# Count all possible N digit numbers that satisfy the given condition

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 if x + y = 9 then 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 Nth 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 ` `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; ` `} `

## 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 `

## 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 `

## 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 `

## PHP

 ` `

Output:

```90
```

Time Complexity: O(N)

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Improved By : ihritik, AnkitRai01