// Package diffsquares provides a utility for the determination of the
// difference between the square of sums and sum of squares on a given
// number range.
package diffsquares

// SquareOfSums returns the square of the sum of the range of natural
// numbers from 1 to the given value.
func SquareOfSums(n int) int {
	s := n * (n + 1) / 2
	return s * s
}

// SumOfSquares returns the sum of the squares of the range of natural
// numbers from 1 to the given value.
func SumOfSquares(n int) int {
	return n * (n + 1) * (2*n + 1) / 6

}

// Difference returns the difference between the sum of squares and the
// square of sums for the range of natural number up to the given value.
func Difference(n int) int {
	return SquareOfSums(n) - SumOfSquares(n)
}

// Iteration 2 benchmark:
//
// PASS
// BenchmarkSquareOfSums-12	2000000000	         1.04 ns/op
// BenchmarkSumOfSquares-12	2000000000	         1.89 ns/op
// BenchmarkDifference-12  	1000000000	         2.75 ns/op
//
// benchmark                    old ns/op     new ns/op     delta
// BenchmarkSquareOfSums-12     55.6          1.04          -98.13%
// BenchmarkSumOfSquares-12     63.4          1.89          -97.02%
// BenchmarkDifference-12       114           2.75          -97.59%