Project Euler #125: Palindromic sums

Question

The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 6² + 7² + 8² + 9² + 10² + 11² + 12².
There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that 1 = 0² + 1² has not been included as this problem is concerned with the squares of positive integers.
Find the sum of all the numbers less than 10⁸ that are both palindromic and can be written as the sum of consecutive squares.

Answer :

Hacker Rank Problem

Solution

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import java.math.BigInteger; import java.util.Arrays; import java.util.HashSet; import java.util.Scanner; public class Solution { public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner in = new Scanner(System.in); int numberOfTestCases = in.nextInt(); for (int i = 0; i < numberOfTestCases; i++) { int n = in.nextInt(); int d = in.nextInt(); double sqrtLimit = Math.sqrt(n); BigInteger sum = BigInteger.valueOf(0); HashSet<Long> set = new HashSet<Long>(); for (int j = 1; j <= sqrtLimit; j++) { long number = j * j; int k = j + d; //System.out.println("j " + j); while (number < n && k <= sqrtLimit) { number += k * k; if (number == reverse(number) && number < n) { //System.out.println(k + ":" + number); if (!set.contains(number)) { //System.out.println("addedToSet: " + number); sum = sum.add(BigInteger.valueOf(number)); set.add(number); } } k += d; } } System.out.println(sum); } } public static long reverse(long number) { long reverse = 0; while (number != 0) { reverse = reverse * 10 + number % 10; number /= 10; } return reverse; } }