit can have 9 values from 0 to 9. 
Second and third digit can have value 2 and first digit 
can have  8 values from 1 to 9 except 2. 
First and third digit can have values 2 and second digit 
can have 9 values from 0 to 9 except 2. 
Hence total valid combination with two occurrence of 2 = 9 + 8 + 9 = 26.
For all three digits to be 2 – 
There can be only 1 combination.
Hence total possible numbers with at least one occurrence of 2 =  225 + 26 + 1 = 252.

Input: N = 9, K = 8
Output: 555626232

Approach: The problem can be solved based on the following mathematical idea:

Find the difference between count of unique N digit numbers possible and count of all unique N digit numbers with no occurrence of digit K.

Follow the steps mentioned below to implement this idea:

• Find the count of all N digits numbers = 9 x 10^N-1, Leftmost place can be any digit from 1-9, other digits can have any value from between 0 and 9.
• Find the count of all N digits number with no occurrence of K = 8 x 9^n-1, Leftmost place can be any digit from 1 to 9 except K and other digits can have any value between 0 to 9 except K.  
• Total count of N digit numbers with at least one occurrence of K 
  = Count of all N digits numbers –  Count of all N digit numbers with no occurrence of K.

Below is the implementation of the above approach:

C++

// C++ Code to Implement the approach
// Function to find the total possible numbers
#include <iostream>
#include <bits/stdc++.h>
using namespace std;

// Function to find the total possible numbers
void required_numbers(int n, int k)
{
    int t, h, r;

    // Find all n digits numbers
    t = 9 * pow (10, (n - 1));

    // Find n digits number in which no k occurs
    h = 8 * pow (9, (n - 1));

    // Calculate the required value as
    // the difference of the above two values
    r = t - h;
    cout << r;
}

// Driver code
int main()
{

    int N, K;
    N = 3;
    K = 2;

    // Function call
    required_numbers(N, K);
    return 0;
}

// This code is contributed by ANKITKUMAR34.

Java

// Java Code to implement the approach
// Function to find the total possible numbers
import java.io.*;
import java.util.*;

class GFG {
    // Function to find the total possible numbers
    public static void required_numbers(int n, int k)
    {
        // Find all n digits numbers
        int t = 9 * (int)Math.pow(10, (n - 1));

        // Find n digits number in which no k occurs
        int h = 8 * (int)Math.pow(9, (n - 1));

        // Calculate the required value as
        // the difference of the above two values
        int r = t - h;
        System.out.print(r);
    }
    public static void main(String[] args)
    {
        int N = 3;
        int K = 2;

        // Function call
        required_numbers(N, K);
    }
}

// This code is contributed by Rohit Pradhan

Python3

# Python Code to Implement the approach


# Function to find the total possible numbers
def required_numbers(n, k):

    # Find all n digits numbers
    t = 9 * 10 ** (n - 1)

    # Find n digits number in which no k occurs
    h = 8 * 9 ** (n - 1)

    # Calculate the required value as
    # the difference of the above two values
    r = t - h
    return(r)


# Driver code
if __name__ == '__main__':
    N = 3
    K = 2

    # Function call
    print(required_numbers(N, K))

C#

// C# Code to implement the approach
// Function to find the total possible numbers

using System;

public class GFG {

    // Function to find the total possible numbers
    public static void required_numbers(int n, int k)
    {
        // Find all n digits numbers
        int t = 9 * (int)Math.Pow(10, (n - 1));

        // Find n digits number in which no k occurs
        int h = 8 * (int)Math.Pow(9, (n - 1));

        // Calculate the required value as
        // the difference of the above two values
        int r = t - h;

        Console.WriteLine(r);
    }

    public static void Main(string[] args)
    {
        int N = 3;
        int K = 2;

        // Function call
        required_numbers(N, K);
    }
}

// This code is contributed by AnkThon

Javascript

<script>
// javascript Code to implement the approach
// Function to find the total possible numbers

   // Function to find the total possible numbers
    function required_numbers(n , k)
    {
        // Find all n digits numbers
        var t = 9 * parseInt(Math.pow(10, (n - 1)));

        // Find n digits number in which no k occurs
        var h = 8 * parseInt(Math.pow(9, (n - 1)));

        // Calculate the required value as
        // the difference of the above two values
        var r = t - h;
        document.write(r);
    }

var N = 3;
var K = 2;

// Function call
required_numbers(N, K);
// This code contributed by shikhasingrajput
</script>

Output

252

Time Complexity: O(1).
Auxiliary Space: O(1).