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C++ program to find all numbers less than n, which are palindromic in base 10 and base 2.

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Find all numbers less than n, which are palindromic in base 10 as well as base 2. Examples:

33 is Palindrome in its decimal representation.
100001(binary equivalent of 33) in Binary is a Palindrome.

313 is Palindrome in its decimal representation.
100111001 (binary equivalent of 313) in Binary is a Palindrome.

Brute Force: We check all the numbers from 1 to n whether their decimal representation is palindrome or not. Further, if a number is palindromic in base 10 then we check for its binary representation. If we found both representations a palindrome then we print it. Efficient Approach: We start from 1 and create palindromes of odd digit and even digit up to n and check whether its binary representation is palindrome or not. Note: This will reduce the number of operations as we should check only for decimal palindrome instead of checking all numbers from 1 to n. This approach uses two methods: int createPalindrome(int input, int b, bool isOdd): The palindrome creator takes in an input number and a base b as well as a boolean telling if the palindrome should have an even or odd number of digits. It takes the input number, reverses it, and appends it to the input number. If the result should have an odd number of digits, it chops off a digit of the reversed part. bool IsPalindrome(int number, int b) It takes the input number and calculates its reverse according to base b. Return result whether the number is equal to its reverse or not. 


// A C++ program for finding numbers which are
// decimal as well as binary palindrome
#include <iostream>
using namespace std;
// A utility to check if  number is palindrome on base b
bool IsPalindrome(int number, int b)
    int reversed = 0;
    int k = number;
    // calculate reverse of number
    while (k > 0) {
        reversed = b * reversed + k % b;
        k /= b;
    // return true/false depending upon number is palindrome or not
    return (number == reversed);
// A utility for creating palindrome
int createPalindrome(int input, int b, bool isOdd)
    int n = input;
    int palin = input;
    // checks if number of digits is odd or even
    // if odd then neglect the last digit of input in finding reverse
    // as in case of odd number of digits middle element occur once
    if (isOdd)
        n /= b;
    // creates palindrome by just appending reverse of number to itself
    while (n > 0) {
        palin = palin * b + (n % b);
        n /= b;
    return palin;
// Function to print decimal and binary palindromic number
void findPalindromic(int n)
    int number;
    for (int j = 0; j < 2; j++) {
        bool isOdd = (j % 2 == 0);
        // Creates palindrome of base 10 upto n
        // j always decides digits of created palindrome
        int i = 1;
        while ((number = createPalindrome(i, 10, isOdd)) < n) {
            // if created palindrome of base 10 is
            // binary palindrome
            if (IsPalindrome(number, 2))
                cout << number << " ";
// Driver Program to test above function
int main()
    int n = 1000;
    return 0;


1 3 5 7 9 313 585 717 33 99

Time Complexity: O(N*log(N)), as we are using the while loop to loop over N times in the worst case and IsPalindrome cost O(log(N)) which is nested inside the while loop. 

Auxiliary Space: O(1), as we are not using any extra space.

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Last Updated : 11 Sep, 2023
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