Dado un valor n, encuentre el número palíndromo más grande que es producto de dos números de n dígitos.
Ejemplos:
Input : n = 2 Output : 9009 9009 is the largest number which is product of two 2-digit numbers. 9009 = 91*99. Input : n = 3 Output : 906609
A continuación se muestran los pasos para encontrar el número requerido.
- Encuentre un límite inferior en números de n dígitos. Por ejemplo, para n = 2, lower_limit es 10.
- Encuentre un límite superior en números de n dígitos. Por ejemplo, para n = 2, límite_superior es 99.
- Considere todos los pares de números siempre que el número se encuentre en el rango [límite_inferior, límite_superior]
A continuación se muestra la implementación de los pasos anteriores.
C++
// C++ problem to find out the
// largest palindrome number which
// is product of two n digit numbers
#include <bits/stdc++.h>
using namespace std;
// Function to calculate largest
// palindrome which is product of
// two n-digits numbers
int larrgestPalindrome(int n)
{
int upper_limit = pow(10,n) - 1;
// largest number of n-1 digit.
// One plus this number is lower
// limit which is product of two numbers.
int lower_limit = 1 + upper_limit / 10;
// Initialize result
int max_product = 0;
for (int i = upper_limit;
i >= lower_limit;
i--)
{
for (int j = i; j >= lower_limit; j--)
{
// calculating product of
// two n-digit numbers
int product = i * j;
if (product < max_product)
break;
int number = product;
int reverse = 0;
// calculating reverse of
// product to check whether
// it is palindrome or not
while (number != 0)
{
reverse = reverse * 10 +
number % 10;
number /= 10;
}
// update new product if exist
// and if greater than previous one
if (product == reverse &&
product > max_product)
max_product = product;
}
}
return max_product;
}
// Driver code
int main()
{
int n = 2;
cout << larrgestPalindrome(n);
return 0;
}
Java
// Java problem to find out the
// largest palindrome number
// which is product of two
// n digit numbers.
import java.lang.Math;
class GFG
{
// Function to calculate largest
// palindrome which isproduct of
// two n-digits numbers
static int larrgestPalindrome(int n)
{
int upper_limit = (int)Math.pow(10, n) - 1;
// largest number of n-1 digit.
// One plus this number
// is lower limit which is
// product of two numbers.
int lower_limit = 1 + upper_limit / 10;
// Initialize result
int max_product = 0;
for (int i = upper_limit; i >= lower_limit; i--)
{
for (int j = i; j >= lower_limit; j--)
{
// calculating product of two
// n-digit numbers
int product = i * j;
if (product < max_product)
break;
int number = product;
int reverse = 0;
// calculating reverse of product
// to check whether it is
// palindrome or not
while (number != 0)
{
reverse = reverse * 10 + number % 10;
number /= 10;
}
// update new product if exist and if
// greater than previous one
if (product == reverse && product > max_product)
max_product = product;
}
}
return max_product;
}
// Driver code
public static void main (String[] args)
{
int n = 2;
System.out.print(larrgestPalindrome(n));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python problem to find # out the largest palindrome # number which is product of # two n digit numbers. # Function to calculate largest # palindrome which is # product of two n-digits numbers def larrgestPalindrome(n): upper_limit = (10**(n))-1 # largest number of n-1 digit. # One plus this number # is lower limit which is # product of two numbers. lower_limit = 1 + upper_limit//10 max_product = 0 # Initialize result for i in range(upper_limit,lower_limit-1, -1): for j in range(i,lower_limit-1,-1): # calculating product of # two n-digit numbers product = i * j if (product < max_product): break number = product reverse = 0 # calculating reverse of # product to check # whether it is palindrome or not while (number != 0): reverse = reverse * 10 + number % 10 number =number // 10 # update new product if exist and if # greater than previous one if (product == reverse and product > max_product): max_product = product return max_product # Driver code n = 2 print(larrgestPalindrome(n)) # This code is contributed # by Anant Agarwal.
C#
// C# problem to find out the
// largest palindrome number
// which is product of two
// n digit numbers.
using System;
class GFG
{
// Function to calculate largest
// palindrome which isproduct of
// two n-digits numbers
static int larrgestPalindrome(int n)
{
int upper_limit = (int)Math.Pow(10, n) - 1;
// largest number of n-1 digit.
// One plus this number
// is lower limit which is
// product of two numbers.
int lower_limit = 1 + upper_limit / 10;
// Initialize result
int max_product = 0;
for (int i = upper_limit; i >= lower_limit; i--)
{
for (int j = i; j >= lower_limit; j--)
{
// calculating product of two
// n-digit numbers
int product = i * j;
if (product < max_product)
break;
int number = product;
int reverse = 0;
// calculating reverse of product
// to check whether it is
// palindrome or not
while (number != 0)
{
reverse = reverse * 10 + number % 10;
number /= 10;
}
// update new product if exist and if
// greater than previous one
if (product == reverse && product > max_product)
max_product = product;
}
}
return max_product;
}
// Driver code
public static void Main ()
{
int n = 2;
Console.Write(larrgestPalindrome(n));
}
}
// This code is contributed by nitin mittal.
PHP
<?php
// PHP problem to find out
// the largest palindrome
// number which is product
// of two n digit numbers
// Function to calculate
// largest palindrome which
// is product of two n-digit numbers
function larrgestPalindrome($n)
{
$upper_limit = 0;
// Loop to calculate upper bound
// (largest number of n-digit)
for ($i = 1; $i <= $n; $i++)
{
$upper_limit *= 10;
$upper_limit += 9;
}
// largest number of n-1 digit
// One plus this number
// is lower limit which is
// product of two numbers.
$lower_limit = 1 + (int)($upper_limit / 10);
// Initialize result
$max_product = 0;
for ($i = $upper_limit;
$i >= $lower_limit;
$i--)
{
for ($j = $i;
$j >= $lower_limit;
$j--)
{
// calculating product of
// two n-digit numbers
$product = $i * $j;
if ($product < $max_product)
break;
$number = $product;
$reverse = 0;
// calculating reverse of
// product to check whether
// it is palindrome or not
while ($number != 0)
{
$reverse = $reverse * 10 +
$number % 10;
$number = (int)($number / 10);
}
// update new product if exist
// and if greater than previous one
if ($product == $reverse &&
$product > $max_product)
$max_product = $product;
}
}
return $max_product;
}
// Driver code
$n = 2;
echo(larrgestPalindrome($n));
// This code is contributed by Ajit.
?>
Javascript
<script>
// Javascript problem to find out the
// largest palindrome number
// which is product of two
// n digit numbers.
// Function to calculate largest
// palindrome which isproduct of
// two n-digits numbers
function larrgestPalindrome(n)
{
let upper_limit = Math.pow(10, n) - 1;
// largest number of n-1 digit.
// One plus this number
// is lower limit which is
// product of two numbers.
let lower_limit = 1 +
parseInt(upper_limit / 10, 10);
// Initialize result
let max_product = 0;
for (let i = upper_limit; i >= lower_limit; i--)
{
for (let j = i; j >= lower_limit; j--)
{
// calculating product of two
// n-digit numbers
let product = i * j;
if (product < max_product)
break;
let number = product;
let reverse = 0;
// calculating reverse of product
// to check whether it is
// palindrome or not
while (number != 0)
{
reverse = reverse * 10 + number % 10;
number = parseInt(number / 10, 10);
}
// update new product if exist and if
// greater than previous one
if (product == reverse &&
product > max_product)
max_product = product;
}
}
return max_product;
}
let n = 2;
document.write(larrgestPalindrome(n));
</script>
Producción :
9009
Complejidad de tiempo: O(n*n*log 10 (producto)), donde n es el límite superior calculado y el producto es (producto de dos números de n dígitos)
Espacio auxiliar: O(1), ya que no se requiere espacio adicional |
El enfoque utilizado en esta publicación es simple y directo. Por favor comente si encuentra un mejor enfoque.
Este artículo es una contribución de Shivam Pradhan (anuj_charm) . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA