# 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 = 2Output:252Explanation: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 = 8Output:555626232

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

Find the difference between count of unique

Ndigit numbers possible and count of all uniqueNdigitnumbers with no occurrence of digitK.

Follow the steps mentioned below to implement this idea:** **

- Find the count of all
**N**digits numbers**= 9 x 10**, Leftmost place can be any digit from 1-9, other digits can have any value from between 0 and 9.^{N-1} - Find the count of all
**N**digits number with no occurrence of**K**=**8 x 9**, Leftmost place can be any digit from 1 to 9 except^{n-1}**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).