Probability that a N digit number is palindrome

Given an integer N, the task is to find the probability that a number with a number of digits as N is a palindrome. 
The number may have leading zeros.
Examples:

Input: N = 5 
Output: 1 / 100
Input: N = 6 
Output: 1 / 1000 

Solution: 

• As leading zeroes are allowed a total number of N digit numbers is 10^N.
• A number is a palindrome when the first N/2 digits match with the last N/2 digits in reverse order.
• For an even number of digits, we can pick the first N/2 digits and then duplicate them to form the rest of N/2 digits so we can choose (N)/2 digits.
• For an odd number of digits, we can pick first (N-1)/2 digits and then duplicate them to form the rest of (N-1)/2 digits so we can choose (N+1)/2 digits.
• So the probability that an N digit number is palindrome is 10^{ceil( N / 2 )} / 10^N or 1 / 10^{floor( N / 2 )}

Below is the implementation of the approach:  

C++

// C++ code of above approach
#include <bits/stdc++.h>
using namespace std;

// Find the probability that a
// n digit number is palindrome
void solve(int n)
{
    int n_2 = n / 2;

    // Denominator
    string den;
    den = "1";

    // Assign 10^(floor(n/2)) to
    // denominator
    while (n_2--)
        den += '0';

    // Display the answer
    cout << 1 << "/" << den << "\n";
}

// Driver code
int main()
{

    int N = 5;

    solve(N);

    return 0;
}

Java

// Java code of above approach
import java.util.*;

class GFG
{

// Find the probability that a
// n digit number is palindrome
static void solve(int n)
{
    int n_2 = n / 2;

    // Denominator
    String den;
    den = "1";

    // Assign 10^(floor(n/2)) to
    // denominator
    while (n_2-- > 0)
        den += '0';

    // Display the answer
    System.out.println(1 + "/" + den);
}

// Driver code
public static void main(String[] args)
{
    int N = 5;

    solve(N);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 code of above approach

# Find the probability that a
# n digit number is palindrome
def solve(n) :

    n_2 = n // 2;

    # Denominator
    den = "1";

    # Assign 10^(floor(n/2)) to
    # denominator
    while (n_2) :
        den += '0';

        n_2 -= 1

    # Display the answer
    print(str(1) + "/" + str(den))

# Driver code
if __name__ == "__main__" :

    N = 5;

    solve(N);

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

// Find the probability that a
// n digit number is palindrome
static void solve(int n)
{
    int n_2 = n / 2;

    // Denominator
    String den;
    den = "1";

    // Assign 10^(floor(n/2)) to
    // denominator
    while (n_2-- > 0)
        den += '0';

    // Display the answer
    Console.WriteLine(1 + "/" + den);
}

// Driver code
public static void Main(String[] args)
{
    int N = 5;

    solve(N);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
    // Javascript implementation of the approach

    // Find the probability that a
    // n digit number is palindrome
    function solve(n)
    {
        let n_2 = parseInt(n / 2, 10);

        // Denominator
        let den;
        den = "1";

        // Assign 10^(floor(n/2)) to
        // denominator
        while (n_2-- > 0)
            den += '0';

        // Display the answer
        document.write(1 + "/" + den + "</br>");
    }

    let N = 5;

    solve(N);

// This code is contributed by divyeshrabadiya07.
</script>

Output:

1/100

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.