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

• Last Updated : 23 Jun, 2022

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:
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 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 A but 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



## Javascript



Output:

90

Time Complexity: O(N)

Auxiliary Space: O(1)

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