lc-go/solutions/1/q172/solution.go
2026-01-07 18:04:39 +09:00

21 lines
345 B
Go

package q172
// Note: actually, counting 5 alone is enough since 2 certainly occurs more than 5.
func trailingZeroes(n int) int {
count2, count5 := 0, 0
for i := 2; i <= n; i++ {
num := i
for num%2 == 0 {
num /= 2
count2++
}
for num%5 == 0 {
num /= 5
count5++
}
}
return min(count2, count5)
}
var _ = trailingZeroes