Count all palindrome which is square of a palindrome


Given two positive integers L and R (represented as strings) where 1\leq L \leq R\leq10^{18}. The task is to find the total number of super-palindromes in the inclusive range [L, R].
A palindrome is called super-palindrome if it is a palindrome and also square of a palindrome.
Examples:

Input: L = "4", R = "1000"
Output: 4
Explanation: 4, 9, 121, and 484 are super-palindromes.

Input : L = "100000", R = "10000000000"
Output : 11

Approach:
Lets say P = R^{2} is a super-palindrome.
Now since R is a palindrome, the first half of the digits of R can be used to determine R up-to two possibilities. Let i be the first half of the digits in R. For eg. if i = 123, then R = 12321 or R = 123321. Thus we can iterate through these all these digits. Also each possibility can have either odd or even number of digits in R.
Thus we iterate through each i upto 105 and create the associated palindrome R, and check whether R2 is a palindrome.
Also we will handle the odd and even palindromes separately, and break whenever out palindrome goes beyond R.
Now since P \leq 10^{18}, and R \leq (10^{18})^{\frac{1}{2}} \equiv 10^{9} and  R = i||i^{'} (on Concatenation), where i is reverse of i (in both ways), so our LIMIT will not be greater than i\leq10^{5}.
Below is the implementation of above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the 
// above approach
#include <bits/stdc++.h> 
using namespace std;
  
// check if a number is a palindrome
bool ispalindrome(int x)
{
    int ans = 0;
    int temp = x;
    while (temp > 0)
    {
        ans = 10 * ans + temp % 10;
        temp = temp / 10;
    }
    return ans == x;
}
  
// Function to return required count 
// of palindromes
int SuperPalindromes(int L, int R)
{
    // Range [L, R]
  
    // Upper limit
    int LIMIT = 100000;
  
    int ans = 0;
  
    // count odd length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
        string s = to_string(i); // if s = '1234'
  
        string rs = s.substr(0, s.size() - 1);
        reverse(rs.begin(), rs.end());
  
        // then, t = '1234321'
        string p = s + rs; 
        int p_sq     = pow(stoi(p), 2);
        if (p_sq > R)
            break;
        if (p_sq >= L and ispalindrome(p_sq))
            ans = ans + 1;
    }
  
    // count even length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
        string s = to_string(i); // if s = '1234'
  
        string rs = s;
        reverse(rs.begin(), rs.end());
        string p = s + rs; // then, t = '12344321'
        int p_sq = pow(stoi(p), 2);
        if (p_sq > R)
            break;
        if (p_sq >= L and ispalindrome(p_sq))
            ans = ans + 1;
    }
  
    // Return count of super-palindromes
    return ans;
}
  
// Driver Code
int main()
{
    string L = "4";
    string R = "1000";
      
    // function call to get required answer
    printf("%d\n", SuperPalindromes(stoi(L), 
                                  stoi(R)));
    return 0;
  
// This code is contributed 
// by Harshit Saini

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the
// above approach
import java.lang.*;
  
class GFG
{
  
// check if a number is a palindrome
public static boolean ispalindrome(int x)
{
    int ans = 0;
    int temp = x;
    while (temp > 0)
    {
        ans = 10 * ans + temp % 10;
        temp = temp / 10;
    }
    return ans == x;
}
  
// Function to return required 
// count of palindromes
public static int SuperPalindromes(int L, 
                                   int R)
{
    // Range [L, R]
  
    // Upper limit
    int LIMIT = 100000;
  
    int ans = 0;
  
    // count odd length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
  
        // if s = '1234'
        String s = Integer.toString(i);
  
        StringBuilder rs = new StringBuilder();
        rs.append(s.substring(0,
                     Math.max(1, s.length() - 1)));
        String srs = rs.reverse().toString();
  
        // then, t = '1234321'
        String p = s + srs;
        int p_sq = (int)(Math.pow(
                         Integer.parseInt(p), 2));
        if (p_sq > R)
        {
            break;
        }
        if (p_sq >= L && ispalindrome(p_sq))
        {
            ans = ans + 1;
        }
    }
  
    // count even length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
  
        // if s = '1234'
        String s = Integer.toString(i); 
  
        StringBuilder rs = new StringBuilder();
        rs.append(s);
        rs = rs.reverse();
  
        String p = s + rs; // then, t = '12344321'
        int p_sq = (int)(Math.pow(
                         Integer.parseInt(p), 2));
        if (p_sq > R)
        {
            break;
        }
        if (p_sq >= L && ispalindrome(p_sq))
        {
            ans = ans + 1;
        }
    }
  
    // Return count of super-palindromes
    return ans;
}
  
// Driver program
public static void main(String [] args)
{
    String L = "4";
    String R = "1000";
  
    // function call to get required answer
    System.out.println(SuperPalindromes(
       Integer.parseInt(L), Integer.parseInt(R)));
}
}
  
// This code is contributed 
// by Harshit Saini 

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python implementation of the above approach
  
# check if a number is a palindrome
def ispalindrome(x):
    ans, temp = 0, x
    while temp > 0:
        ans = 10 * ans + temp % 10
        temp = temp // 10
    return ans == x
  
# Function to return required count of palindromes
def SuperPalindromes(L, R):
    # Range [L, R]
    L, R = int(L), int(R)
  
    # Upper limit
    LIMIT = 100000
  
    ans = 0
  
    # count odd length palindromes
    for i in range(LIMIT):
        s = str(i)  # if s = '1234'
        p = s + s[-2::-1# then, t = '1234321'
        p_sq = int(p) ** 2
        if p_sq > R:
            break
        if p_sq >= L and ispalindrome(p_sq):
            ans = ans + 1
  
    # count even length palindromes
    for i in range(LIMIT):
        s = str(i)  # if s = '1234'
        p = s + s[::-1# then, t = '12344321'
        p_sq = int(p) ** 2
        if p_sq > R:
            break
        if p_sq >= L and ispalindrome(p_sq):
            ans = ans + 1
  
    # Return count of super-palindromes
    return ans
  
# Driver program
L = "4"
R = "1000"
  
# function call to get required answer
print(SuperPalindromes(L, R))
  
# This code is written by
# Sanjit_Prasad

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the 
// above approach
using System;
  
class GFG
{
  
// check if a number is a palindrome
static bool ispalindrome(int x)
{
    int ans = 0;
    int temp = x;
    while (temp > 0)
    {
        ans = 10 * ans + temp % 10;
        temp = temp / 10;
    }
    return ans == x;
}
  
// utility function used for 
// reversing a string
static string Reverse( string s )
{
    char[] charArray = s.ToCharArray();
    Array.Reverse( charArray );
    return new string( charArray );
}
  
// Function to return required 
// count of palindromes
static int SuperPalindromes(int L, int R)
{
    // Range [L, R]
  
    // Upper limit
    int LIMIT = 100000;
  
    int ans = 0;
  
    // count odd length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
  
        // if s = '1234'
        string s = i.ToString();
  
        string rs = s.Substring(0,
                       Math.Max(1, s.Length - 1));
        rs = Reverse(rs);
  
        // then, t = '1234321'
        string p = s + rs;
        int p_sq = (int)(Math.Pow(
                            Int32.Parse(p), 2));
        if (p_sq > R)
        {
            break;
        }
        if (p_sq >= L && ispalindrome(p_sq))
        {
            ans = ans + 1;
        }
    }
  
    // count even length palindromes
    for (int i = 0 ;i < LIMIT; i++)
    {
  
        // if s = '1234'
        string s = i.ToString();
  
        string rs = Reverse(s);
  
        string p = s + rs; // then, t = '12344321'
        int p_sq = (int)(Math.Pow(
                            Int32.Parse(p), 2));
        if (p_sq > R)
        {
            break;
        }
        if (p_sq >= L && ispalindrome(p_sq))
        {
            ans = ans + 1;
        }
    }
  
    // Return count of super-palindromes
    return ans;
}
  
// Driver Code
public static void Main()
{
    string L = "4";
    String R = "1000";
  
    // function call to get required answer
    Console.WriteLine(SuperPalindromes(
            Int32.Parse(L), Int32.Parse(R)));
}
}
  
// This code is contributed 
// by Harshit Saini 

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php 
// PHP implementation of the 
// above approach
  
// check if a number is a palindrome
function ispalindrome($x)
{
    $ans = 0;
    $temp = $x;
  
    while($temp > 0)
    {     
        $ans = (10 * $ans) +
               ($temp % 10);
        $temp = (int)($temp / 10);
    }
  
    return $ans == $x;
}
  
// Function to return required
// count of palindromes
function SuperPalindromes($L, $R)
{
    // Range [L, R]
    $L = (int)$L;
    $R = (int)$R;
  
    // Upper limit
    $LIMIT = 100000;
  
    $ans = 0;
  
    // count odd length palindromes
    for($i = 0 ;$i < $LIMIT; $i++)
    {
  
        $s = (string)$i; // if s = '1234'
        $rs = substr($s, 0, strlen($s) - 1);
        $p = $s.strrev($rs); // then, t = '1234321'
        $p_sq = (int)$p ** 2;
        if ($p_sq > $R)
        {
            break;
        }
        if ($p_sq >= $L and ispalindrome($p_sq))
        {
            $ans = $ans + 1;
        }
    
  
    // count even length palindromes
    for($i = 0 ;$i < $LIMIT; $i++)
    {
        $s = (string)$i; // if s = '1234'
        $p = $s.strrev($s); // then, t = '12344321'
          
        $p_sq = (int)$p ** 2;
  
        if ($p_sq > $R)
        {
            break;
        }
        if ($p_sq >= $L and ispalindrome($p_sq))
        {
            $ans = $ans + 1;
        }
    }
  
    // Return count of super-palindromes
    return $ans;
}
  
  
// Driver Code
$L = "4";
$R = "1000";
  
// function call to get required answer
echo SuperPalindromes($L, $R);
  
// This code is contributed 
// by Harshit Saini 
?>

chevron_right



Output:

4

Time Complexity: O(N*log(N)) where N is upper limit and the log(N) term comes from checking if a candidate is palindrome.



My Personal Notes arrow_drop_up