Largest palindrome which is product of two n-digit numbers

04 Jun, 2022
Given a value n, find out the largest palindrome number which is product of two n digit numbers.
Examples : 
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
Below are steps to find the required number. 
Find a lower limit on n digit numbers. For example, for n = 2, lower_limit is 10.Find an upper limit on n digit numbers. For example, for n = 2, upper_limit is 99.Consider all pairs of numbers where ever number lies in range [lower_limit, upper_limit]
Find a lower limit on n digit numbers. For example, for n = 2, lower_limit is 10.
Find an upper limit on n digit numbers. For example, for n = 2, upper_limit is 99.
Consider all pairs of numbers where ever number lies in range [lower_limit, upper_limit]
Below is the implementation of above steps. 
C++
Java
Python3
C#
PHP
Javascript
// 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 numbersint 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 codeint main(){    int n = 2;    cout << larrgestPalindrome(n);    return 0;}
// 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.
# 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 = 2print(larrgestPalindrome(n)) # This code is contributed# by Anant Agarwal.
// 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 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 numbersfunction 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.?>
<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>
Output :
9009
Time Complexity: O(n*n*log10(product)), where n is the upper limit calculated and product is (product of two n-digit numbers)Auxiliary Space: O(1), as no extra space is required|
The approach used in this post is simple and straightforward. Please comment if you find a better approach.This article is contributed by Shivam Pradhan(anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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