## Hackerrank – Problem description

The problem description – Hackerrank.

## Solution

We need to find out if the product of two 3-figured number is a palindrome and it is smaller than exercise upper constraint.

My solution starts with initializing two numbers as `999`

. I continually decremented second number by `1`

, until it is equal to `100`

. I calculated the product for each pair. If the product is palindrome and compare it with actual maximum of products.

Repeat the steps until both numbers are equal to `100`

. The result is actual maximum palindrome number

`999`

does not exceed exercise time limit.^{2}

I created solution in:

All solutions are also available on my GitHub profile.

## Java

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import java.util.Scanner; public class LargestPalindromeProduct { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int tests = Integer.parseInt(scanner.nextLine()); for (int i = 0; i < tests; i++) { int number = Integer.parseInt(scanner.nextLine()); System.out.println(palindromeNumber(number)); } scanner.close(); } private static int palindromeNumber(int number) { // int root = (int) Math.sqrt(number); int max = 0; for (int j = 999; j >= 100; j--) { for (int k = 999; k >= 100; k--) { int product = j * k; if(product < number && isPalindrome(String.valueOf(product))) { if(product > max) { max = product; } } } } return max; } private static boolean isPalindrome(String text) { for (int i = 0; i < text.length() / 2; i++) { if(text.charAt(i) != text.charAt(text.length() - 1 - i)) { return false; } } return true; } } |