Given an integer **N**, the task is to count all possible **N digit numbers** such that **A + reverse(A) = 10 ^{N} – 1** where

**A**is an N digit number and reverse(A) is reverse of A.

**A**shouldn’t have any leading 0s.

**Examples:**

Input:N = 2

Output:9

All possible 2 digit numbers are 90, 81, 72, 63, 54, 45, 36, 27 and 18.

Input:N = 4

Output:90

**Approach:** First we have to conclude that if N is odd then there is no number which will satisfy the given condition, let’s prove it for **N = 3**,

,

so and .

which is impossible as it is a floating point number.

Now Find answer for when **N is even**. For example, N=4,

and now ifx + y = 9then the number of pairs which satisfy this condition are 10.

(0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)

Now, 1st and N^{th}digit cannot have the pair (0, 9) as there shouldn’t be any leading 0s in Abut for all the remaining N/2-1 pairs there can be 10 pairs.

So the answer is , As N is large so we will print 9 followed by N/2-1 number of 0s.

Below is the implementation of the above approach:

## C++

`// C++ implementation of above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the count of required numbers ` `string getCount(` `int` `N) ` `{ ` ` ` ` ` `// If N is odd then return 0 ` ` ` `if` `(N % 2 == 1) ` ` ` `return` `0; ` ` ` ` ` `string result = ` `"9"` `; ` ` ` ` ` `for` `(` `int` `i = 1; i <= N / 2 - 1; i++) ` ` ` `result += ` `"0"` `; ` ` ` ` ` `return` `result; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` ` ` `int` `N = 4; ` ` ` `cout << getCount(N); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java implementation of above approach ` `class` `GFG ` `{ ` ` ` `// Function to return the count of required numbers ` ` ` `static` `String getCount(` `int` `N) ` ` ` `{ ` ` ` ` ` `// If N is odd then return 0 ` ` ` `if` `(N % ` `2` `== ` `1` `) ` ` ` `return` `"0"` `; ` ` ` ` ` `String result = ` `"9"` `; ` ` ` `for` `(` `int` `i = ` `1` `; i <= N / ` `2` `- ` `1` `; i++) ` ` ` `result += ` `"0"` `; ` ` ` `return` `result; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main(String []args) ` ` ` `{ ` ` ` ` ` `int` `N = ` `4` `; ` ` ` `System.out.println(getCount(N)); ` ` ` `} ` `} ` ` ` `// This code is contributed by ihritik ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 implementation of above approach ` ` ` `# Function to return the count of required numbers ` `def` `getCount(N): ` ` ` ` ` `# If N is odd then return 0 ` ` ` `if` `(N ` `%` `2` `=` `=` `1` `): ` ` ` `return` `"0"` ` ` ` ` `result ` `=` `"9"` ` ` ` ` `for` `i ` `in` `range` `(` `1` `, N ` `/` `/` `2` `): ` ` ` `result ` `=` `result ` `+` `"0"` ` ` ` ` `return` `result ` ` ` `# Driver Code ` `N ` `=` `4` `print` `(getCount(N)) ` ` ` `# This code is contributed by ihritik ` |

*chevron_right*

*filter_none*

## C#

`// C# implementation of above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to return the count of required numbers ` ` ` `static` `string` `getCount(` `int` `N) ` ` ` `{ ` ` ` ` ` `// If N is odd then return 0 ` ` ` `if` `(N % 2 == 1) ` ` ` `return` `"0"` `; ` ` ` `string` `result = ` `"9"` `; ` ` ` `for` `(` `int` `i = 1; i <= N / 2 - 1; i++) ` ` ` `result += ` `"0"` `; ` ` ` `return` `result; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` ` ` `int` `N = 4; ` ` ` `Console.WriteLine(getCount(N)); ` ` ` `} ` `} ` ` ` `// This code is contributed by ihritik ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP implementation of above approach ` ` ` `// Function to return the count of ` `// required numbers ` `function` `getCount(` `$N` `) ` `{ ` ` ` ` ` `// If N is odd then return 0 ` ` ` `if` `(` `$N` `% 2 == 1) ` ` ` `return` `0; ` ` ` ` ` `$result` `= ` `"9"` `; ` ` ` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$N` `/ 2 - 1; ` `$i` `++) ` ` ` `$result` `.= ` `"0"` `; ` ` ` ` ` `return` `$result` `; ` `} ` ` ` `// Driver Code ` `$N` `= 4; ` `echo` `getCount(` `$N` `); ` ` ` `// This code is contributed by Ryuga ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

90

**Time Complexity:** O(N)

## Recommended Posts:

- Count of N digit numbers possible which satisfy the given conditions
- Count of sub-sequences which satisfy the given condition
- Count triplet pairs (A, B, C) of points in 2-D space that satisfy the given condition
- Count index pairs which satisfy the given condition
- Count of indices in an array that satisfy the given condition
- Pairs from an array that satisfy the given condition
- Number of strings that satisfy the given condition
- Check if a cycle of length 3 exists or not in a graph that satisfy a given condition
- Number of cells in a matrix that satisfy the given condition
- Append two elements to make the array satisfy the given condition
- Find the Number of Permutations that satisfy the given condition in an array
- Mimimum number of leaves required to be removed from a Tree to satisfy the given condition
- Count of N-digit numbers having digit XOR as single digit
- Count of numbers in the range [L, R] which satisfy the given conditions
- Count n digit numbers not having a particular digit
- Count of Numbers in Range where first digit is equal to last digit of the number
- Count numbers in a range with digit sum divisible by K having first and last digit different
- Find numbers a and b that satisfy the given conditions
- Count of Fibonacci pairs which satisfy the given equation
- Count of unordered pairs (x, y) of Array which satisfy given equation

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.