Double Base Palindrome

Double base Palindrome as the name suggest is a number which is Palindrome in 2 bases. One of the base is 10 i.e. decimal and another base is k.(which can be 2 or others). 
Note : The palindromic number, in either base, may not include leading zeros. 
Example : The decimal number, 585 = 1001001001₂ (binary), is palindromic in both bases. 

A Palindrome is a word, phrase, number, or other sequence of characters which reads the same backward as forward, such as madam or 12321.

Find the sum of all numbers less than n which are palindromic in base 10 and base k.

Examples:

Input :  10 2
Output : 25
Explanation : (here n = 10 and k = 2)
              1 3 5 7 9 (they are all palindrome 
              in base 10 and 2) so sum is :
              1 + 3 + 5 + 7 + 9 = 25

Input :  100 2
Output : 157
Explanation : 1 + 3 + 5 + 7 + 9 + 33 + 99 = 157

Method 1 : This method is simple. For every number less than n : 

• Check if it is a palindrome in base 10
• If yes, then convert it into base k
• Check if it is palindrome in base k
• If yes, then add it in sum.

This method is quite lengthy as it checks for every number whether it is a palindrome or not. So, for number as large as 1000000, it checks for every number. 
If k = 2, then a palindrome in base 2 can only be odd number, which might reduce the comparisons to 1000000 / 2 = 500000 (which is still large).

Below is the implementation of the above approach :  

C++

// CPP Program for Checking double base
// Palindrome.
#include <bits/stdc++.h>
using namespace std;

// converts number to base k by changing
// it into string.
string integer_to_string(int n, int base)
{
    string str;
    while (n > 0) {
        int digit = n % base;
        n /= base;
        str.push_back(digit + '0');
    }
    return str;
}

// function to check for palindrome
int isPalindrome(int i, int k)
{
    int temp = i;

    // m stores reverse of a number
    int m = 0;
    while (temp > 0) {
        m = temp % 10 + m * 10;
        temp /= 10;
    }

    // if reverse is equal to number
    if (m == i) {

        // converting to base k
        string str = integer_to_string(m, k);
        string str1 = str;

        // reversing number in base k
        reverse(str.begin(), str.end());

        // checking palindrome in base k
        if (str == str1) {
            return i;
        }
    }
    return 0;
}

// function to find sum of palindromes
void sumPalindrome(int n, int k){

    int sum = 0;
    for (int i = 1; i < n; i++) {
        sum += isPalindrome(i, k);
    }
    cout << "Total sum is " << sum;
}

// driver function
int main()
{
    int n = 100;
    int k = 2;

    sumPalindrome(n, k);
    return 0;
}

Java

// Java Program for Checking double base
// Palindrome.
import java.util.*;

class GFG{
// converts number to base k by changing
// it into string.
static String integer_to_string(int n, int base)
{
    String str="";
    while (n > 0) {
        int digit = n % base;
        n /= base;
        str+=(char)(digit+'0');
    }
    return str;
}

// function to check for palindrome
static int isPalindrome(int i, int k)
{
    int temp = i;

    // m stores reverse of a number
    int m = 0;
    while (temp > 0) {
        m = temp % 10 + m * 10;
        temp /= 10;
    }

    // if reverse is equal to number
    if (m == i) {

        // converting to base k
        String str = integer_to_string(m, k);
        StringBuilder sb = new StringBuilder(str);
        String str1=sb.reverse().toString();
        if (str.equals(str1)) {
            return i;
        }
    }
    return 0;
}

// function to find sum of palindromes
static void sumPalindrome(int n, int k){

    int sum = 0;
    for (int i = 1; i < n; i++) {
        sum += isPalindrome(i, k);
    }
    System.out.println("Total sum is "+sum);
}

// driver function
public static void main(String[] args)
{
    int n = 100;
    int k = 2;

    sumPalindrome(n, k);
}
}
// This code is contributed by mits

Python3

# Python3 Program for Checking
# double base Palindrome.

# converts number to base
# k by changing it into string.
def integer_to_string(n, base):

    str = "";
    while (n > 0):
        digit = n % base;
        n = int(n / base);
        str = chr(digit + ord('0')) + str;
    return str;

# function to check for palindrome
def isPalindrome(i, k):
    temp = i;

    # m stores reverse of a number
    m = 0;
    while (temp > 0):
        m = (temp % 10) + (m * 10);
        temp = int(temp / 10);

    # if reverse is equal to number
    if (m == i):

        # converting to base k
        str = integer_to_string(m, k);
        str1 = str;

        # reversing number in base k
        # str=str[::-1];

        # checking palindrome
        # in base k
        if (str[::-1] == str1):
            return i;
    return 0;

# function to find sum of palindromes
def sumPalindrome(n, k):

    sum = 0;
    for i in range(n):
        sum += isPalindrome(i, k);
    print("Total sum is", sum);

# Driver code
n = 100;
k = 2;

sumPalindrome(n, k);

# This code is contributed
# by mits

C#

// C# Program for Checking double base
// Palindrome.
using System;

class GFG{
// converts number to base k by changing
// it into string.
static string integer_to_string(int n, int base1)
{
    string str="";
    while (n > 0) {
        int digit = n % base1;
        n /= base1;
        str+=(char)(digit+'0');
    }
    return str;
}

// function to check for palindrome
static int isPalindrome(int i, int k)
{
    int temp = i;

    // m stores reverse of a number
    int m = 0;
    while (temp > 0) {
        m = temp % 10 + m * 10;
        temp /= 10;
    }

    // if reverse is equal to number
    if (m == i) {

        // converting to base k
        string str = integer_to_string(m, k);
        char[] ch = str.ToCharArray();
        Array.Reverse(ch);
        string str1=new String(ch);;
        if (str.Equals(str1)) {
            return i;
        }
    }
    return 0;
}

// function to find sum of palindromes
static void sumPalindrome(int n, int k){

    int sum = 0;
    for (int i = 1; i < n; i++) {
        sum += isPalindrome(i, k);
    }
    Console.WriteLine("Total sum is "+sum);
}

// driver function
 static void Main()
{
    int n = 100;
    int k = 2;

    sumPalindrome(n, k);
}
}
// This code is contributed by mits

PHP

<?php
// PHP Program for Checking
// double base Palindrome.

// converts number to base
// k by changing it into string.
function integer_to_string($n, $base)
{
    $str = "";
    while ($n > 0)
    {
        $digit = $n % $base;
        $n = (int)($n / $base);
        $str = ($digit + '0') . $str;
    }
    return $str;
}

// function to check
// for palindrome
function isPalindrome($i, $k)
{
    $temp = $i;

    // m stores reverse
    // of a number
    $m = 0;
    while ($temp > 0)
    {
        $m = ($temp % 10) + ($m * 10);
        $temp = (int)($temp / 10);
    }

    // if reverse is
    // equal to number
    if ($m == $i)
    {

        // converting to base k
        $str = integer_to_string($m, $k);
        $str1 = $str;

        // reversing number in base k
        // $str=strrev($str);

        // checking palindrome
        // in base k
        if (strcmp(strrev($str), $str1) == 0)
        {
            return $i;
        }
    }
    return 0;
}

// function to find
// sum of palindromes
function sumPalindrome($n, $k)
{

    $sum = 0;
    for ($i = 0; $i < $n; $i++)
    {
        $sum += isPalindrome($i, $k);
    }
    echo "Total sum is " . $sum;
}

// Driver code
$n = 100;
$k = 2;

sumPalindrome($n, $k);

// This code is contributed
// by mits
?>

Javascript

<script>

// Javascript program for checking
// double base Palindrome.

// Converts number to base k by changing
// it into string.
function integer_to_string(n, base1)
{
    let str="";
    while (n > 0)
    {
        let digit = n % base1;
        n = parseInt(n / base1, 10);
        str += String.fromCharCode(digit +
                                   '0'.charCodeAt());
    }
    return str;
}

// Function to check for palindrome
function isPalindrome(i, k)
{
    let temp = i;

    // m stores reverse of a number
    let m = 0;
    while (temp > 0)
    {
        m = temp % 10 + m * 10;
        temp = parseInt(temp / 10, 10);
    }

    // If reverse is equal to number
    if (m == i)
    {

        // Converting to base k
        let str = integer_to_string(m, k);
        let ch = str.split('');
        ch.reverse();
        let str1 = ch.join("");

        if (str == str1)
        {
            return i;
        }
    }
    return 0;
}

// Function to find sum of palindromes
function sumPalindrome(n, k)
{
    let sum = 0;
    for(let i = 1; i < n; i++)
    {
        sum += isPalindrome(i, k);
    }
    document.write("Total sum is " + sum + "</br>");
}

// Driver code
let n = 100;
let k = 2;

sumPalindrome(n, k);

// This code is contributed by decode2207

</script>

Output

Total sum is 157

Time Complexity: O(n*log(n))
Auxiliary Space: O(log(n))

Method 2 : This method is a little complex to understand but more advance than method 1. Rather than checking palindrome for two bases. This method generates palindrome in given range. 
Suppose we have a palindrome of the form 123321 in base k, then the first 3 digits define the palindrome. However, the 3 digits 123 also define the palindrome 12321. So the 3-digit number 123 defines a 5-digit palindrome and a 6 digit palindrome. From which follows that every positive number less than kⁿ generates two palindromes less than k²ⁿ . This holds for every base k. Example : let’s say k = 10 that is decimal. Then for n = 1, all numbers less than 10ⁿ have 2 palindrome, 1 even length and 1 odd length in 10²ⁿ. These are 1, 11 or 2, 22 or 3, 33 and so on. So, for 1000000 we generate around 2000 and for 10⁸ we generate around 20000 palindromes. 

• Start from i=1 and generate odd palindrome of it.
• Check if this generated odd palindrome is also palindrome in base k
• If yes, then add this number to sum.
• Repeat the above three steps by changing i=i+1 until the last generated odd palindrome has crossed limit.
• Now, again start from i=1 and generate even palindrome of it.
• Check if this generated even palindrome is also palindrome in base k
• If yes, then add this number to sum.
• Repeat the above three steps by changing i=i+1 until the last generated even palindrome has crossed limit.

Below is the implementation of the above approach :  

C++

// CPP Program for Checking double
// base Palindrome.
#include <bits/stdc++.h>
using namespace std;

// generates even and odd palindromes
int makePalindrome(int n, bool odd)
{
    int res = n;
    if (odd)
        n = n / 10;
    while (n > 0) {
        res = 10 * res + n % 10;
        n /= 10;
    }
    return res;
}

// Check if a number is palindrome
// in base k
bool isPalindrome(int n, int base)
{
    int reversed = 0;
    int temp = n;
    while (temp > 0) {
        reversed = reversed * base +
                   temp % base;
        temp /= base;
    }
    return reversed == n;
}

// function to print sum of Palindromes
void sumPalindrome(int n, int k){

    int sum = 0, i = 1;

    int p = makePalindrome(i, true);

    // loop for odd generation of
    // odd palindromes
    while (p < n) {
        if (isPalindrome(p, k))
            sum += p;
        i++;

        // cout << p << " ";
        p = makePalindrome(i, true);
    }

    i = 1;

    // loop for generation of
    // even palindromes
    p = makePalindrome(i, false);
    while (p < n) {
        if (isPalindrome(p, k))
            sum += p;
        i++;
        p = makePalindrome(i, false);
    }

    // result of all palindromes in
    // both bases.
    cout << "Total sum is " << sum
         << endl;
}

// driver code
int main()
{
    int n = 1000000, k = 2;
    sumPalindrome(n ,k);
    return 0;
}

Java

// Java Program for Checking double
// base Palindrome.

public class GFG {

// generates even and odd palindromes
    static int makePalindrome(int n, boolean odd) {
        int res = n;
        if (odd) {
            n = n / 10;
        }
        while (n > 0) {
            res = 10 * res + n % 10;
            n /= 10;
        }
        return res;
    }

// Check if a number is palindrome
// in base k
    static boolean isPalindrome(int n, int base) {
        int reversed = 0;
        int temp = n;
        while (temp > 0) {
            reversed = reversed * base
                    + temp % base;
            temp /= base;
        }
        return reversed == n;
    }

// function to print sum of Palindromes
    static void sumPalindrome(int n, int k) {

        int sum = 0, i = 1;

        int p = makePalindrome(i, true);

        // loop for odd generation of
        // odd palindromes
        while (p < n) {
            if (isPalindrome(p, k)) {
                sum += p;
            }
            i++;

            // cout << p << " ";
            p = makePalindrome(i, true);
        }

        i = 1;

        // loop for generation of
        // even palindromes
        p = makePalindrome(i, false);
        while (p < n) {
            if (isPalindrome(p, k)) {
                sum += p;
            }
            i++;
            p = makePalindrome(i, false);
        }

        // result of all palindromes in
        // both bases.
        System.out.println("Total sum is " + sum);
    }

// driver code
    public static void main(String[] args) {
        int n = 1000000, k = 2;
        sumPalindrome(n, k);

    }
}

Python3

# Python3 Program for Checking double
# base Palindrome.

# Function generates even and
# odd palindromes
def makePalindrome(n, odd):

    res = n;
    if (odd):
        n = int(n / 10);
    while (n > 0):
        res = 10 * res + n % 10;
        n = int(n / 10);
    return res;

# Check if a number is palindrome
# in base k
def isPalindrome(n, base):
    reversed = 0;
    temp = n;
    while (temp > 0):
        reversed = reversed * base + temp % base;
        temp = int(temp / base);

    return reversed == n;

# function to print sum of Palindromes
def sumPalindrome(n, k):

    sum = 0;
    i = 1;

    p = makePalindrome(i, True);

    # loop for odd generation of
    # odd palindromes
    while (p < n):
        if (isPalindrome(p, k)):
            sum += p;
        i += 1;

        p = makePalindrome(i, True);

    i = 1;

    # loop for generation of
    # even palindromes
    p = makePalindrome(i, False);
    while (p < n):
        if (isPalindrome(p, k)):
            sum += p;
        i += 1;
        p = makePalindrome(i, False);

    # result of all palindromes in
    # both bases.
    print("Total sum is", sum);

# Driver code
n = 1000000;
k = 2;
sumPalindrome(n, k);

# This code is contributed by mits

C#

// C# Program for Checking double
// base1 Palindrome.

public class GFG {

// generates even and odd palindromes
    static int makePalindrome(int n, bool odd) {
        int res = n;
        if (odd) {
            n = n / 10;
        }
        while (n > 0) {
            res = 10 * res + n % 10;
            n /= 10;
        }
        return res;
    }

// Check if a number is palindrome
// in base1 k
    static bool isPalindrome(int n, int base1) {
        int reversed = 0;
        int temp = n;
        while (temp > 0) {
            reversed = reversed * base1 + temp % base1;
            temp /= base1;
        }
        return reversed == n;
    }

// function to print sum of Palindromes
    static void sumPalindrome(int n, int k) {

        int sum = 0, i = 1;

        int p = makePalindrome(i, true);

        // loop for odd generation of
        // odd palindromes
        while (p < n) {
            if (isPalindrome(p, k)) {
                sum += p;
            }
            i++;

            p = makePalindrome(i, true);
        }

        i = 1;

        // loop for generation of
        // even palindromes
        p = makePalindrome(i, false);
        while (p < n) {
            if (isPalindrome(p, k)) {
                sum += p;
            }
            i++;
            p = makePalindrome(i, false);
        }

        // result of all palindromes in
        // both base1s.
        System.Console.WriteLine("Total sum is " + sum);
    }

// driver code
    public static void Main() {
        int n = 1000000, k = 2;
        sumPalindrome(n, k);

    }
}

// This code is contributed by mits

PHP

<?php
// PHP Program for Checking double
// base Palindrome.

// Function generates even and
// odd palindromes
function makePalindrome($n, $odd)
{
    $res = $n;
    if ($odd)
    {
        $n = (int)($n / 10);
    }
    while ($n > 0)
    {
        $res = 10 * $res + $n % 10;
        $n = (int)($n / 10);
    }
    return $res;
}

// Check if a number is palindrome
// in base k
function isPalindrome($n, $base)
{
    $reversed = 0;
    $temp = $n;
    while ($temp > 0)
    {
        $reversed = $reversed * $base +
                        $temp % $base;
        $temp = (int)($temp / $base);
    }
    return $reversed == $n;
}

// function to print sum of Palindromes
function sumPalindrome($n, $k)
{
    $sum = 0; $i = 1;

    $p = makePalindrome($i, true);

    // loop for odd generation of
    // odd palindromes
    while ($p < $n)
    {
        if (isPalindrome($p, $k))
        {
            $sum += $p;
        }
        $i++;

        $p = makePalindrome($i, true);
    }

    $i = 1;

    // loop for generation of
    // even palindromes
    $p = makePalindrome($i, false);
    while ($p < $n)
    {
        if (isPalindrome($p, $k))
        {
            $sum += $p;
        }
        $i++;
        $p = makePalindrome($i, false);
    }

    // result of all palindromes in
    // both bases.
    echo("Total sum is " . $sum);
}

// Driver code
$n = 1000000; $k = 2;
sumPalindrome($n, $k);

// This code is contributed
// by Mukul Singh.
?>

Javascript

<script>
// javascript Program for Checking var
// base Palindrome.

// generates even and odd palindromes
    function makePalindrome(n, odd) {
        var res = n;
        if (odd) {
            n = parseInt(n / 10);
        }
        while (n > 0) {
            res = 10 * res + n % 10;
            n = parseInt(n / 10);
        }
        return res;
    }

// Check if a number is palindrome
// in base k
    function isPalindrome(n , base) {
        var reversed = 0;
        var temp = n;
        while (temp > 0) {
            reversed = reversed * base
                    + temp % base;
            temp = parseInt(temp/base);
        }
        return reversed == n;
    }

// function to print sum of Palindromes
    function sumPalindrome(n , k) {

        var sum = 0, i = 1;

        var p = makePalindrome(i, true);

        // loop for odd generation of
        // odd palindromes
        while (p < n) {
            if (isPalindrome(p, k)) {
                sum += p;
            }
            i++;

            // cout << p << " ";
            p = makePalindrome(i, true);
        }

        i = 1;

        // loop for generation of
        // even palindromes
        p = makePalindrome(i, false);
        while (p < n) {
            if (isPalindrome(p, k)) {
                sum += p;
            }
            i++;
            p = makePalindrome(i, false);
        }

        // result of all palindromes in
        // both bases.
        document.write("Total sum is " + sum);
    }

// driver code
var n = 1000000, k = 2;
sumPalindrome(n, k);

// This code contributed by Princi Singh
</script>

Output

Total sum is 872187

Time Complexity: O(n*log(n))
Auxiliary Space: O(1)

This article is contributed by Shubham Rana. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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