# Check if number is palindrome or not in base B

• Last Updated : 17 Mar, 2021

Given an integer N, the task is to check if Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

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(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 ``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

 ``
Output:
`Yes`

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

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