Check if number is palindrome or not in base B

Given an integer N, the task is to check if 

(N in base B) is palindrome or not.

Examples:

Input: N = 5, B = 2 
Output: Yes 
Explanation:
(5)₁₀ = (101)₂ which is palindrome. Therefore, the required output is Yes.
Input: N = 4, B = 2 
Output: No 

Approach: The problem can be solved by checking if the decimal value of the reverse of 

is equal to N or not. Follow the steps below to solve the problem. 

1. Initialize the variable, rev = 0 to store the reverse of N.
2. Extract the digits of

1. by N % B.
2. For each digit of

1. Update rev= rev * B + N % B
2. Finally, check if N is equal to rev or not

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if N in
// base B is palindrome or not
int checkPalindromeB(int N, int B)
{
    // Stores the reverse of N
    int rev = 0;

    // Stores the value of N
    int N1 = N;

    // Extract all the digits of N
    while (N1) {
        // Generate its reverse
        rev = rev * B + N1 % B;
        N1 = N1 / B;
    }

    return N == rev;
}

// Driver Code
int main()
{
    int N = 5, B = 2;
    if (checkPalindromeB(N, B)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

Java

// Java program to implement
// the above approach
class GFG{

// Function to check if N in
// base B is palindrome or not
static boolean checkPalindromeB(int N,
                                int B)
{

    // Stores the reverse of N
    int rev = 0;

    // Stores the value of N
    int N1 = N;

    // Extract all the digits of N
    while (N1 > 0)
    {

        // Generate its reverse
        rev = rev * B + N1 % B;
        N1 = N1 / B;
    }
    return N == rev;
}

// Driver code
public static void main(String[] args)
{
    int N = 5, B = 2;

    if (checkPalindromeB(N, B))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("No");
    }
}
}

// This code is contributed by Dewanti

Python3

# Python3 program to implement
# the above approach

# Function to check if N in
# base B is palindrome or not
def checkPalindromeB(N, B):

    # Stores the reverse of N
    rev = 0;

    # Stores the value of N
    N1 = N;

    # Extract all the digits of N
    while (N1 > 0):

        # Generate its reverse
        rev = rev * B + N1 % B;
        N1 = N1 // B;

    return N == rev;

# Driver code
if __name__ == '__main__':
    N = 5; B = 2;

    if (checkPalindromeB(N, B)):
        print("Yes");
    else:
        print("No");

# This code is contributed by Princi Singh

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to check if N in
// base B is palindrome or not
static bool checkPalindromeB(int N,
                             int B)
{

    // Stores the reverse of N
    int rev = 0;

    // Stores the value of N
    int N1 = N;

    // Extract all the digits of N
    while (N1 > 0)
    {

        // Generate its reverse
        rev = rev * B + N1 % B;
        N1 = N1 / B;
    }
    return N == rev;
}

// Driver code
public static void Main(String[] args)
{
    int N = 5, B = 2;

    if (checkPalindromeB(N, B))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to check if N in
// base B is palindrome or not
function checkPalindromeB(N, B)
{
    // Stores the reverse of N
    var rev = 0;

    // Stores the value of N
    var N1 = N;

    // Extract all the digits of N
    while (N1) {
        // Generate its reverse
        rev = rev * B + N1 % B;
        N1 = parseInt(N1 / B);
    }

    return N == rev;
}

// Driver Code
var N = 5, B = 2;
if (checkPalindromeB(N, B)) {
    document.write("Yes");
}
else {
    document.write("No");
}


</script>

Output:

Yes

Time Complexity:O(log_BN)
Auxiliary Space:O(1)