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Check if number is palindrome or not in base B
  • Last Updated : 17 Mar, 2021
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Given an integer N, the task is to check if 

N_B
 

(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 

N_B
 

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

N_B

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

N_B

  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(logBN) 
Auxiliary Space:O(1)

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