Super Palindromes in C++

Suppose we have a positive integer N, that is said to be a superpalindrome if it is a palindrome, and it is also the square of a palindrome. Now consider we have two positive integers L and R we have to find the number of superpalindromes in the inclusive range of [L, R].

So, if the input is like L = 5 and R = 500, then the output will be 3, the superpalindromes are 9, 121, 484.

To solve this, we will follow these steps −

• Define a function helper(), this will take x, m, M, lb, ub,

• if x > ub, then −

  • return

• if x >= lb and (x * x) is palindrome, then −

  • (increase ans by 1)

• for initialize i := 1, when m + 2 * i <= M, update (increase i by 1), do:

  • W := 10^(m + 2 * i - 1)

  • w := 10^i

  • for initialize z := 1, when z <= 9, update (increase z by 1), do −

    • helper(z * W + x * w, m + 2 * i, M, lb, ub)

• From the main method, do the following −

• lb := square root of L, ub := square root of R

• M := perform log of ub base 10 + 1

• for initialize z := 0, when z <= 9, update (increase z by 1), do−

  • helper(z, 1, M, lb, ub)

  • helper(11 * z, 2, M, lb, ub)

• return ans

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   int ans = 0;
   public:
   int superpalindromesInRange(string L, string R){
      long double lb = sqrtl(stol(L)), ub = sqrtl(stol(R));
      int M = log10l(ub) + 1;
      for (int z = 0; z <= 9; z++) {
         helper(z, 1, M, lb, ub);
         helper(11 * z, 2, M, lb, ub);
      }
      return ans;
   }
   private:
   void helper(long x, int m, int M, long double lb, long double ub){
      if (x > ub)
      return;
      if (x >= lb && is_palindrome(x * x))
      ans++;
      for (int i = 1; m + 2 * i <= M; i++) {
         long W = powl(10, m + 2 * i - 1) + 1;
         long w = powl(10, i);
         for (int z = 1; z <= 9; z++)
         helper(z * W + x * w, m + 2 * i, M, lb, ub);
      }
   }
   bool is_palindrome(long x){
      if (x == 0)
      return true;
      if (x % 10 == 0)
      return false;
      long left = x, right = 0;
      while (left >= right) {
         if (left == right || left / 10 == right)
         return true;
         right = 10 * right + (left % 10), left /= 10;
      }
      return false;
   }
};
main(){
   Solution ob;
   cout << (ob.superpalindromesInRange("5", "500"));
}

Input

"5", "500"

Output

3