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Count of N-digit Palindrome numbers

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  • Last Updated : 10 Mar, 2022
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Given an integer N, the task is to find the count of N-digit Palindrome numbers.
Examples: 
 

Input: N = 1 
Output:
{1, 2, 3, 4, 5, 6, 7, 8, 9} are all the possible 
single digit palindrome numbers.
Input: N = 2 
Output:
 

 

Approach: The first digit can be any of the 9 digits (not 0) and the last digit will have to be same as the first in order for it to be palindrome, the second and the second last digit can be any of the 10 digits and same goes for the rest of the digits. So, for any value of N, the count of N-digit palindromes will be 9 * 10(N – 1) / 2.
Below is the implementation of the above approach: 
 

C++




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

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the count
// of N-digit palindrome numbers
static int nDigitPalindromes(int n)
{
    return (9 * (int)Math.pow(10,
           (n - 1) / 2));
}
 
// Driver code
public static void main(String []args)
{
    int n = 2;
 
    System.out.println(nDigitPalindromes(n));
}
}
 
// This code is contributed by Code_Mech

Python3




# Python3 implementation of the approach
 
# Function to return the count
# of N-digit palindrome numbers
def nDigitPalindromes(n) :
 
    return (9 * pow(10, (n - 1) // 2));
 
# Driver code
if __name__ == "__main__" :
 
    n = 2;
 
    print(nDigitPalindromes(n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
     
class GFG
{
     
// Function to return the count
// of N-digit palindrome numbers
static int nDigitPalindromes(int n)
{
    return (9 * (int)Math.Pow(10,
           (n - 1) / 2));
}
 
// Driver code
public static void Main(String []args)
{
    int n = 2;
 
    Console.WriteLine(nDigitPalindromes(n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the count
// of N-digit palindrome numbers
function nDigitPalindromes(n)
{
    return (9 * Math.pow(10, parseInt((n - 1) / 2)));
}
 
// Driver code
var n = 2;
document.write(nDigitPalindromes(n));
 
</script>

Output: 

9

 

Time Complexity: O(log n)

Auxiliary Space: O(1)


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