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.Â
CPP
// 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 << " ";             i++;         }     } }   // Driver Program to test above function int main() {     int n = 1000;     findPalindromic(n);     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.Â
Auxiliary Space: O(1), as we are not using any extra space.
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