# Super Palindromes in C++

C++Server Side ProgrammingProgramming

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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
Updated on 04-Jun-2020 09:03:27