Largest palindrome which is product of two ndigit numbers (original) (raw)

Last Updated : 22 Aug, 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.

1. Find a lower limit on n digit numbers. For example, for n = 2, lower_limit is 10.
2. Find an upper limit on n digit numbers. For example, for n = 2, upper_limit is 99.
3. Consider all pairs of numbers wherever number lies in range [lower_limit, upper_limit]

Below is the implementation of above steps.

Try It Yourself

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

JavaScript

`

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.