Count of all N digit numbers such that num + Rev(num) = 10^N – 1

Given an integer N, the task is to find the count of all N digit numbers such that num + Rev(num) = 10N – 1

Examples:

Input: N = 2
Output: 9
All possible numbers are
18 + 81 = 99
27 + 72 = 99
36 + 45 = 99
45 + 54 = 99
54 + 45 = 99
63 + 54 = 99
72 + 27 = 99
81 + 18 = 99
90 + 09 = 99

Input: N = 4
Output: 90



Approach There are 2 cases:
If n is odd then answer will be 0.

Let n = 3 then num = d1d2d3 and rev(num) = d3d2d1
num + rev(num) should be 999 since n = 3.
So the below equations must be satisfied
d1 + d3 = 9 … (1)
d2 + d2 = 9 … (2)
d3 + d1 = 9 … (3)
Considering equation 2:
d2 + d2 = 9
2 * d2 = 9
d2 = 4.5 which is not possible because the digit of a number should always be a whole number.
Therefore If n is odd then answer will be 0.

If n is even then answer will be 9 * 10(N / 2 – 1).

Let n = 4 then num = d1d2d3d4 and rev(num) = d4d3d2d1
So the below equations should be satisfied
d1 + d4 = 9 … (1)
d2 + d3 = 9 … (2)
d3 + d2 = 9 … (3)
d4 + d1 = 9 … (4)

Considering equation 1: d1 + d4 = 9. It can be true in 9 ways:
(1 + 8), (2 + 7), (3 + 6), (4 + 5), (5 + 4), (6 + 3), (7 + 2), (8 + 1) and (9 + 0)
Similarly other equations will also have 9 solutions + 1 more solution since the remaining digits are not the first and last digit of the number and we can take sum of the form (0 + 9).
And since half of the equations are same
Therefore, if n is even then answer will be 9 * 10(N / 2 – 1).

Below is the implementation of the above approach

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the
// count of such numbers
int countNumbers(int n)
{
  
    // If n is odd
    if (n % 2 == 1)
        return 0;
  
    return (9 * pow(10, n / 2 - 1));
}
  
// Driver code
int main()
{
    int n = 2;
    cout << countNumbers(n);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG {
  
    // Function to return the
    // count of such numbers
    static int countNumbers(int n)
    {
  
        // If n is odd
        if (n % 2 == 1)
            return 0;
  
        return (9 * (int)Math.pow(10, n / 2 - 1));
    }
  
    // Driver code
    public static void main(String args[])
    {
        int n = 2;
        System.out.print(countNumbers(n));
    }
}

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Python3

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# Python3 implementation of the approach 
  
# Function to return the 
# count of such numbers 
def countNumbers(n): 
  
    # If n is odd 
    if n % 2 == 1:
        return 0
  
    return (9 * pow(10, n // 2 - 1)) 
  
# Driver code 
if __name__ == "__main__":
  
    n = 2
    print(countNumbers(n)) 
  
# This code is contributed 
# by Rituraj Jain

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C#

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// C# implementation of the approach
using System;
class GFG {
  
    // Function to return the
    // count of such numbers
    static int countNumbers(int n)
    {
  
        // If n is odd
        if (n % 2 == 1)
            return 0;
  
        return (9 * (int)Math.Pow(10, n / 2 - 1));
    }
  
    // Driver code
    public static void Main()
    {
        int n = 2;
        Console.WriteLine(countNumbers(n));
    }
}

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PHP

Output:

9

Time Complexity: O(1)



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Improved By : rituraj_jain, Code_Mech