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

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,

d_{3}d_{2}d_{1}+d_{1}d_{2}d_{3} = 999,
so d_{3}+d_{1}=9 and 2*d_{2}=9.
d_{2}=4.5 which is impossible as it is a floating point number.

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

d_{4}d_{3}d_{2}d_{1}+d_{1}d_{2}d_{3}d_{4} = 9999
d_{4}+d_{1}=9 and d_{3}+d_{2}=9 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 9*10^{(N/2-1)}, 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

90

Time Complexity: O(N)



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Improved By : ihritik, AnkitRai01