Go – CodeBlog.ch

A lot of tech companies ask interesting questions when they hire employees. One of these questions is: How many trailing zeros are there in 1000! (1 × 2 × 3 × .. × 1000).
As you might imagine, the actual result of 1000! is a rather large number, something you can hardly do in your mind, at least unless you’re some kind of savant.

However, there’s a solution you can come up with that doesn’t need a calculator or anything digital. Spoiler alert: There are certain multiplies that give you a zero at the end. Google knows the answer if you’re wondering how to solve this riddle with a paper and a pen.

I’ve been playing around with Google’s language “Go” for a while and wondered if I could solve this riddle using a more brute-force like solution. The code is rather simple, there’s a factorial function I needed to add because the result is too big for a data type like double or int. I then call that function and count the number of zeros using a simple regex pattern. Something you can do when you’re in an interview but it works as well and is also pretty fast.

package main
 
import "fmt"
import "math/big"
import "regexp"
 
func main() {
	c := int64(1000)
	f := big.NewInt(c)
	s := factorial(f)
	r := regexp.MustCompile(`[0]+$`)
 
	l := r.FindString(s.String())   
	fmt.Printf("There are %d zeros at the end of %d!\n", len(l), c);
}
 
func factorial(n *big.Int) (result *big.Int) {
	result = new(big.Int)
 
	switch n.Cmp(&big.Int{}) {
	case -1, 0:
		result.SetInt64(1)
	default:
		result.Set(n)
		var one big.Int
		one.SetInt64(1)
		result.Mul(result, factorial(n.Sub(n, &one)))
	}
	return
}

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How many trailing Zeros does the result of 1000! have? Brute-force solution with Google Go

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