Visit the palindrome-products exercise on Exercism to read the full instructions and download the exercise files.
Dig Deeper
Factors First
Factors first
public static class PalindromeProducts
{
public static (int, Pairs) Largest(int min, int max)
{
var largestProduct = int.MinValue;
var factors = new List<(int, int)>();
foreach(var (a, b) in ReversedFactorPairsInRange(min, max))
{
var product = a * b;
if (a == b && product < largestProduct)
break;
if (IsPalindrome(product))
{
if (product > largestProduct)
{
largestProduct = product;
factors.Clear();
}
if (product == largestProduct)
{
factors.Add((a, b));
}
}
}
if (factors.Count == 0)
throw new ArgumentException("No palindromes in the range");
return (largestProduct, factors.ToArray());
}
public static (int, IEnumerable<(int,int)>) Smallest(int min, int max)
{
var smallestProduct = int.MaxValue;
var factors = new List<(int, int)>();
foreach(var (a, b) in FactorPairsInRange(min, max))
{
if (a > smallestProduct)
break;
var product = a * b;
if (IsPalindrome(product))
{
if (product < smallestProduct)
{
smallestProduct = product;
factors.Clear();
}
if (product == smallestProduct)
{
factors.Add((a, b));
}
}
}
if (factors.Count == 0)
throw new ArgumentException("No palindromes in the range");
return (smallestProduct, factors.ToArray());
}
private static IEnumerable<(int, int)> ReversedFactorPairsInRange(int min, int max)
{
for (var a = max; a >= min; --a)
for (var b = a; b >= min; --b)
yield return (b, a);
}
private static IEnumerable<(int, int)> FactorPairsInRange(int min, int max)
{
for (var a = min; a <= max; ++a)
for (var b = a; b <= max; ++b)
yield return (a, b);
}
private static bool IsPalindrome(int number)
=> number.ToString().Equals(
new String(number.ToString().Reverse().ToArray())
);
}
Solving the problem
We are given a range of numbers between min and max and a task to generating all possible pairs of such numbers.
IEnumerable<(int, int)> FactorPairsInRange(int min, int max)
{
for (var a = min; a <= max; ++a)
for (var b = a; b <= max; ++b)
yield return (a, b);
}
The above function does it. Both numbers are returned as elements of a single tuple.
The algorithm has a very characteristic shape of nested for loops.
It has quadratic complexity O(n^2). If the range is 1 to 10 it will generate 55 pairs.
If the range is 10 to 100 it will generate 4095 pairs.
We can calculate the exact number using the expression (n(n+1))/2 where n is the count of possible numbers.
Finding the smallest palindrome and its factors
With the pairs generated we can no look for pairs that multiplied result in a palindrom number.
To find all the unique pairs resulting in the smallest palindrome product we will need two variables:
var smallestProduct = int.MaxValue;
var factors = new List<(int, int)>();
We will then iterate over the factor pairs (a, b) calculating the their product.
If the product is a palindrome and it is smaller than the smallestProduct we will say it is the new smallestProduct, and we will remove the factors previously collected.
Then, if the product is a palindrome and it is the same as the smallestProduct we will capture the factors.
Otherwise, we will move on to examin the next pair.
foreach(var (a, b) in FactorPairsInRange(min, max)) {
var product = a * b;
if (IsPalindrome(product))
{
if (product < smallestProduct)
{
smallestProduct = product;
factors.Clear();
}
if (produt == smallestProduct)
{
factors.Add((a, b));
}
}
}
If at the end of the loop there are no factors then there are no palindrome numbers with factors in the given range.
if (results.Count == 0)
throw new ArgumentException("No palindromes in the range");
But if there are such pairs, we can return the smallest palindrome and its factors
return (smallestProduct, factors.ToArray());
Additionally we can stop the loop as soon as we know that the small of the numbers in a given pair is greater than the smallestProduct.
This is because even though we look at the factor pairs independently, we know they are ordered by first and then the second and neither of them can be less than 1.
If the first value is greater than smallestProduct then the product of the first and second must be greater too.
If we put it all together we get:
(int, IEnumerable<(int,int)>) Smallest(int min, int max)
{
var smallestProduct = int.MaxValue;
var factors = new List<(int, int)>();
foreach(var (a, b) in FactorPairsInRange(min, max)) {
if (a > smallestProduct)
break;
var product = a * b;
if (IsPalindrome(product))
{
if (product < smallestProduct)
{
smallestProduct = product;
factors.Clear();
}
if (product == smallestProduct)
{
factors.Add((a, b));
}
}
}
if (factors.Count == 0)
throw new ArgumentException("No palindromes in the range");
return (smallestProduct, factors.ToArray());
}
To get the largest palindrome we can follow a very similar algorithm but returning the pairs of numbers in a descending order.
Source: Exercism csharp/palindrome-products